Srinivas

Referring to the AP Inter 2nd Year Maths Study Material Chapter 13 Probability Exercise 13b Solutions makes it easier to understand complex problems.

AP Inter 2nd Year Maths Probability Solutions Exercise 13b

I.

Question 1.
If P(A) = \(\frac{3}{5}\) and P(B) = \(\frac{1}{5}\), find P (A ∩ B) if A and B are independent events.
Solution:
Given that A and B are independent events P(A) = \(\frac{3}{5}\) and P(B) = \(\frac{1}{5}\)
∴ P(A ∩ B) = P(A) P(B) = \(\frac{3}{5}\) × \(\frac{1}{5}\) = \(\frac{3}{25}\)

Question 2.
Two cards are drawn at random in succession and without replacement from a pack of 52 playing cards. Find the probability that both the cards are black.
Solution:
There are 26 black cards in a deck of 52 cards.
Let A,B be the events of drawing black cards in the successive draws.
The probability of getting a black card in the first draw is P(A) = \(\frac{26}{52}\)
The probability of getting a black card in the second draw is P(B|A) = \(\frac{25}{51}\)
Since the drawn card is not replaced, A and B are dependent events.
∴ Probability of getting both the cards black P(A ∩ B) = P(A)P(B|A) = \(\frac{26}{52}\) × \(\frac{25}{52}\) = \(\frac{25}{102}\)

Question 3.
Let E and F be events with P(E) = \(\frac{3}{5}\), P(F) = \(\frac{3}{10}\) and P(E ∩ F) = \(\frac{1}{5}\). Are E and F independent?
Solution:
Given P(E) = \(\frac{3}{5}\), P(F) = \(\frac{3}{10}\) and P(E ∩ F) = \(\frac{1}{5}\)
Now P(E) P(F) = \(\frac{3}{5}\) × \(\frac{3}{10}\) = \(\frac{9}{50}\) ≠ \(\frac{1}{5}\)
P(E)P(F) ≠ P(E ∩ F)
∴ E and F are not independent.

II.

Question 1.
A box of oranges is inspected by examining three randomly selected oranges drawn without replacement. If all the three oranges are good, the box is approved for sale, otherwise, it is rejected. Find the probability that a box containing 15 oranges out of which 12 are good and 3 are bad ones will be approved for sale.
Solution:
Let A, B, C be the respective events that the first, second, and the third drawn orange is good.
Probability that first drawn orange is good, P(A) = \(\frac{12}{13}\)
The oranges are not replaced.
Probability of getting second orange is good, P (B) = \(\frac{11}{14}\)
Probability of getting third orange is good, P(C) = \(\frac{10}{13}\)
Since the box is approved for sale, if all the three oranges are good.
∴ Probability of getting all the oranges good
P(A ∩ B ∩ C) = P(A)P(B)P(C) = \(\frac{12}{15}\) × \(\frac{11}{14}\) × \(\frac{10}{13}\) = \(\frac{44}{91}\)

Question 2.
A fair coin and an unbiased die are tossed. Let A be the event head appears on the coin’ and B be the event 43 on the die’. Check whether A and B are independent events or not.
Solution:
The sample space is given by,
S = {(H, 1), (H, 2) (H, 3), (H, 4), (H, 5), (H, 6)
(T, 1),(T, 2),(T, 3),(T, 4),(T, 5),(T, 6)}
Let A: Head appears on the coin
⇒ P(A) = \(\frac{6}{12}\) = \(\frac{1}{2}\)
A = {(H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6)}
B: 3 on the die ⇒ B = {(H, 3),(T, 3)} ⇒ P(B) = \(\frac{2}{12}\) = \(\frac{1}{6}\)
Now, A ∩ B = {(H, 3)} ⇒ P(A ∩ B) = \(\frac{1}{12}\)
Hence, P(A)P(B) = \(\frac{1}{2}\) × \(\frac{2}{6}\) = P(A ∩ B)
∴ A and B are independent events.

Question 3.
A die marked 1, 2, 3 In red and 4, 5, 6 in green is tossed. Let A be the event, ‘the number ¡s even,’ and B be the event, ‘the number Is red’. Are A and B independent?
Solution:
When a die is tossed, the sample space is S = {1, 2, 3, 4, 5, 6}
Here, 1, 2, 3 are red in colour and 4, 5, 6 are green.
Let A: the number is even {2, 4, 6} ⇒ P(A) = \(\frac{3}{6}\) = \(\frac{1}{2}\)
B: the number is red = {1, 2, 3} ⇒ P(B) = \(\frac{3}{6}\) = \(\frac{1}{2}\)
∴ A ∩ B = {2}
P(A ∩ B) = \(\frac{1}{6}\)
Now P(A) P(B) = \(\frac{1}{2}\) × \(\frac{1}{2}\) = \(\frac{1}{4}\) ≠ \(\frac{1}{6}\)
∴ P(A)P(B) ≠ P(AB)
∴ A and B are not independent events.

Question 4.
Given that the events A and B are such that P(A) = \(\frac{1}{2}\), P(A ∪ B) = \(\frac{3}{5}\) and P(B) = p, Find p if they are (i) mutually exclusive (ii) independent.
Solution:
Given that P(A) = \(\frac{1}{2}\), P(A ∪ B) = \(\frac{3}{5}\) and P(B) = p
(i) When A and B are mutually exclusive,, A ∩ B = Φ
∴ P(A ∩ B) = 0
We know that P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
⇒ \(\frac{3}{5}\) = \(\frac{1}{2}\) + p = 0
⇒ p = \(\frac{3}{5}\) – \(\frac{1}{2}\) = \(\frac{1}{10}\)

(ii) When A and B are independent, P(A ∩ B) = P (A) P(B) = \(\frac{1}{2}\)p
We know that P(A ∪ B) = P(A)+P(B)-P(A ∩ B)
⇒ \(\frac{3}{5}\) = \(\frac{1}{2}\) + p – \(\frac{1}{2}\)p
⇒ \(\frac{3}{5}\) = \(\frac{1}{2}\) + \(\frac{p}{2}\)
⇒ \(\frac{p}{2}\) = \(\frac{3}{5}\) – \(\frac{1}{2}\) = \(\frac{1}{10}\)
⇒ \(\frac{p}{2}\) = \(\frac{2}{10}\) = \(\frac{1}{5}\)

Question 5.
Let A and B be independent events with P(A) = 0.3 and P(B) = 0.4. Find
(i) P(A ∩ B)
(ii) P(A ∪ B)
(iii) P (A|B)
(iv) P (B|A)
Solution:
Given P (A) = 0.3 and P(B) = 0.4.
(i) If A and B are independent events, then P(A ∩ B) = P(A) × P(B) = 0.3 × 0.4 = 0.12
(ii) P(A ∪ B) = P(A) + P(B) – P (A ∩ B) ⇒ P(A ∪ B) = 0.3 + 0.4 – 0.12 = 0.58
(iii) P(A|B) = \(\frac{P(A \cap B)}{P(B)}=\frac{0.12}{0.4}\) = 0.3
(iv) P(B|A) = \(\frac{P(A \cap B)}{P(A)}=\frac{0.12}{0.3}\) = 0.4

Question 6.
If A and B are also events such that P(A) = \(\frac{1}{4}\), P(B) = \(\frac{1}{2}\) and P(A ∩ B) = \(\frac{1}{8}\), find P(not A and not B).
Solution:
Given that P(A) = \(\frac{1}{4}\), P(B) = \(\frac{1}{2}\) and P(A ∩ B) = \(\frac{1}{8}\)
P(not on Aand not on B) = P(A’ ∩ B’)
P(not on A and not on B) = P[(A ∪ B)’] [∵ A’ ∩ B = (A ∪ B)’]
= 1 – P(A ∪ B) = 1 – [P(A) + P(B) – P(AB)]
= 1 – [\(\frac{1}{4}\) + \(\frac{1}{2}\) – \(\frac{1}{8}\)] = 1 – \(\frac{5}{8}\)
= \(\frac{3}{8}\)

Question 7.
Events A and B are such that P(A) = \(\frac{1}{2}\), P(B) = \(\frac{7}{12}\) and P(not A or not B) = \(\frac{1}{4}\). State whether A and B are independent ?
Solution:
Given that P(A) = \(\frac{1}{2}\),P(B) = \(\frac{7}{12}\) and P (not A or not B) = \(\frac{1}{4}\),
⇒ P(A’ ∪ B’) = \(\frac{1}{4}\) ⇒ P[(A ∩ B)’] = \(\frac{1}{4}\) [∵ A’ ∪ B’ = (A ∩ B)’]
⇒ 1 – P(A ∩ B) = \(\frac{1}{4}\) ⇒ P(A ∩ B) = \(\frac{3}{4}\) …………. (1)
But P(A)P(B) = \(\frac{1}{2}\) × \(\frac{7}{12}\) = \(\frac{7}{24}\) ………….. (2)
Here, \(\frac{3}{4}\) ≠ \(\frac{7}{24}\)
∴ P(A ∩ B) ≠ P(A)P(B)
∴ A and B are not independent events.

Question 8.
Given two independent events A and B such that P(A) = 0.3, P(B) = 0.6. Find
(i) P(A and B)
(ii) P(A and not B)
(iii) P(A or B)
(iv) P(neither A nor B)
Solution:
Given that P(A) = 0.3, P(B) = 0.6 are independent events.
(i) P(A and B) = P(A ∩ B)=P(A)P(B) = 0.3 × 0.6 = 0.18
(ii) P(A and not B) = P(A ∩ B) = P(A)’ – P(A ∩ B) = 0.3 – 0.18 = 0.12
(iii) P(A or B) = P(A ∪ B)=P(A)+ P (B)- P(A ∩ B)=0.3+0.6 – 0.18 = 0.72
(iv) P (neither A nor B) = P(A ∩ B ) = P [(A ∪ B)’] [∵ A’ ∩ B’ =(A ∪ B)’]
= 1 – P(A ∪ B) = 1 – 0.72 = 0.28

Question 9.
A die is tossed thrice. Find the probability of getting an odd number at least once.
Solution:
Probability of getting an odd number in a single throw of a die = \(\frac{3}{6}\) = \(\frac{1}{2}\)
Probability of getting an even number = \(\frac{3}{6}\) = \(\frac{1}{2}\)
Probability of getting an even number three times = \(\frac{1}{2}\) × \(\frac{1}{2}\) × \(\frac{1}{2}\) = \(\frac{1}{8}\)
= 1 – Probability of getting an odd number in none of the throws
= 1 – Probability of getting an even number thrice = 1 – \(\frac{1}{8}\) = \(\frac{7}{8}\)

Question 10.
Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls. Find the probability that
(i) both balls are red.
(ii) first ball is black and second is red.
(iii) one of them is black and other is red.
Solution:
Given that total number of balls = 18,
Number of red balls = 8,
Number of black balls = 10
(i) Probability of getting a red ball in the first draw = \(\frac{8}{18}\) = \(\frac{4}{9}\) . The ball is replaced after the first draw.
So, probability of getting a red ball in the second draw = \(\frac{8}{18}\) = \(\frac{4}{9}\)
∴ Probability of getting both the balls red = \(\frac{4}{9}\) × \(\frac{4}{9}\) = \(\frac{16}{81}\)

(ii) Probability of getting first ball black = \(\frac{10}{18}\) = \(\frac{5}{9}\).The ball is replaced after the first draw.
So, Probability of getting second ball as red = \(\frac{8}{18}\) = \(\frac{4}{9}\)
∴ Probability of getting first ball as black and second ball as red = \(\frac{5}{9}\) × \(\frac{4}{9}\) = \(\frac{20}{81}\)

(iii) Probability of getting first ball as red = \(\frac{8}{18}\) = \(\frac{4}{9}\).The ball is replaced after the first draw.
So, Probability of getting second ball as black = \(\frac{10}{18}\) = \(\frac{4}{9}\)
∴ Probability of getting first ball as black and second ball as red = \(\frac{4}{9}\) × \(\frac{5}{9}\) = \(\frac{20}{81}\)

Now probability that one of them is black and other is red = Probability of getting first ball black and second red + Probability of getting first ball red and second ball black
= \(\frac{20}{81}\)+ \(\frac{20}{81}\) = \(\frac{40}{81}\)

Question 11.
Probability of solving specific problem independently by A and B are \(\frac{1}{2}\) and \(\frac{1}{3}\) respectively. If both try to solve the problem independently, find the probability that (i) the problem is solved(ii) exactly one of them solves the problem.
Solution:
Probability of solving the problem by A is P(A) = \(\frac{1}{2}\)
Probability of solving the problem by B is P(B) = \(\frac{1}{3}\)
Since the problem is solved independently by A and B, we have
P(AB) = P(A) P(B) = \(\frac{1}{2}\) × \(\frac{1}{3}\) = \(\frac{1}{6}\)
Now P(A’) = 1 – P(A) = 1 – \(\frac{1}{2}\) = \(\frac{1}{2}\);
P(B’) = 1 – P(B) = 1 – \(\frac{1}{3}\) = \(\frac{2}{3}\)
(i) Probability that the problem is solved is
P(A ∪ B) = P(A) – P(B) – P(AB) = \(\frac{1}{2}\) + \(\frac{1}{3}\) – \(\frac{1}{6}\) = \(\frac{4}{6}\) = \(\frac{2}{3}\)

(ii) Probability that exactly one of them solves the problem
= P(A)P(B’) + P(B)P(A’) = \(\frac{1}{2}\) × \(\frac{2}{3}\) + \(\frac{1}{2}\) × \(\frac{1}{3}\) = \(\frac{1}{6}\) + \(\frac{1}{6}\) = \(\frac{1}{2}\)

III.

One card is drawn at random from a well shuffled deck of 52 cards. In which of the following cases are the events E and F independent ?
(i) E: ‘the card drawn ¡s a spade’
F: ‘the card drawn is an ace’
(ii) F : ‘the card drawn is black’
F: ‘the card drawn Is a king’ .
(iii) E : ‘the card drawn is a king or queen’
F : ‘the card drawn is a queen or jack’.
Solution:
(i) In a deck of 52 cards, 13 cards are spades and 4 cards are aces.
∴ P(E) = P (the card drawn is a spade) =\(\frac{13}{52}\) = \(\frac{1}{4}\)
Also P(F) = P(the card drawn in an ace) = \(\frac{4}{52}\) = \(\frac{1}{13}\)
In the deck of cards, an ace of spades is only one.
P(EF) = P (the card drawn is spade and an ace) = \(\frac{1}{52}\)
Now P(E) P(F) \(\frac{1}{4}\) × \(\frac{1}{13}\) = \(\frac{1}{52}\) ≠ P(EF)
∴ the events E and F are independent.

(ii) In a deck of 52 cards, 26 cards are black and 4 cards are kings.
26 1
∴ P(E) = P (the card drawn is a black) = \(\frac{26}{52}\) = \(\frac{1}{2}\)
Also P(F) = P (the card drawn in an ace) = \(\frac{4}{52}\) = \(\frac{1}{13}\)
In the pack of 52 cards, 2 cards are black as well as kings.
P(EF) = P( the card drawn is black king) = \(\frac{2}{52}\) = \(\frac{1}{26}\)
Now P(E)P(F) = \(\frac{1}{2}\) × \(\frac{1}{13}\) = \(\frac{1}{26}\) = P(EF)
∴ the given events E and F are independent.

(iii) In a deck of 52 cards, 4 cards are kings, 4 cards are queens, and 4 cards are jacks.
∴ P(E) = P(the card drawn is a king or a queen) = \(\frac{8}{52}\) = \(\frac{2}{13}\)
Also P(F) = P(the card drawn in a queen or a jack) = \(\frac{8}{52}\) = \(\frac{3}{13}\)
There are 4 cards which are king or queen and queen or jack.
P(EF) = P(the card drawn is king or a queen, or queen or a jack) = \(\frac{4}{52}\) = \(\frac{1}{13}\)
Now P(E)P(F) = \(\frac{2}{13}\) × \(\frac{2}{13}\) = \(\frac{4}{1}\) ≠ \(\frac{1}{13}\)
P(E)P(F) ≠ P(EF)
∴ the given events E and F are not independent.

Question 2.
In a hostel, 60% of the students read Hindi newspaper, 40% read English newspaper and 20% read both Hindi and English newspapers. A student is selected at random.
(a) Find the probability that she reads neither Hindi nor English newspapers.
(b) If she reads Hindi newspaper, find the probability that she reads English newspaper.
(c) If she reads English newspaper, find the probability that she reads Hindi newspaper.
Solution:
Let H denote the students who read Hindi newspaper and
E denote the students who read English newspaper.
Given that, P (H ) = 60% = \(\frac{60}{100}\) = \(\frac{3}{5}\)
P(E) = 40% = \(\frac{40}{100}\) = \(\frac{2}{5}\)
P(H ∩ E) = 20% = \(\frac{20}{100}\) = \(\frac{1}{5}\)

(i) Probability that a student reads neither Hindi nor English newspaper
P(H’ ∪ E’) = 1 – P(H ∪ E)
= 1 – [P(H) + P(E) – P(H ∩ E)]
= 1 – (\(\frac{3}{5}\) + \(\frac{2}{5}\) – \(\frac{1}{5}\))
= 1 – \(\frac{4}{5}\) = \(\frac{1}{5}\)

(ii) Probability that a randomly chosen student reads English newspaper, if she reads Hindi newspaper, is given by P(E|H)
P(E|H) = \(\frac{P(E \cap H)}{P(H)}\)
= \(\frac{\frac{1}{5}}{\frac{3}{5}}\) = \(\frac{1}{3}\)

(iii) Probability that a randomly chosen student reads Hindi newspaper, if she reads English newspaper, is given by P(H|E).
P(H|E) = \(\frac{P(H \cap E)}{P(E)}\)
= \(\frac{\frac{1}{5}}{\frac{2}{5}}\) = \(\frac{1}{2}\)

Referring to the AP Inter 2nd Year Maths Study Material Chapter 9 Differential Equations Exercise 9e Solutions makes it easier to understand complex problems.

AP Inter 2nd Year Maths Differential Equations Solutions Exercise 9e

II.

Question 1.
Find the general solution of \(\frac{d y}{d x}\) + 2y = sinx
Solution:
Given D.E. is \(\frac{d y}{d x}\) + 2y = sin x
This is in the form \(\frac{d y}{d x}\) + Py = Q where p = 2 and Q = sin x
IF = \(e^{\int P d x}=e^{\int 2 d x}=e^{2 x}\)
G.S: y(I.F) = \(\int(Q \times I . F) \cdot d x+C \Rightarrow y e^{2 x}=\int \sin x e^{2 x} d x+C\) …..(1)
Let I = \(\int \sin x \cdot e^{2 x} d x \Rightarrow I=\sin x \int e^{2 x} d x-\int\left(\frac{d}{d x}(\sin x) \int e^{2 x} d x\right) d x\)
⇒ I = \(\sin x \cdot \frac{e^{2 x}}{2}-\int\left(\cos x \cdot \frac{e^{2 x}}{2}\right) d x\)
⇒ I = \(\frac{e^{2 x} \sin x}{2}-\frac{1}{2}\left[\cos x \cdot \int e^{2 x}-\int\left(\frac{d}{d x}(\cos x) \cdot \int e^{2 x} d x\right) d x\right]\)
⇒ I = \(\frac{e^{2 x} \sin x}{2}-\frac{1}{2}\left[\cos x \cdot \frac{e^{2 x}}{2}-\int\left[(-\sin x) \cdot \frac{e^{2 x}}{2}\right] d x\right]\)
⇒ I = \(\frac{e^{2 x} \sin x}{2}-\frac{e^{2 x} \cos x}{4}-\frac{1}{4} \int\left(\sin x \cdot e^{2 x}\right) d x\)
⇒ I = \(\frac{e^{2 x}}{4}(2 \sin x-\cos x)-\frac{1}{4} I \Rightarrow \frac{5}{4} I=\frac{e^{2 x}}{4}(2 \sin x-\cos x)\)
⇒ I = \(\frac{e^{2 x}}{5}\)(2sin x – cos x)
(1) ⇒ ye^2x = \(\frac{e^{2 x}}{5}\)(2sin x – cos x) + C ⇒ y = \(\frac{1}{5}\)(2sin x – cos x) + Ce^-2x

Question 2.
Find the general solution of \(\frac{d y}{d x}\) + 3y = e^-2x
Solution:
Given D.E. is in the form \(\frac{d y}{d x}\) + Py = Q where, p = 3 and Q = e^-2x
IF = \(e^{\int P d x}=e^{\int 3 d x}=e^{3 x}\)
GS: y(I.F) = ∫(Q × F)dx + C
⇒ ye^3x = ∫(e^-2x x e^3x) + C ⇒ ye^3x = ∫e^xdx + C ⇒ ye^3x = e^x + C ⇒ y = e^-2x + Ce^-3x

Question 3.
Find the general solution of \(\frac{d y}{d x}+\frac{y}{x}\) = x²
Solution:
Given D.E. is in the form \(\frac{d y}{d x}\) + Py = Q where, p = \(\frac{1}{x}\) and Q = x²
IF = \(e^{\int P d x}=e^{\int \frac{1}{x} d x}=e^{\log x}\) = x
GS: y(I.F) = \(\int(\mathrm{Q} \times \mathrm{IF}) \mathrm{dx}+\mathrm{C}\)
∴ yx = ∫(x².x)dx + C ⇒ yx = ∫ x³dx + C ⇒ xy = \(\frac{x^4}{4}\) + C

Question 4.
Find the general solution of \(\frac{d y}{d x}\) + (sec x)y = tan x,(0 ≤ x < \(\frac{\pi}{2}\))
Solution:
Given D.E. is in the form \(\frac{d y}{d x}\) + Py = Q where, P = sec x and Q = tan x
IF = \(\) = sec x + tan x
GS: y(I.F) = ∫(Q × IF)dx + C ⇒ y(sec x + tan x) = ∫tan x(sec x + tan x)dx + C
⇒ y(sec x + tan x) = ∫secx tan x dx + ∫tan² dx + C
⇒ y(sec x+ tan x) = sec x + ∫(sec²x – 1)dx + C ⇒ y(sec x + tan x) = sec x + tan x – x + C

Question 5.
Find the general solution of cos² x\(\frac{d y}{d x}\) + y = tan x(0 ≤ x < \(\frac{\pi}{2}\))
Solution:
Given D.E. is cos² x\(\frac{d y}{d x}\) + y = tan x(0 ≤ x < \(\frac{\pi}{2}\)) ⇒ \(\frac{d y}{d x}+\frac{y}{\cos ^2 x}=\frac{\tan x}{\cos ^2 x}\)
⇒ \(\frac{d y}{d x}\) + (sec² x)y = sec² x.tan x which is in the form of \(\frac{d y}{d x}\) + Py = Q
where, P = sec² x and Q = sec² x. tan x IF = \(\)
G.S: y(IF) = ∫(Q × IF)dx + C ⇒ ye^{tan x} = ∫(sec² x tan x e^{tan x})dx + C
⇒ ye^{tan x} = e^{tan x}(tan x – 1) + C ⇒ y = (tan x -1) + Ce^{-tan x}

Question 6.
Find the general solution of \(x \frac{d y}{d x}\) + 2y = x² log x
Solution:
Given D.E. is \(x \frac{d y}{d x}\) + 2y = x² log x ⇒ \(\frac{d y}{d x}+\frac{2}{x} y\) = x log x
\(\frac{d y}{d x}\) + PY = Q(where, P = \(\frac{2}{x}\) and Q = x log x)
I.F = \(e^{\int P d x}=e^{\int \frac{2}{x} d x}=e^{2 \log x}=e^{\log x^2}=x^2\)
G.S: y(IF) = ∫(Q × I.F)dx + C
∴ y.x² = ∫(x logx.x²)dx + C ⇒ x² y = ∫(x³logx)dx + C
⇒ x²y = log x ∫x³dx – ∫[\(\frac{d}{d x}\)(log x)∫x³ dx]dx + C
⇒ x²y = \(\log x \cdot \frac{x^4}{4}-\int\left(\frac{1}{x} \cdot \frac{x^4}{4}\right) d x+C \Rightarrow x^2 y=\frac{x^4 \log x}{4}-\frac{1}{4} \int x^3 d x+C\)
⇒ x²y = \(\frac{x^4 \log x}{4}-\frac{1}{4} \frac{x^4}{4}+C\)
⇒ x²y = \(\frac{1}{16} x^4(4 \log x-1)+C \Rightarrow y=\frac{1}{16} x^2(4 \log x-1)+C x^{-2}\)

Question 7.
Find the general solution of x log x \(\frac{d y}{d x}\) + y = \(\frac{2}{x}\)log x
Solution:
Given D.E. is x log x\(\frac{d y}{d x}\) + y = \(\frac{2}{x}\)log x ⇒ \(\frac{d y}{d x}+\frac{y}{x \log x}=\frac{2}{x^2}\), which is in the form of
\(\frac{d y}{d x}\) + Py = Q (where, P = \(\frac{1}{x \log x}\) and Q = \(\frac{2}{x^2}\))
IF = \(\) = e^{log(log x)} = log x
G.S: y(IF) = ∫(Q × I.F)dx + C ⇒ y log x = \(\int\left(\frac{2}{x^2} \log x\right) d x+C\)

∴ y log x = \(\frac{-2}{x}\)(1 + log x) + C is the required general solution of the given D.E.

Question 8.
Find the general solution of (1 + x²)dy + 2xydx = cot xdx, (x ≠ 0)
Solution:
Given D.E. is (1 + x²)dy + 2xydx = cot xdx ⇒ \(\frac{d y}{d x}+\frac{2 x y}{\left(1+x^2\right)}=\frac{\cot x}{\left(1+x^2\right)}\), which is in the form of \(\frac{d y}{d x}\) + Py = Q(where, P = \(\frac{2 x}{\left(1+x^2\right)}\) and Q = \(\frac{\cot x}{\left(1+x^2\right)}\))
IF = \(\mathrm{e}^{\int \mathrm{Pdx}}=\mathrm{e}^{\int \frac{2 \mathrm{x}}{1+\mathrm{x}^2} \mathrm{dx}}=\mathrm{e}^{\log \left(1+\mathrm{x}^2\right)}\) = 1 + x²
G.S: y(IF) = ∫(Q × IF)dx + C
∴ y(1 + x²) = \(\int\left(\frac{\cot x}{1+x^2} \times\left(1+x^2\right)\right) d x+C=\int \cot x d x+C\)
⇒ y(1 + x²) = log|sin x| + C

Question 9.
Find the general solution of x\(\frac{d y}{d x}\) + y – x + xy cotx = 0, (x ≠ 0)
Solution:
Given D.E. is x\(\frac{d y}{d x}\) + y – x + xy cotx = 0, (x ≠ 0)
⇒ \(\frac{d y}{d x}+\frac{y}{x}-1+y \cot x=0 \Rightarrow \frac{d y}{d x}+y\left(\frac{1}{x}+\cot x\right)-1=0\)
⇒ \(\frac{d y}{d x}+\left(\frac{1}{x}+\cot x\right) y=1\) which is in the form of
\(\frac{d y}{d x}\) + Py = Q(where, P = (\(\frac{1}{x}\) + cot x) and Q = 1)
IF = \(\mathrm{e}^{\int \mathrm{Pdx}}=\mathrm{e}^{\int\left(\frac{1}{\mathrm{x}}+\cot \mathrm{x}\right) \mathrm{dx}}=\mathrm{e}^{\log \mathrm{x}+\log (\sin \mathrm{x})}=\mathrm{e}^{\log (\mathrm{x} \sin \mathrm{x})}\) = x sin x
G.S: y(IF) = ∫(Q × IF)dx + C
⇒ y(x sin x) = (1 × x sin x)dx + C
⇒ y(x sin x) = (x sin x)dx + C
⇒ y(x sin x) = \(x \int \sin x d x-\int\left[\frac{d}{d x}(x) \cdot \int \sin x d x\right]+C\)
⇒ y(x sin x) = x(-cos x) – \(\int 1 \cdot(-\cos x) d x+C\)
⇒ y(x sin x) = -x cos x + sin x + C
⇒ y = \(\frac{-x \cos x}{x \sin x}+\frac{\sin x}{x \sin x}+\frac{C}{x \sin x}\)
⇒ y = \(-\cot x+\frac{1}{x}+\frac{C}{x \sin x} \Rightarrow y=\frac{1}{x}-\cot x+\frac{C}{x \sin x}\)

Question 10.
Find the general solution of (x + y)\(\frac{d y}{d x}\) = 1
Solution:
Given D.E is (x + y)\(\frac{d y}{d x}\) = 1 ⇒ \(\frac{d y}{d x}=\frac{1}{x+y} \Rightarrow \frac{d x}{d y}=x+y \Rightarrow \frac{d x}{d y}-x=y\) which is in the form of \(\frac{d x}{d y}\) + P₁x = Q₁(where, P₁ = -1 and Q₁ = y)
IF = \(\mathrm{e}^{\int P_1 d y}=\int \mathrm{e}^{-1 \mathrm{dy}}\) = e^-y
G.S: x(I.F) = ∫(Q × IF)dx + C

Question 11.
Find the general solution of ydx + (x – y²)dy = 0
Solution:
Given D.E is ydx + (x – y²)dy = 0 ⇒ ydx – (y² -x)dy = 0
⇒ \(\frac{d x}{d y}=\frac{y^2-x}{y}=y-\frac{x}{y} \Rightarrow \frac{d x}{d y}+\frac{x}{y}=y\), which is in the form of
\(\frac{d x}{d y}\) + P₁x = Q₁(where, P₁ = \(\frac{1}{y}\) and Q₁ = y)
IF = \(e^{\int P_1 d y}=\int e^{\frac{1}{y} d y}=e^{\log y}\) = y
G.S: x(I.F) = ∫(Q₁ × I.F)dy + C ⇒ xy = ∫(y.y)dy + C
⇒ xy = ∫y²dy + C
⇒ xy = \(\frac{y^3}{3}\) + C ⇒ x = \(\frac{y^2}{3}+\frac{C}{y}\)

Question 12.
Find the general solution of (x + 3y²)\(\frac{d y}{d x}\) = y (y > 0)
Solution:
Given D.E is (x + 3y²)\(\frac{d y}{d x}\) = y ⇒ \(\frac{d y}{d x}=\frac{y}{x+3 y^2}\)
⇒ \(\frac{d x}{d y}=\frac{x+3 y^2}{y}=\frac{x}{y}+3 y \Rightarrow \frac{d x}{d y}-\frac{x}{y}=3 y\), which is in the form of
\(\frac{d x}{d y}\) + P₁x = Q₁(where, P₁ = \(-\frac{1}{y}\) and Q₁ = 3y)
IF = \(e^{\int P_i d y}=e^{-\int \frac{d y}{y}}=e^{-\log y}=e^{\log y^{-1}}=\frac{1}{y}\)
G.S: x(I.F) = ∫(Q₁ × I.F)dy + C
∴ \(x \times \frac{1}{y}=\int\left(3 y \times \frac{1}{y}\right) d y+C \Rightarrow \frac{x}{y}=3 y+C \Rightarrow x=3 y^2+C y\)

III.

Question 1.
Find the particular solution of the differential equation
\(\frac{d y}{d x}\) + 2y tan x = sin x; y = 0 when x = \(\frac{\pi}{3}\)
Solution:
Given D.E. is \(\frac{d y}{d x}\) + 2y tan x = sin x, which is in the form of
\(\frac{d y}{d x}\) + Py = Q(where, P = 2 tan x and Q = sin x)
IF = \(e^{\int P d x}=e^{\int 2 \tan x d x}=e^{2 \log |\sec x|}=e^{\log \left(\sec ^2 x\right)}=\sec ^2 x\)
G.S: y(IF) = \(\int(Q \times I . F) d x+C \Rightarrow y\left(\sec ^2 x\right)=\int\left(\sin x \cdot \sec ^2 x\right) d x+C\)
∴ y sec² x = ∫(sec x. tan x)dx + C ⇒ y sec² x = sec x + C
we have y = 0 at x = \(\frac{\pi}{3}\)
⇒ 0 × sec²\(\frac{\pi}{3}\) = sec\(\frac{\pi}{3}\) + C ⇒ 0 = 2 + C ⇒ C = -2
∴ ysec² x = sec x – 2 ⇒ y = cos x – 2cos² x, which is the required particular solution.

Question 2.
Find the particular solution of the differential equation.
(1 + x²)\(\frac{d y}{d x}\) + 2xy = \(\frac{1}{1+x^2}\); y = 0 when x = 1
Solution:
Given D.E is (1 + x²)\(\frac{d y}{d x}\) + 2xy = \(\frac{1}{1+x^2}\) ⇒ \(\frac{d y}{d x}+\frac{2 x y}{1+x^2}=\frac{1}{\left(\left(1+x^2\right)\right)^2}\) which is in the form of
\(\frac{d y}{d x}\) + Py = Q(where, P = \(\frac{2 x}{1+x^2}\) and Q = \(\frac{1}{\left(1+x^2\right)^2}\))
IF = \(e^{\int P d x}=e^{\int \frac{2 x}{1+x^2} d x}=e^{\log \left(1+x^2\right)}\) = 1 + x²
G.S: y(IF) = ∫(Q × I.F)dx + C
∴ \(y\left(1+x^2\right)=\int\left[\frac{1}{\left(1+x^2\right)^2} \cdot\left(1+x^2\right)\right] d x+C \Rightarrow y\left(1+x^2\right)=\int \frac{1}{1+x^2} d x+C\)
⇒ y(1 + x²) = tan^-1 x + C …………(1)
we have y = 0 at x = 1 ⇒ C = \(-\frac{\pi}{4}\); y(1 + x²) = tan^-1 x – \(\frac{\pi}{4}\) is the required solution.

Question 3.
Find the particular solution of the differential equation.
\(\frac{d y}{d x}\) – 3y cot x = sin 2x; y = 2 when x = \(\frac{\pi}{2}\)
Solution:
\(\frac{d y}{d x}\) – 3y cot x = sin 2x which is in the form of \(\frac{d y}{d x}\) + Py = Q(where, P = -3cot x and Q = sin 2x)

we have y = 2 at x = \(\frac{\pi}{2}\) ⇒ 2 = -2 + C ⇒ C = 4
∴ y = -2sin² x + 4sin³ x ⇒ y = 4sin³x – 2sin²x is the required particular solution.

Question 4.
Find the equation of a curve passing through the origin given that the slope of the tangent to the curve at any point (x, y) is equal lo the sum of the coordinates of the point.
Solution:
Let F(x, y)be the curve passing through origin.
At (x, y), slope of curve will be \(\frac{d y}{d x}\)
∴ \(\frac{d y}{d x}\) = x + y ⇒ \(\frac{d y}{d x}\) – y = x, which is in the form of \(\frac{d y}{d x}\) + Py = Q (where, P = -1 and Q = x)
IF = \(e^{\int P d x}=e^{\int(-1) d x}=e^{-x}\)
∴ ye^-x = \(\int x e^{-x} d x+C \Rightarrow y e^{-x}=x \int e^{-x} d x-\int\left[\frac{d}{d x}(x) \cdot \int e^{-x} d x\right] d x+C\)
⇒ \(y e^{-x}=-x e^{-x}+\int e^{-x} d x+C \Rightarrow y e^{-x}=-x e^{-x}+\left(-e^{-x}\right)+C \Rightarrow y e^{-x}=-e^{-x}(x+1)+C\)
⇒ x + y + 1 = ce^-x ………..(1)
As the curve passes through origin, O(0, 0) from (1) we have 0 + 0 + 1 = C.e ⇒ C = 1
(1) ⇒ x + y + 1 = e^x which is the required equation of the curve.

Question 5.
Find the equation of a curve passing through the point (0, 2) given that the sum of the coordinates of any point on the curve exceeds the magnitude of the slope of the tangent to the curve at that point by 5.
Solution:
F(x, y) be the curve and let (x, y) be a point on the curve.
Slope of the tangent to curve at (x, y)\(\frac{d y}{d x}\)
Given that \(\frac{d y}{d x}\) + 5 = x + y ⇒ \(\frac{d y}{d x}\) – y = x – 5 which is in the form of \(\frac{d y}{d x}\) + Py = Q(where, P = -1 and Q = x – 5)
IF = \(\mathrm{e}^{\int \mathrm{Pdx}}=\mathrm{e}^{\int(-1) \mathrm{dx}}=\mathrm{e}^{-\mathrm{x}}\)
G.S: \(y(I . F)=\int(Q \times I F) d x+C \Rightarrow y e^{-x}=\int(x-5) e^{-x} d x+C\)
= \(\int(x-5) e^{-x} d x=(x-5) \int e^{-x} d x-\int\left[\frac{d}{d x}(x-5) \int e^{-x} d x\right] d x\)
= \((x-5)\left(-e^{-x}\right)-\int\left(-e^{-x}\right) d x=(5-x) e^{-x}-\left(e^{-x}\right)=(4-x) e^{-x}\)
⇒ ye^-x = (4 – x)e^-x + C ⇒ y = (4 – x) + Ce^-x ………….(1)
Given that the curve passes through (0, 2)
⇒ 2 = 4 + c ⇒ c = -2
(1) ⇒ y = 4 – x – 2e^x

Referring to the AP Inter 2nd Year Maths Study Material Chapter 6 Application of Derivatives Exercise 6d Solutions makes it easier to understand complex problems.

AP Inter 2nd Year Maths Application of Derivatives Solutions Exercise 6d

II.

Question 1.
Show that the function given by f(x) = \(\frac{\log x}{x}\), has maximum at x = e.
Solution:

∴ by second derivative test, f is the maximum at x = e.

Question 2.
The to equal sides of an isosceles triangle with fixed base h are decreasing at the rate of 3 cm per second. how fast is the area decreasing when the two equal sides are equal to the base ?
Solution:
Let ∆ABC be isosceles where BC is the base of fixed length.
Also, let the length of the two equal sides of ∆ABC be a. Draw AD ⊥ BC.

Now in ∆ADC by applying the Pythagoras theorem, we have: AD = \(\sqrt{a^2-\frac{b^2}{4}}\)
Area of triangle, A = \(\frac{1}{2} b \sqrt{a^2-\frac{b^2}{4}}\)
The rate of change of the area with respect to time (t) is given by,
\(\frac{\mathrm{dA}}{\mathrm{dt}}=\frac{1}{2} b \cdot \frac{2 a}{2 \sqrt{a^2-\frac{b^2}{4}}} \frac{d a}{d t}=\frac{a b}{\sqrt{4 a^2-b^2}} \frac{d a}{d t}\)
It is given that the two equal sides of the triangle are decreasing at the rate of 3 cm per second.
∴ \(\frac{\mathrm{da}}{\mathrm{dt}}\) = -3cm/s
Hence, \(\frac{\mathrm{dA}}{\mathrm{dt}}=\frac{-3 \mathrm{ab}}{\sqrt{4 \mathrm{a}^2-\mathrm{b}^2}}\)
When a = b, we have \(\frac{\mathrm{dA}}{\mathrm{dt}}=\frac{-3 b^2}{\sqrt{4 a^2-b^2}}=\frac{-3 b^2}{\sqrt{3 b^2}}=-\sqrt{3} b\)
Hence, if the two equal sides are equal to the base, then the area of the triangle is decreasing at the rate of \(\sqrt{3} \mathrm{~b}\) cm² / s.

Question 3.
Find the intervals in which the function f given by f(x) = \(\frac{4 \sin x-2 x-x \cos x}{2+\cos x}\) is (i) increasing (ii) decreasing
Solution:

Now, f'(x) = 0 ⇒ cos x = 0 or cos x = 4
But cos x ≠ 4
Hence,cos x = 0 ⇒ x = \(\frac{\pi}{2}\), \(\frac{3 \pi}{2}\)
Now x = \(\frac{\pi}{2}\)and x = \(\frac{3\pi}{2}\) divide(0, 2π) into three disjoint intervals i.e.,,
(0, \(\frac{\pi}{2}\)), (\(\frac{\pi}{2}\), \(\frac{3\pi}{2}\)) and (\(\frac{3\pi}{2}\), 2π)
In intervals, (0, \(\frac{\pi}{2}\)) and (\(\frac{3\pi}{2}\), 2π), f'(x) > 0
Thus, f(x) is increasing for 0 < x < \(\frac{\pi}{2}\) and \(\frac{3\pi}{2}\) < x < 2π
In the interval (\(\frac{\pi}{2}\), \(\frac{3\pi}{2}\)) , f'(x) < 0
Thus, f(x) isdecreasing for \(\frac{\pi}{2}\) < x < \(\frac{3\pi}{2}\)

Question 4.
Find the intervals in which function f given by f(x) = x³ + \(\frac{1}{x^3}\), x ≠ 0
(i) increasing
(ii) decreasing
Solution:
Given that f(x) = x³ + \(\frac{1}{x^3}\) ⇒ f'(x) = 3x² – \(\frac{3}{x^4}\) = \(\frac{3 x^6-3}{x^4}\)

Now, f'(x) = 0 ⇒ 3x⁶ – 3 = 0
⇒ x⁶ = 1
⇒ x = ±1
Now, the points x = 1 and x = -1 divide the real line into three disjoint intervals
i.e., ,(-∞, -1), (-1, 1) and (1, ∞)
In intervals (-∞, -1) and (1, ∞) i.e., when x < -1 and x > 1, f'(x) > 0
Thus, when x < -1 and x > 1, f is increasing.
In interval (-1, 1) i.e., -1 < x < 1, f'(x) < 0.
Thus, when -1 < x < 1, f is decreasing.

Question 5.
Find the points at which the function f given by f (x) = (x – 2)⁴ (x + 1)³ has
(i) local maxima
(ii) local minima
(iii) point of inflexion
Solution:
The given function is f (x) = (x – 2)⁴ (x + 1)³
f'(x) = 4(x – 2)³ (x + 1)³ + 3(x + 1)² (x – 2)⁴
= (x – 2)³ (x + 1)² [4(x + 1) + 3(x – 2)] = (x – 2)³ (x + 1)² (7x – 2)
Now, f'(x) = 0 ⇒ x = -1, x = \(\frac{2}{7}\), x = 2
For values of close to \(\frac{2}{7}\) and to the left of \(\frac{2}{7}\), f'(x) > 0
Also, for values of x close to \(\frac{2}{7}\) and to the right of \(\frac{2}{7}\), f'(x) < 0.
Thus, x = \(\frac{2}{7}\) is the point of local maxima.
Now, for values of x close to 2 and to the left of 2, f'(x) < 0 Also, for values of close to 2 and to the right of 2, f'(x) > 0.
Thus, x = 2 is the point of local minima.
Now, as the value of varies through -1, f'(x) does not change its sign.
Thus, x = -1 is the point of inflexion.

Question 6.
Find the absolute maximum and minimum values of the function f given by f (x) = cos² x + sin x, x ∈ [0, π]
Solution:
Given that f (x) = cos² x + sin x, x ∈ [0, π]
f'(x) = 0 ⇒ -2sin x cos x + cos x = 0
⇒ cos x = 2 sin x cos x ⇒ cos x (2 sin x – 1) = 0
⇒ sin x = \(\frac{1}{2}\) or cos x = 0 ⇒ x = \(\frac{\pi}{6}\) or \(\frac{\pi}{2}\) ∵ x ∈ [0, π]
Now we evaluate the value of f at critical points x = \(\frac{\pi}{6}\), \(\frac{\pi}{2}\) and at the, end points of the interval [0, π] i.e., at x = 0 and x = π, we have.
(i) f\(\left(\frac{\pi}{6}\right)\) = c0s²\(\left(\frac{\pi}{6}\right)\) + sin \(\left(\frac{\pi}{6}\right)\) = \(\left(\frac{\sqrt{3}}{2}\right)^2+\frac{1}{2}=\frac{5}{4}\)
(ii) f(0) = cos²(0) + sin(0) = 1 + 0 = 1
(iii) f(π) = cos²(π) + sin(π) = (-1)² + 0 = 1
(iv) f\(\left(\frac{\pi}{2}\right)\) = cos²\(\left(\frac{\pi}{2}\right)\) + sin\(\left(\frac{\pi}{2}\right)\) + sin\(\left(\frac{\pi}{2}\right)\) = 0 + 1 = 1
Hence, the absolute maximum value of f is 5/4 occurring at x = π/6 and the absolute minimum value of f is 1 occurring at x = 0, π/2, π.

Question 7.
Let f be a function defined on [a, b] such that f'(x) > 0. for all x ∈ (a, b). Then prove that f is an increasing function on (a, b).
Solution:
Let x₁, x₂ ∈ (a, b) such that
Consider the sub-interval [x₁, x₂]
Since f(x) is differentiable on (a, b) and [x₁, x₂] ⊂ (a, b).
∴ f(x) is continuous on [x₁, x₂] and differentiable on (x₁, x₂).
By the Lagrange’s mean value theorem, there exists x ∈ (x₁, x₂) such that
f'(c) = \(\frac{f\left(x_2\right)-f\left(x_1\right)}{x_1-x_2}\) ……………. (1)
Since, f(x) > 0 for all x ∈ (a, b), so in particular, f'(c) > 0 ⇒ \(\frac{f\left(x_2\right)-f\left(x_1\right)}{x_1-x_2}\) > 0 [Using (1)]
⇒ f(x₂) – f(x₁) > 0
⇒ f(x₂) > f(x₁)
⇒ f(x₁) < f(x₂)
Since , x₁, x₂ are arbitrary points in (a, b).
∴ x₁ < x₂ ⇒ f(x₁) < f(x₂) for all x₁, x₂ ∈ (a, b).
Hence, f(x) is increasing on (a, b).

III.

Question 1.
Find the maximum area of an isosceles triangle inscribed in the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}\) = 1 with its vertex at one end of the major axis.

Solution:
The given ellipse is \(\frac{x^2}{a^2}+\frac{y^2}{b^2}\) = 1
Let the major axis be along the x-axis.
Let ABC be the triangle inscribed in the ellipse where vertex C is at (a, 0)
Since the ellipse is symmetrical with respect to the x- axis and y-axis, we can assume the
coordinates of A to be (-x₁, y₁) and the coordinates of B to be(-x₁, -y₁)
Now, we have y₁ = ±\(\frac{b}{a} \sqrt{a^2-x_1^2}\)
Coordinates of A are (-x₁, \(\frac{b}{a} \sqrt{a^2-x_1^2}\)) and the coordinates of B are (-x₁, \(\frac{b}{a} \sqrt{a^2-x_1^2}\))
As the point (x₁, y₁) lies on the ellipse, the area of triangle ABC (A) is given by,


Thus, the area is the maximum when x₁ = \(\frac{\mathrm{a}}{2}\).
Hence, Maximum area of the triangle is given by,
A = \(b \sqrt{a^2-\frac{a^2}{4}}+\left(\frac{a}{2}\right) \frac{b}{a} \sqrt{a^2-\frac{a^2}{4}}=a b \frac{\sqrt{3}}{2}+\left(\frac{a}{2}\right) \frac{b}{a} \times \frac{a \sqrt{3}}{2}\)
= \(\frac{a b \sqrt{3}}{2}+\frac{a b \sqrt{3}}{4}=\frac{3 \sqrt{3}}{4} a b\)

Question 2.
A tank with rectangular base and rectangular sides, open at the top is to be constructed so that its depth is 2 m and volume is 8 m³. If building of tank costs Rs 70 per sq metres for the base and Rs 45 per square metre for sides. What is the cost of least expensive tank?
Solution:
Let l, b, and h represent the length, breadth, and height of the tank respectively.
Then, we have height , h = 2m and volume of the tank, V = 8m³
Volume of the tank V = lbh
⇒ 8 = l × b × 2
⇒ lb = 4
⇒ b = \(\frac{4}{l}\)
Now, area of the base, lb = 4
Area of the 4 walls, A = 2h(l + b) = 4(l + \(\frac{4}{l}\)) = 4(l + 4l^-1)
⇒ \(\frac{\mathrm{dA}}{\mathrm{dl}}=4\left(1-\frac{4}{l^2}\right)\)
Now, \(\frac{\mathrm{dA}}{\mathrm{dl}}\) = 0 ⇒ (1 – \(\frac{4}{l^2}\)) = 0
⇒ l² = 4
⇒ l = ± 2
However, the length cannot be negative,
∴ we have l = 2
Hence, b = \(\frac{4}{l}\) = \(\frac{4}{2}\) = 2
Now, \(\frac{\mathrm{d}^2 \mathrm{~A}}{\mathrm{dl}^2}=\frac{32}{l^3}\)
When, l = 2
Then, \(\frac{\mathrm{d}^2 \mathrm{~A}}{\mathrm{~dl}^2}=\frac{32}{8}\) = 4 > 0
Thus, by second derivative test, the area is the minimum when l = 2
We have l = b = h = 2
∴ Cost of building the base in ₹ is 70(lb) = 70(4) = Rs 280
Cost of building the walls in ₹ is 2h(l + b) × 45 = 2 × 2(2 + 2) × 45 = Rs 720
Required total cost is ₹ is 280 + 720 = Rs 1000
Thus, the total cost of the tank will be ₹ 1000.

Question 3.
The sum of the perimeter of a circle and square is k, where k is some constant. Prove that the sum of their areas is least when the side of square is double the radius of the circle.
Solution:
Let V be the radius of the circle and ‘a’ be the side of the square .
Then, we have 2πr + 4a = k ⇒ a = \(\frac{\mathrm{k}-2 \pi \mathrm{r}}{4}\)
The sum of the areas of the circle and the square (A) is given by,
A = πr² + a² = πr² + \(\frac{(\mathrm{k}-2 \pi \mathrm{r})^2}{16}\)
Now, \(\frac{\mathrm{dA}}{\mathrm{dr}}\) = 0
⇒ 2πr – \(\frac{\pi(\mathrm{k}-2 \pi \mathrm{r})}{4}\) = 0 ⇒ 2πr = \(\frac{\pi(\mathrm{k}-2 \pi \mathrm{r})}{4}\)
⇒ 8r = k – 2πr
⇒ (8 + 2π)r = k
⇒ r = \(\frac{\mathrm{k}}{(8+2 \pi)}\)
⇒ r = \(\frac{\mathrm{k}}{2(4+\pi)}\) ………… (1)
When, r = \(\frac{\mathrm{k}}{2(4+\pi)}\) ⇒ \(\frac{\mathrm{d}^2 \mathrm{~A}}{\mathrm{dr}^2}\) > 0
The sum of the areas is least when, r = \(\frac{\mathrm{k}}{2(4+\pi)}\)
When r = \(\frac{\mathrm{k}}{2(4+\pi)}\)
We have a = \(\frac{\mathrm{k}-2 \pi\left[\frac{\mathrm{k}}{2(4+\pi)}\right]}{4}=\frac{\mathrm{k}(4+\pi)-\pi \mathrm{k}}{4(4+\pi)}\)
= \(\frac{4 \mathrm{k}}{4(4+\pi)}=\frac{\mathrm{k}}{4+\pi}\)
= 2r [From (1)]
Hence, it has been proved that the sum of their areas is least when the side of the square is double the radius of the circle.

Question 4.
A window is in the form of a rectangle surmounted by a semicircular opening. The total perimeter of the window is 10 m. Find the dimensions of the window to admit maximum light through the whole opening.

Solution:
Let x and y be the length and breadth of the rectangular window.
Radius of the semicircular opening be x/2.
It is given that the perimeter of the window is 10m.

∴ by second derivative test, the area is the maximum when length is \(\frac{20}{\pi+4}\) m
Now, y =5 – \(\frac{20}{\pi+}\left(\frac{2+\pi}{4}\right)=5-\frac{5(2+\pi)}{\pi+4}=\frac{10}{\pi+4}\)
Hence, the required dimensions of the window to admit maximum light is given by length \(\frac{20}{\pi+4}\) m and breadth \(\frac{10}{\pi+4}\) m.

Question 5.
A point on the hypotenuse of a triangle is at distance a and b from the sides of the triangle. Show that the minimum length of the hypotenuse is \(\left(a^{\frac{2}{3}}+b^{\frac{2}{3}}\right)^{\frac{3}{2}}\)
Solution:
Let ∆ABC be right-angled at B, AB = x, BC = y and ZC = 0
Also, let P be a point on the hypotenuse of the triangle such that P is at a distance of a and b from the sides AB and BC respectively.

We have, AC = \(\sqrt{x^2+y^2}\) Now, PC = b cosecθ; AP = asecθ
Hence, AC = AP + PC ⇒ AC = bcosec0 + asec0 (1)
Therefore, \(\frac{\mathrm{d}(\mathrm{AC})}{\mathrm{d} \theta}\) = -b cosec θ cot θ + a sec θ tan θ
Now, \(\frac{\mathrm{d}(\mathrm{AC})}{\mathrm{d} \theta}\) = 0 ⇒ -b cosec θ cot θ + a sec θ tan θ = 0
⇒ a sec θ tan θ = b cosec θ cot θ dθ

Thus, the maximum length of the hypotenuse is \(\left(a^{\frac{2}{3}}+b^{\frac{2}{3}}\right)^{\frac{3}{2}}\) proved.

Question 6.
Show that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius r is \(\frac{4 r}{3}\).

Solution:
A sphere of fixed radius (r) is given.
Let R and h be the radius and the height of the cone respectively.
The volume V of the cone is given by V = \(\frac{1}{3}\) πR² h
Now, from the right ABCD , we have: BC = \(\sqrt{\mathrm{r}^2-\mathrm{R}^2}\) ⇒ h = r + \(\sqrt{\mathrm{r}^2-\mathrm{R}^2}\)


Hence, it can be seen that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius 4π/3

Question 7.
Show that the height of the cylinder of maximum volume that can be inscribed in a sphere of radius R is \(\frac{2 R}{\sqrt{3}}\). Also find the maximum volume.

Solution:
A sphere of fixed radius (R) is given.
Let r and h be the radius and the height of the cylinder respectively.
From the given figure, we have h = 2\(\sqrt{R^2-r^2}\)
The volume (V) of the cyclinder is given by, V = πr²h = 2πr²\(\sqrt{R^2-r^2}\)
∴ V = 2πr²\(\sqrt{R^2-r^2}\)

Hence, the volume of the cylinder is the maximum when the height of the cylinder is \(\frac{2 R}{\sqrt{3}}\).
∴ Maximum volume of cylinder = πr²h
= 2πr²\(\sqrt{R^2-r^2}\) = 2π\(\left(\frac{2 \mathrm{R}^2}{3}\right) \sqrt{\mathrm{R}^2-\left(\frac{2 \mathrm{R}^2}{3}\right)}\) = \(\frac{4 \pi}{3 \sqrt{3}} R^3\)

Question 8.
Show that height of the cylinder of greatest volume which can be inscribed in a right circular cone of height h and semi vertical angle a is one-third that of the cone and the greatest volume of cylinder is \(\frac{4}{27}\) πh³ tan² α.

Solution:
The given right circular cone of fixed height h and semi-vertical angle a can be drawn as:
Here, a cylinder of radius R and height H is inscribed in the cone.
Then ∠GAO = α, OG = r, OA = h; OE = r and CE = H
We have, r = h tanα
Now, since ∆AOG is similar to ∆ CEQ we have:

Thus, the height of the cylinder is one-third the height of the cone when the volume of the cylinder is the greatest.
Thus, the maximum volume of the cylinder can be obtained as
\(\pi\left(\frac{2 \mathrm{~h}}{3} \tan \alpha\right)^2\left(\frac{\mathrm{~h}}{3}\right)=\pi\left(\frac{4 \mathrm{~h}^2}{9} \tan ^2 \alpha\right)\left(\frac{\mathrm{h}}{3}\right)=\frac{4}{27} \pi \mathrm{~h}^3 \tan ^2 \alpha\)
Hence, the given result is proved.

Referring to the AP Inter 2nd Year Maths Study Material Chapter 9 Differential Equations Exercise 9d Solutions makes it easier to understand complex problems.

AP Inter 2nd Year Maths Differential Equations Solutions Exercise 9d

I.

Question 1.
Show that the differential equatin y’ = \(\frac{x+y}{x}\) is homogenous and solve it.
Solution:
Given D.E. is y’ = \(\frac{x+y}{x} \Rightarrow \frac{d y}{d x}=\frac{x+y}{x}\) …………(1)
Let F(x, y) = \(\frac{x+y}{x}\)
Then F(λx, λy) = \(\frac{\lambda \mathrm{x}+\lambda \mathrm{y}}{\lambda \mathrm{x}}=\frac{\lambda(\mathrm{x}+\mathrm{y})}{\lambda(\mathrm{x})}=\lambda^0 \mathrm{~F}(\mathrm{x}, \mathrm{y})\)
Given Differential Equation is a homogenous equation.
Let y = vx ⇒ \(\frac{d y}{d x}\) = v + x\(\frac{d v}{d x}\) (1) v + x\(\frac{d v}{d x}\) = \(\frac{x+v x}{x}\)
⇒ \(v+x \frac{d v}{d x}=1+v \Rightarrow x \frac{d v}{d x}=1 \Rightarrow d v=\frac{d x}{x} \Rightarrow \int d v=\int \frac{d x}{x} \Rightarrow v=\log |x|+C\)
⇒ \(\frac{y}{x}=\log |x|+C \Rightarrow y=x \log |x|+C x\)
This is the required general solution to the given D.E.

Question 2.
Show that the differential equation \(x \frac{d y}{d x}-y+x \sin \left(\frac{y}{x}\right)=0\) is homogenous and solve it.
Solution:
Given D.E. is \(x \frac{d y}{d x}-y+x \sin \left(\frac{y}{x}\right)=0 \Rightarrow x \frac{d y}{d x}=y-x \sin \left(\frac{y}{x}\right) \Rightarrow \frac{d y}{d x}=\frac{y-x \sin \left(\frac{y}{x}\right)}{x}\) ………..(1)
Let F(x, y) = \(\frac{y-x \sin \left(\frac{y}{x}\right)}{x}\)
Then F(λx, λy) = \(\frac{\lambda \mathrm{y}-\lambda \mathrm{x} \sin \left(\frac{\lambda \mathrm{y}}{\lambda \mathrm{x}}\right)}{\lambda \mathrm{x}}=\frac{\lambda\left(\mathrm{y}-\mathrm{x} \sin \left(\frac{\mathrm{y}}{\mathrm{x}}\right)\right)}{\lambda(\mathrm{x})}=\lambda^0 \mathrm{~F}(\mathrm{x}, \mathrm{y})\)
∴ Given differential equation is a homogenous equation.
Let y = vx ⇒ \(\frac{d y}{d x}=\frac{d}{d x}(v x) \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}\)
Substituting this in eqn (1), we have v + x\(x \frac{d v}{d x}=\frac{v x-x \sin v}{x}\)
⇒ \(v+x \frac{d v}{d x}=v-\sin v \Rightarrow \frac{d v}{\sin v}=-\frac{d x}{x} \Rightarrow {cosec} v d v=-\frac{d x}{x}\)
log|cosec v – cot v| = -log x + log C = \(\log \frac{C}{x}\)
⇒ \({cosec}\left(\frac{y}{x}\right)-\cot \left(\frac{y}{x}\right)=\frac{C}{x} \Rightarrow \frac{1}{\sin \left(\frac{y}{x}\right)}-\frac{\cos \left(\frac{y}{x}\right)}{\sin \left(\frac{y}{x}\right)}=\frac{C}{x}\)
⇒ \(x\left[1-\cos \left(\frac{y}{x}\right)\right]=C \sin \left(\frac{y}{x}\right)\)
This is the required general solution to the given D.E.

III.

Question 1.
Show that the differential equation (x² + xy)dy = (x² + y²) dx is homogenous and solve it.
Solution:
Given D.E. is (x² + xy)dy = (x² + y²)dx and it can be written as \(\frac{d y}{d x}=\frac{x^2+y^2}{x^2+x y}\) ……..(1)
Let F(x, y) = \(\frac{x^2+y^2}{x^2+x y}\)
Then F(λx, λy) = \(\frac{(\lambda x)^2+(\lambda y)^2}{(\lambda x)^2+(\lambda x)(\lambda y)}=\frac{\lambda^2\left(x^2+y^2\right)}{\lambda^2\left(x^2+x y\right)}=\lambda^0 \mathrm{~F}(\mathrm{x}, \mathrm{y})\)
∴ Given differential equation is a homogenous equation.

This is the required general solution to the given D.E.

Question 2.
Show that the differential equation (x – y)dy – (x + y) dx = 0 is homogenous and solve it.
Solution:
Given D.E. (x – y)dy – (x + y)dx = 0 ⇒ \(\frac{d y}{d x}=\frac{x+y}{x-y}\) ………..(1)
Let F(x, y) = \(\frac{x+y}{x-y}\)
Then F(λx, λy) = \(\frac{\lambda x+\lambda y}{\lambda x-\lambda y}=\frac{\lambda(x+y)}{\lambda(x-y)}=\lambda^0 \mathrm{~F}(x, y)\)
∴ Given differential equation is a homogenous equation.

This is the required general solution to the given D.E.

Question 3.
Show that the differential equation (x² – y²)dy + 2xy dy = 0 is homogenous and solve it.
Solution:
Given D.E. is (x² – y²)dx + 2xydy = 0 ⇒ \(\frac{d y}{d x}=-\frac{\left(x^2-y^2\right)}{2 x y}\) ……………….(1)
Let F(x, y) = \(-\frac{\left(x^2-y^2\right)}{2 x y}\)
Then F(λx, λy) = \(\left[\frac{(\lambda x)^2-(\lambda y)^2}{2(\lambda x)(\lambda y)}\right]=\frac{-\lambda^2\left(x^2-y^2\right)}{\lambda^2(2 x y)}=\lambda^0 F(x, y)\)
∴ Given differential equation is a homogenous equation.

This is the required general solution to the given D.E.

Question 4.
Show that the differential equation x²\(\frac{d y}{d x}\) = x² – 2y² + xy is homogenous and solve it.
Solution:
Given D.E. is x²\(\frac{d y}{d x}\) = x² – 2y² + xy ⇒ \(\frac{d y}{d x}=\frac{x^2-2 y^2+x y}{x^2}\) ……..(1)
Let F(x, y) = \(\frac{x^2-2 y^2+x y}{x^2}\)
Then F(λx, λy) = \(\frac{(\lambda x)^2-2(\lambda y)^2+(\lambda x)(\lambda y)}{(\lambda x)^2}=\frac{\lambda^2\left(x^2-2 y^2+x y\right)}{\lambda^2\left(x^2\right)}=\lambda^0 F(x, y)\)
∴ Given differential equation is a homogenous equation.

This is the required general solution to the given D.E.

Question 5.
Show that the differential equation xdy – ydx = \(\sqrt{x^2+y^2} d x\) is homogenous and solve it.
Solution:
Given D.E. is xdy – ydx = \(\sqrt{x^2+y^2} d x\) ⇒ xdy = ydx + \(\sqrt{x^2+y^2} d x\)dx
⇒ xdy = \(\left(y+\sqrt{x^2+y^2}\right) d x \Rightarrow \frac{d y}{d x}=\frac{y+\sqrt{x^2+y^2}}{x}\) = F(x, y) …….(1). This is a homogenous D.E
Then F(λx, λy) = \(\frac{\lambda y+\sqrt{(\lambda x)^2+(\lambda y)^2}}{\lambda x}=\lambda\left(\frac{y+\sqrt{x^2+y^2}}{\lambda x}\right)=\lambda^0 F(x, y)\)
Let y = vx ⇒ \(\frac{d y}{d x}=v+x \frac{d v}{d x}\)

This is the required general solution to the given D.E.

Question 6.
Show that the differential equation \(\left\{x \cos \left(\frac{y}{x}\right)+y \sin \left(\frac{y}{x}\right)\right\} y d x=\left\{y \sin \left(\frac{y}{x}\right)-x \cos \left(\frac{y}{x}\right)\right\} x d y\) is homogenous and solve it.
Solution:


This is the required general solution to the given D.E.

Question 7.
Show that the differential equation ydx + x log\(\left(\frac{y}{x}\right)\)dy – 2xdy = 0 is homogenous and solve it.
Solution:
Given D.E. is ydx + x log\(\left(\frac{y}{x}\right)\)dy – 2xdy = 0


This is the required general solution to the given D.E.

Question 8.
Show that the differential equation \(\left(1+e^{\frac{x}{y}}\right) d x+e^{\frac{x}{y}}\left(1-\frac{x}{y}\right) d y=0\) is homogenous and solve it.
Solution:

∴ Given D.E. is a homogenous equation.
Here, the derivative in the H.D.E. is \(\frac{d x}{d y}\). So take x = vy ………..(2) Diff w.r.t, we get
\(\frac{d x}{d y}=v+y \frac{d v}{d y}\)
Let x = vy ⇒ \(\frac{d}{d y}(x)=\frac{d}{d y}(v y) \Rightarrow \frac{d x}{d y}=v+y \frac{d v}{d y}\)
Substituting this in eqn (1), we have

This is the required general solution to the given D.E.

Question 9.
Find the particular solution of (x + y)dy + (x – y)dx = 0; y = 1 rhen x = 1
Solution:
Given D.E. is (x + y)dy + (x – y)dx = 0 ⇒ (x + y)dy = -(x – y)dx ⇒ \(\frac{d y}{d x}=-\frac{(x-y)}{x+y}\) ……….(1)
Let F(x, y) = \(\frac{-(x-y)}{x+y}\)
Then F(λx, λy) = \(\frac{-(\lambda x-\lambda y)}{\lambda x+\lambda y}=\frac{-\lambda(x-y)}{\lambda(x+y)}=\lambda^0 \mathrm{~F}(x, y)\)
∴ Given D.E. is a homogenous equation.
Let y = vx ⇒ \(\frac{d}{d x}(y)=\frac{d}{d x}(v x) \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}\)
Substituting this in eqn (1), we have

Also from the given data in the problem, we have y = 1 at x = 1
Substitute this value in (2), we have
⇒ log 2 + 2 tan^-1 1 = 2k ⇒ log 2 + 2 × \(\frac{\pi}{4}\) = 2k ⇒ \(\frac{\pi}{2}\) + log 2 = 2k
log(x² + y²) + 2tan^-1 \(\frac{y}{x}=\frac{\pi}{2}\) + log 2

Question 10.
Find the particular solution of x²dy + (xy + y²)dx = 0; y = 1 when x = 1
Solution:
Given D.E. is x²dy + (xy + y²)dx = 0 ⇒ x²dy = -(xy + y²) dx ⇒ \(\frac{d y}{d x}=\frac{-\left(x y+y^2\right)}{x^2}\) ………..(1)
F(x, y) = \(\frac{-\left(x y+y^2\right)}{x^2}\)
Then F(λx, λy) = \(\frac{\left[\lambda x \cdot \lambda y+(\lambda y)^2\right]}{(\lambda x)^2}=\frac{-\lambda^2\left(x y+y^2\right)}{\lambda^2\left(x^2\right)}=\lambda^0 F(x, y)\)
∴ Given D.E. is a homogenous equation.

Also from the given data in the problem, we have
⇒ \(\frac{1}{1+2}\) = C² ⇒ C² = \(\frac{1}{3}\)
\(\frac{x^2 y}{y+2 x}=\frac{1}{3}\) ⇒ y + 2x = 3x²y

Question 11.
Find the particular solution of \(\left[x \sin ^2\left(\frac{y}{x}\right)-y\right]\)dx + xdy = 0; y = \(\frac{\pi}{4}\) when x = 1
Solution:

∴ Given D.E. is a homogenous equation.
Let y = vx ⇒ \(\frac{d y}{d x}\) = v + x\(\frac{d v}{d x}\)
(1) ⇒ v + x\(\frac{d v}{d x}=\frac{-\left[x \sin ^2 v-v x\right]}{x}\)

Question 12.
Find the particular solution of \(\frac{d y}{d x}-\frac{y}{x}+{cosec}\left(\frac{y}{x}\right)=0\); y = 0 when x = 1
Solution:
Given D.E. is \(\frac{d y}{d x}-\frac{y}{x}+\csc \left(\frac{y}{x}\right)=0 \Rightarrow \frac{d y}{d x}=\frac{y}{x}-{cosec}\left(\frac{y}{x}\right)\)
Let F(x, y) = \(\frac{y}{x}-{cosec}\left(\frac{y}{x}\right)\)
Then F(λx, λy) = \(\frac{\lambda y}{\lambda x}-{cosec}\left(\frac{\lambda y}{\lambda x}\right)=\frac{y}{x}-{cosec}\left(\frac{y}{x}\right)=\lambda^0 F(x, y)\)
Let y = vx ⇒ \(\frac{d y}{d x}=v+x \frac{d v}{d x}\)
(1) ⇒ v + x \(\frac{d v}{d x}=v-{cosecv} \Rightarrow-\frac{d v}{{cosec} v}=\frac{d x}{x} \Rightarrow-\sin v d v=\frac{d x}{x}\)
⇒ cos v = log x + log C = log|Cx| ⇒ cos\(\left(\frac{y}{x}\right)\) = log |Cx|
We have y = 0 at x = 1
⇒ cos(0) = log C ⇒ C = e¹ = e
∴ cos\(\left(\frac{\mathrm{y}}{\mathrm{x}}\right)\) = log|(ex)|

Question 13.
Find the particular solution of 2xy + y² – 2x²\(\frac{d y}{d x}\) = 0; y = 2 when x = 1
Solution:

Referring to the AP Inter 2nd Year Maths Study Material Chapter 6 Application of Derivatives Exercise 6c Solutions makes it easier to understand complex problems.

AP Inter 2nd Year Maths Application of Derivatives Solutions Exercise 6c

I.

Question 1.
Find the maximum and minimum value of f (x) = | x + 2 | – 1
Solution:
Given that f(x) = |x + 2| – 1
It is obvious that | x + 2 | ≥ 0 for every x ∈ R
∴ f (x) = | x + 2 | – 1 ≥ -1 for every x ∈ R
The minimum value of f is attained when | x + 2 | = 0 ⇒ x + 2 = 0 ⇒ x = -2
∴ minimum value of f = f(-2) =|-2 + 1| – 1 = -1
Hence, the function f does not have a maximum value.

Question 2.
Find the maximum and minimum value of g (x) = – | x + 1 | + 3
Solution:
Given that g (x) = -| x + 1 | + 3
It is obvious that | x + 1 | ≥ 0 for every x ∈ R
∴ f (x) = – | x + 1 | + 3 ≤ 3 for every x ∈ R
The minimum value of g is attained when | x + 1 | = 0 ⇒ x + 1 = 0 ⇒ x = -1
∴ maximum value of g = g(-1) = |-1 + 1| + 3 = 3
Hence, the function g does not have a minimum value.

Question 3.
Find the maximum and minimum value of h (x) = sin (2x) + 5
Solution:
Given that h (x) = sin (2x) + 5
We know that -1 ≤ sin2x ≤ 1 ⇒ – 1 + 5 ≤ sin2x ≤ 1 + 5 ⇒ 4 ≤ sin2x + 5 ≤ 6
Hence, the maximum and minimum values of h are 6 and 4 respectively.

Question 4.
Find the maximum and minimum value of f(x) = | sin 4x + 3 |
Solution:
Given that f(x) = | sin4x + 3|
We know that —1 ≤ sin4x ≤ 1 ⇒ 2 ≤ sin4x+3 ≤ 4 ⇒ 2 ≤ |sin4x + 3| ≤ 4
Hence, the maximum and minimum values of f are 4 and 2 respectively.

Question 5.
Find the maximum and minimum value of h (x) = x + 1, x ∈ (- 1, 1)
Solution:
Given that h (x) = x + 1, x ∈ (-1, 1). Now h'(x) = 1 ≠ 0
∴ h(x) has no maximum or no minimum.

Question 6.
Prove that f (x) = e^x do not have maxima or minima.
Solution:
Given that f(x) = e^x ⇒ f ‘(x) = e^x ⇒ f “(x) = e^x
For maxima or minima f(x) = 0 ⇒ e^x = 0.
This equation is not satified for any real x.
∴ f(x) has no maxima or minima on R.

Question 7.
Prove that g (x) = log x do not have maxima or minima.
Solution:
Given that g(x)= log x ⇒ g'(x) = \(\frac{1}{x}\) ⇒ g”(x) = –\(\frac{1}{x^2}\)
For maximum f ‘(x) = 0 ⇒ \(\frac{1}{x}\) = 0 is not satisfied for any real x.
∴ f(x) has no maxima or minima.

Question 8.
Prove that h(x) = x³ + x² + x + 1 do not have maxima or minima.
Solution:
Given that h(x) = x³ + x² + x + 1 ⇒ h'(x) = 3x² + 2x +1
Now, h'(x) = 0 ⇒ 3x² + 2x +1 = 0 ⇒ x = \(\frac{-2 \pm 2 \sqrt{2} i}{6}\) ⇒ x = \(\frac{-1 \pm \sqrt{2} i}{3}\) ≠ R
∴ there does not exist c ∈ R such that h'(c) = 0
Hence, function h does not have maxima or minima.

Question 9.
It is given that at x = 1, the function x⁴ – 62x² + ax + 9 attains its maximum value, on the interval |0, 2|. Find the value of a.
Solution:
Let f(x) = x⁴ – 62x² + ax + 9 ⇒ f'(x) = 4x³ – 124x + a
It is given that function f attains its maximum value on the interval [0, 2] at x = 1.
Hence, f'(1) = 0 ⇒ 4x³ – 124x + a = 0 ⇒ 4 – 124 + a = 0 ⇒ -120 + a = 0 ⇒ a = 120
Thus, the value of a = 120

II.

Question 1.
Find the maximum and minimum values of f (x) = (2x – 1)² + 3
Solution:
Given that f (x) = (2x – 1)² + 3
It is obvious that (2x – 1)² ≥ 0 for every x ∈ R
∴ (2x – 1)² + 3 ≥ 3 for every x ∈ R
The minimum value of f is attained when 2x – 1 = 0 ⇒ 2x – 1 = 0 ⇒ x = \(\frac{1}{2}\)
Hence, minimum value of f is f(\(\frac{1}{2}\)) = (2(\(\frac{1}{2}\)) – 1)² + 3 = 3
Thus, the function f does not have a maximum value

Question 2.
Find the maximum and minimum values of f(x) = 9x² + 12x + 2
Solution:
Given that f (x) = 9x² + 12x + 2
It is obvious that (3x + 2)² ≥ 0 for every x ∈ R
∴ f(x) = (3x + 2)² – 2 ≥ -2 for every x ∈ R
The minimum value off is attained when 3x + 2 = 0 ⇒ 3x + 2 = 0 ⇒ x = –\(\frac{2}{3}\)
Hence, minimum value off is f(-\(\frac{2}{3}\)) = (3(-\(\frac{2}{3}\))+ 2)² – 2 = -2
Thus, the function f does not have a maximum value.

Question 3.
Find the maximum and minimum values of f (x) = -(x – 1)² + 10.
Solution:
Given that f(x) = -(x – 1)² + 10.
It is obvious that (x – 1)² ≥ 0 for every x ∈ R
∴ f(x) = -(x – 1)² + 10 ≤ 10 for every x ∈ R
The maximum value of f is attained when (x – 1) = 0 ⇒ x = 1
∴ Maximum value of f is f(1) = -(1 – 1)² + 10 = 10
Hence, the function f does not have a minimum value.

Question 4.
Find the maximum and minimum values of g (x) = x³ + 1
Solution:
Given that g(x) = x³ ⇒ g'(x) = 3x² > 0 ⇒ g'(x) > 0 ⇒ x = 0
g'(x) = 3x² ≥ 0 for all x.
x = 0 is not a maximum or minimum
Hence, function g neither has a maximum value nor a minimum value.

Question 5.
Find the local maxima and local minima, of f (x) = x² and also find the local maximum and the local minimum values.
Solution:
Given that f(x) = x² ⇒ f ‘(x) = 2x ⇒ f “(x) = 2
For maximum or minimum f'(x) = 0 ⇒ 2x = 0 ⇒ x = 0
Now f “(0) = 2 > 0
∴ f(x) has minimum at x = 0
Point of local minimum is x = 0
Local minimum at x = 0 is f(0) = 0² = 0
∴ Point of local minimum is (0, 0)

Question 6.
Find the local maxima and local minima of g(x) = x³ – 3x and also find the local maximum and the local minimum values.
Solution:
Given that f(x) = x³ – 3x ⇒ f'(x) = 3x² – 3 ⇒ f”(x) = 6x
For maximum or minimum f ‘(x) = 0 ⇒ 3x² – 3 = 0 ⇒ x² – 1 = 0 ⇒ x = ± 1
Now f'(l) = 6(1) = 6 > 0
∴ f(x) has minimum at x = 1
Minimum value is f(1) = 1³ – 3(1) = -2
Also f”(-1) = 6(-1) = -6 < 0
∴ f(x) has maximum value at x = – 1
Maximum value is f(-1) = (-1)³ – 3(-1) = -1 + 3 = 2.

Question 7.
Find the local maxima and local minima of h (x) = sin x + cos x, 0 < x < π/2 and also find the local maximum and the local minimum values.
Solution:
Given that h(x) = sin x + cos x, 0 < x < π/2 ⇒ h'(x) = cos x – sin x
Now h'(x) = 0 ⇒ cos x – sin x ⇒ sin x = cos x ⇒ tan x = 1 ⇒ x = \(\frac{\pi}{4}\) ∈ (0, \(\frac{\pi}{2}\))
Also, h'(x) = -sin x – cos x = -(sin x + cos x)
Hence, h’\(\left(\frac{\pi}{4}\right)\) = –\(\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right)=\frac{-2}{\sqrt{2}}=-\sqrt{2}\) < 0
∴ By second derivative test. x = \(\frac{\pi}{4}\) is a point of local maxima and the local maximum at x = \(\frac{\pi}{4}\)
we have h\(\left(\frac{\pi}{4}\right)\) = sin\(\frac{\pi}{4}\) + cos \(\frac{\pi}{4}\) = \(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\sqrt{2}\)

Question 8.
Find the local maxima and local minima, of f (x) – sin x – cos x, 0 < x < 2π and also find the local maximum and the local minimum values.
Solution:
Given that f (x) = sin x – cos x, 0 Now, f'(x) = 0 ⇒ cos x + sin x = 0 ⇒ sin x = -cos x ⇒ tan x = -1 ⇒ x = \(\frac{3\pi}{4}\), \(\frac{7\pi}{4}\) ∈ (0, 2π)
Also, f”(x ) = – sin x + cos x

Question 9.
Find the local maxima and local minima, of f(x) = x³ – 6x² + 9x + 15 and also find the local maximum and the local minimum values.
Solution:
Given that f(x) = x³ – 6x² + 9x + 15 ⇒ f’(x) = 3xx² – 12x + 9 ⇒ f”(x) = 6x – 12
For maximum or minimum f'(x) = 0 ⇒ 3x² – 12x + 9 = 0 ⇒ 3(x² – 4x + 3) = 0
Now f”(1) = 6(1) – 12 = -6 < 0
∴ f(x) has maximum value at x = 1
Maximum value is f(1) = 1³ – 6(1)² + 9(1) + 15 = 1 – 6 + 9+ 15 = 19
Also f”(3) = 6(3) – 12 = 18 – 12 = 6 > 0
∴ f(x) has minimum value at x = 3
Minimum value is f(3) = 3³ – 6.3² + 9.3 + 15 = 27 – 54 + 27 + 15 = 15

Question 10.
Find the local maxima and local minima, of g(x) = \(\frac{x}{2}+\frac{2}{x}\), (x > 0) and also find the local maximum and the local minimum values.
Solution:
Given that g(x) = \(\frac{x}{2}+\frac{2}{x}\) ⇒ g'(x) = \(\frac{1}{2}-\frac{2}{x^2}\) ⇒ g”(x) = \(\frac{4}{x^3}\)
For max. or min. we have g'(x) = 0
⇒ \(\frac{1}{2}-\frac{2}{x^2}\) = 0
⇒ \(\frac{x^2-4}{2 x^2}\) = 0
⇒ x² – 4 = 0
⇒ x = ±2 = 2 [∵ x > 0]
Now g”(2) = \(\frac{4}{2^3}\) = \(\frac{1}{2}\) > 0
∴ g(x) has minimum value at x = 2
Minimum value is g(2) = \(\frac{2}{2}\) + \(\frac{2}{2}\) = 1 + 1 = 2

Question 11.
Find the local maxima and local minima, of g(x) = \(\frac{1}{x^2+2}\) and also find the local maximum and the local minimum values.
Solution:
Given that g(x) = \(\frac{1}{x^2+2}\) ⇒ g'(x) = \(\frac{-(2 x)}{\left(x^2+2\right)^2}\)
Now g'(x) = 0 ⇒ \(\frac{-(2 x)}{\left(x^2+2\right)^2}\) = 0 ⇒ x = 0
Now, for values close to and to the left of 0, g'(x) > 0
Also, for values close to x = 0 and to the right of 0, g'(x) < 0
Therefore, by first derivative test, x = 0 is a point of local maxima and the local maximum value of g(0) = \(\frac{1}{0+2}\) = \(\frac{1}{2}\)

Question 12.
Find the local maxima and local rninlrnaq of f(x) = x\(\sqrt{1-x}\), x, 0 < x < 1 and also find the local maximum and the local minimum values.
Solution:
Given that f(x) = x\(\sqrt{1-x}\)

Question 13.
Find the absolute maximum value and the absolute minimum value of function f (x) = x³ in the given interval x ∈ [- 2, 2]
Solution:
Given f(x) = x³ ⇒ f'(x) = 3x²
Now f'(x) = 0 ⇒ 3x² = 0
Then we evaluate the value of f at critical point x=0 and at end points of the interval [-2, 2]
(i) f(0) = 0 ……………… (1)
(ii) f(-2) =(-2)³ = -8 ………… (2)
(iii) f(2) = (2)³ =8 ………….. (3)
From (1), (2) & (3) the absolute Minimum value is – 8 & absolute Maximum value is 8

Question 14.
Find the absolute maximum value and the absolute minimum value of function f (x) = sin x + cos x in the given interval x ∈ [0, π]
f(x) = sin x + cos x in the given interval x ∈ [0, π]
Solution:
Given f(x) = sin x + cos x ⇒ f'(x) = cos x – sin x
Now f'(x) = 0 ⇒ cos x – sin x = 0 ⇒ sin x = cos x – 1 ⇒ x = \(\frac{\pi}{4}\) ∈ [0, π]
Now the critical point is \(\frac{\pi}{4}\) and end points of [0, π] are 0, π.
(i) \(f\left(\frac{\pi}{4}\right)=\sin \frac{\pi}{4}+\cos \frac{\pi}{4} \Rightarrow=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2}\) ………….. (1)
(ii) f(0) = sin 0 + cos 0 = 0 + 1 = 1 ……. (2)
(iii) f(π) = sin π + cos π – 0 – 1 = -1 ………….. (3)
From (1), (2) & (3) the absolute Minimum value is -1 & absolute Maximum value is \(\sqrt{2}\)

Question 15.
Find the absolute maximum value and the absolute minimum value of function
f(x) = 4x – \(\frac{1}{2}\)x² in the given interval x ∈ [-2, \(\frac{9}{2}\)]
Solution:
Given f(x) = 4x – \(\frac{1}{2}\)x² and f'(x) = 4 – \(\frac{1}{2}\)(2x) = 4 – x
Now f'(x) = 0 ⇒ 4 – x = 4
Now the critical point is 4 and end points of [-3, 1]
(i) f(4) = 16 – \(\frac{1}{2}\)(16) = 16 – 8 = 8 …………… (1)
(ii) f(-2) = -8 – \(\frac{1}{2}\) (4) = -8 – 2 = -10 ……………. (2)
(iii) f\(\left(\frac{9}{2}\right)=\) = 18 – \(\frac{1}{2}\left(\frac{9}{2}\right)^2\) = 18 – \(\frac{81}{8}\) = 18 – 10.125 = 7.875 …………… (3)
From (1), (2) & (3) the absolute Minimum value is – 10 & absolute Maximum value is 8.

Question 16.
Find the absolute maximum value and the absolute minimum value of function f (x) = (x – 1)² + 3 in the given interval x ∈ [-3, 1]
Solution:
Given f(x) = (x – 1)² + 3 ⇒ f'(x) = 2(x – 1)
Now f'(x) = 0 ⇒ 2 (x – 1) = 0 ⇒ x = 1
Now the critical point is at 1 and end points of [-3, 1]
(i) f(1) = (1 – 1)² + 3 = 3 ………… (1)
(ii) f (-3) = (-3 – 1)² + 3 = 16 + 3 = 19 ……………… (2)
Hence, we conclude that the absolute maximum value is 19 occurring at x = -3.
Also, the absolute minimum value of on [-3, 1] is 3 occurring at x = 1.

Question 17.
Find the maximum profit that a company can make, if the profit function is given by p (x) = 41 – 72x – 18x².
Solution:
Given p(x) = 41 – 72 x – 18x². …………. (1)
⇒ p'(x) = -72 – 36x ⇒ p'(x) = -36 < 0
For maxima or minima, p'(x) = 0 ⇒ -72 – 36x = 0 ⇒ 36x = -72 ⇒ x = -2
Also p”(2) = -36 < 0
∴ The profit f(x) is maximum when x = -2
From (1), the maximum profit is p(2) = 41 – 72 (-2) – 18(4) = 41 + 144 – 72 = 185 – 72 = 113

Question 18.
Find both the maximum value and the minimum value of 3x⁴ – 8x³ + 12x² – 48x + 25 on the interval [0, 3].
Solution:
Given f(x) = 3x⁴ – 8x³ + 12x² – 48x + 25
f'(x) = 12x³ – 24x² + 24x – 48 = 12(x³ – 2x² + 2x – 4)
= 12[x² (x – 2) + 2(x – 2)] = 12(x – 2)(x² + 2)
For maximum or minimum, f'(x) = 0 ⇒ x – 2 = 0 ⇒ x = 2
Hence the critical point is x= 2. Also end points of the interval [0, 3] are 0, 3.
(i) f(2) = 3(2)⁴ – 8(2)³ + 12(2)² – 48(2) + 25 = 48 – 64 + 48 – 96 + 25 = -39 ……… (1)
(ii) f(0) = 3(0)⁴ – 8(0)³ + 12(0)² – 48(0) + 25 = 25 ……………. (2)
(iii) f(3) = 3(3)⁴ – 8(3)³ + 12(3)² – 48(3) + 25 = 243 – 216 + 108 – 144 + 25 = 16 ……… (3)
From (1), (2), (3) Absolute maximum = 25 & Absolute minimum = -39

Question 19.
At what points in the interval [0, 2π], docs tire function sin 2x attain its maximum value?
Solution:
Given f(x) = sin 2x ⇒ f'(x) = 2 cos 2x
For maximum or minimum,
f'(x) = 0 ⇒ 2 cos 2x = 0 ⇒ 2x = \(\frac{\pi}{4}, \frac{3 \pi}{4}, \frac{5 \pi}{4}, \frac{7 \pi}{4}\) ⇒ x = \(\frac{\pi}{4}, \frac{3 \pi}{4}, \frac{5 \pi}{4}, \frac{7 \pi}{4}\)
Hence the critical points are x = \(\frac{\pi}{4}, \frac{3 \pi}{4}, \frac{5 \pi}{4}, \frac{7 \pi}{4}\).
Also end points of the interval [0, 2π] are 0, 2π.
(i) f\(\left(\frac{\pi}{4}\right)\) = sin\(\left(\frac{\pi}{4}\right)\) = 1
(ii) f\(\left(\frac{3\pi}{4}\right)\) = sin\(\frac{3\pi}{2}\) = -1
(iii) f\(\left(\frac{5\pi}{4}\right)\) = sin \(\frac{5\pi}{2}\) = 1
(iv) f\(\left(\frac{7\pi}{4}\right)\) = sin\(\frac{7\pi}{2}\) = -1
(v) f(0) = sin 0 = 0
(vi) f(2π) = sin 2π = 0
Hence, we conclude that the absolute maximum value is occurring at x = \(\frac{\pi}{4}\) and x = \(\frac{5\pi}{4}\)

Question 20.
What is the maximum value of the function sin x + cos x?
Solution:
Let f(x) = sin x + cos x ⇒ f'(x) = cos x – sin x
Now, f”(x) = 0 ⇒ cos x – sinx = 0 ⇒ sin x = cos x ⇒ tan x = 1 ⇒ x = \(\frac{\pi}{4}\), \(\frac{5\pi}{4}\)
Hence, f “(x) = – sin x – cos x = – (sin x + cos x )
Now f”(x) will be negative •
Now, f”(x) will be negative when (sin x + cos x) is positive i.e., when sin x and cos x are both positive.
Also, we know that sin x and cos x both are positive in the first quadrant.
Then, f”(x) will be negative when x ∈ (0, \(\frac{\pi}{2}\))
Thus, we consider x = \(\frac{\pi}{4}\)
\(f^{\prime \prime}\left(\frac{\pi}{4}\right)=-\left(\sin \frac{\pi}{4}+\cos \frac{\pi}{4}\right)=-\left(\frac{2}{\sqrt{2}}\right)=-\sqrt{2}<0\)
By second derivative test, f will be the maximum at x = \(\frac{\pi}{4}\) and the maximum value of f is \(f\left(\frac{\pi}{4}\right)=\sin \frac{\pi}{4}+\cos \frac{\pi}{4}=\frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2}\)

Question 21.
Find the maximum value of 2x³ – 24x + 107 in the interval [1, 3]. Find the maximum value of the same function in [-3, -1].
Solution:
Let f(x) = 2x³ – 24x + 107 ⇒ f (x) = 6x³ – 24 = 6(x³ – 4)
Now, f'(x) = 0 ⇒ 6(x² – 4) = 0 ⇒ x² = 4 ⇒ x = ± 2
Now, we first consider the interval [1, 3].
Then, we evaluate the value of f at the critical point x = 2 ∈ [1, 3] [and at the end points of the interval [1, 3].
Hence, f(2) = 2(2)³ – 24(2)+107 = 16 – 48 + 107 = 75
f(1) = 2(1)³ – 24(1) + 107 = 2 – 24 + 107 = 85
f(3) = 2(3)³ – 24(3) + 107 = 54 – 72 + 107 = 89
Thus, the absolute maximum value of f(x) in the interval [1, 3] is 89 occurring at x = 3.
Next, we consider the interval [-3, -1].
Then, we evaluate the value of f at the critical point x = -2 ∈ [-3, -1] and at the end points of the interval [-3, -1]
Hence, f(-3) = 2(-3)³ – 24(-3) + 107 = -54 + 72 + 107 = 125
f(-1) = 2(-1)³ – 24(-1) + 107 = -2 + 24 + 107 = 129
f(-2) = 2(-2)³ – 24(-2) + 107=—16 + 48 + 107 = 139
Hence, the absolute maximum value of f(x) in the interval [-3, -1] is 139 occurring at x = -2

Question 22.
Find the maximum and minimum values of x + sin 2x on [0, 2π].
Solution:
Let f(x) = x ± sin 2x ⇒ f'(x) = 1 + 2 cos 2x
Now f'(x) = 0 = 1 + 2 cos 2x = 0 ⇒ cos 2x = 0 ⇒ cos 2x = \(\frac{-1}{2}\) = -cos\(\frac{\pi}{3}\) = cos (π – \(\frac{\pi}{3}\)) = cos \(\frac{2\pi}{3}\)
⇒ 2x = 2nπ ± \(\frac{2\pi}{3}\) [[n ∈ Z] ⇒ x = nπ ± \(\frac{\pi}{3}\) [n ∈ Z] ⇒ x = \(\frac{\pi}{3}\), \(\frac{2\pi}{3}\), \(\frac{4\pi}{3}\), \(\frac{5\pi}{3}\) ∈ [0, 2π]
Then, we evaluate the value of fat critical points x = \(\frac{\pi}{3}\), \(\frac{2\pi}{3}\), \(\frac{4\pi}{3}\), \(\frac{5\pi}{3}\) and at the end points of the interval [0, 2π].

(v) f(0) = 0 + sin0 = 0
(vi) f(2π) = 2π + sin 4π = 2π + 0 = 2π
Hence, we conclude that the absolute maximum value of f(x) is 2π occurring at x = 2π and the absolute minimum value of f(x) is 0 occurring at x = 0.

Question 23.
Find two numbers whose sum is 24 and whose product is as large as possible.
Solution:
Let a number be x .
Then, the other number be (24 – x).
Let P(x) denote the product of the two numbers.
Thus, we have: P(x) = x (24 – x) = 24x – x²
∴ P'(x) = 24 – 2x ⇒ P'(x) = -2
Now, P'(x) = 0 ⇒ 24 – 2x = 0 ⇒ 24 = 2x ⇒ x = 12
Also, P'(12) = -2 < 0
By second derivative test, x = 12 is the point of local maxima of P.
Thus, the numbers are 12 and (24 – 12) = 12
Hence, the product of the numbers is the maximum when the numbers are 12 each.

III.

Question 1.
Find two positive numbers x and y such that x + y = 60 and xy³ is maximum.
Solution:
The two numbers are x and y such that x + y = 60 ⇒ y = 60 – x …………. (1)
Let f(x) = xy³ = f(x) = x(60 – x)³ ………………. (1)
⇒ f’(x) = (60 – x)³ – 3x(60 – x)² = (60 – x)²[60 – x – 3x] = (60 – x)²(60 – 4x)
⇒ f”(x) = -2(6o – x)(6o – 4x) – 4(6o – x)² = -2(60 – x)[60 – 4x + 2(60 – x)]
= -2(60 – x)(180 – 6x) = -12(60 – x)(30 – x)
Now. f'(x) = 0 ⇒ x = 60 or x = 15
When x = 60 then, f”(x) = 0
When x = 15 then f”(x) = -12(60 – 15)(30 – 15) = -12 × 45 × 15 < 0
By second derivative test, x1 5 is a point of local maxima of f.
Thus, function xy³ is maximum when x = 15 and y = 60 – 15 = 45
Hence, the required numbers are 15 and 45.

Question 2.
Find two positive numbers x and y such that their sum is 35 and the product x²y⁵ is a maximum.
Solution:
Let a number be x. Then, the other number is y = (35 – x).
Let P(x) = x²y⁵ – Then we have, P(x) = x² (35 – x)⁵
P'(x) = 2x(35 – x)⁵ + x²5(35 – x)⁴ (-1) = 2x (35 – x)⁵ – 5x² (35 – x)⁴
= x(35 – x)⁴[2(35 – x) – 5x] = x(35 – x)⁴ (70 – 7x) = 7x (35 – x)⁴(10 – x)
P”(x) = 7(35 – x)⁴(10 – x) + 7x[-(35 – x)⁴ – 4(35 – x)³(10 – x)]
= 7(35 – x)⁴(10 – x) – 7x(35 – x)⁴ – 28x(35 – x)³ (10 – x)
= 7(35 – x)³[(35 – x)(10 – x) – x(35 – x) – 4x(10 – x)]
= 7(35 – x)³[350 – 45x + x² – 35x + x² – 40x + 4x²]
= 7(35 – x)³(6x² – 120x + 350)
Now P'(x) = 0 ⇒ x = 0, x = 35, x = 10
When, x = 35 then, P'(x) = P (x) = 0 ⇒ y = 35 – 35 = 0
This will make the product x²y⁵ equal to 0.
When, x = 0 then y = 35 – 0 = 35.This will make the product x²y⁵ equal to 0.
Hence,x = 0 and x = 35 cannot be the possible values of x.
When x = 10. Then, p”(x) = 7 (35 – 10)³ (6 × 100 – 120 × 10 + 350) = 7 (25)³ (-250) < 0
By second derivative test, P(x) will be the maximum when x = 10 and y = 35 – 10 = 25
Hence, the required numbers are 10 and 25.

Question 3.
Kind two positive numbers whose sum is 16 and the sum of whose cubes is minimum.
Solution:
Let a number be x . Then, the other number be (16 – x).
Let the sum of the cubes of these numbers be denoted by S(x).
Then, S(x) = x³ + (16 – x)³
∴ S'(x) = 3x² + 3(16 – x)²(-1) = 3x² – 3(16 – x)² ⇒ S'(x) = 6x + 6(16 – x)
Now, S'(x) = 0 ⇒ 3x² – 3(16 – x)² = 0 ⇒ x² – (16 – x)² = 0
⇒ x² – 256 – x² + 32x = 0 ⇒ x = \(\frac{256}{32}\) ⇒ x = 8
Also, S”(8) = 6(8) + 6(16 – 8) = 48 + 48 = 96 > 0
By second derivative test, x=8 is the point of local minima of S.
Thus, the numbers are 8 and (16 – 8) = 8.
Hence, the sum of the cubes of the numbers is the minimum when the numbers are 8 each.

Question 4.
A square piece of tin of side 18 cm is to be made into a box without top, by cutting a square from each corner and folding up the flaps to form the box. What should be the side of the square to be cut off so that the volume of the box is the maximum possible.
Solution:
Let the side of the square to be cut off be x cm.
Then, the length and the breadth of the box will be( 18 – 2x)cm each and the height of the box be x cm.
∴ Volume V (x) of the box is given by,V(x) = lbh = x(18 – 2x)².

Hence, V'(x) = 1 (18 – 2x )² + 2x(18 – 2x)(-2) = (18 – 2x )² – 4x (18 – 2x ) = (18 – 2x)[18 – 2x – 4x]
= (18 – 2x)(18 – 6x) = 6x²(9 – x)(3 – x) = 12(9 – x)(3 – x)
= 12(x² – 12x + 27) = 12(x – 9)(x – 3)
V'(x) = 12(2x – 12) = 24(x – 6) .
Now, V'(x) = 0 ⇒ x = 9, x = 3
If, x = 9 then the length and the breadth will become 0.
Hence, x ≠ 9
When x = 3 then, V”(3)= 24(3 – 6) = -72 < 0
By second derivative test, x = 3 is the point of local maxima of V.
Hence, if we remove a square of side 3 cm from each corner of the square tin and make a box from the remaining sheet, then the volume of the box obtained is the largest possible.

Question 5.
A rectangular sheet of tin 45 cm by 24 cm is to he made into a box without top, by culling off square from each cornet and holding up the flaps. What should he the side of the square to he cut off so that the volume of the box is maximum ?
Solution:
Let the side of the square to be cut off be x cm.
Then, the height of the box is x cm,
the length is (45 – 2x)cm and the breadth (24 – 2x) cm.

Therefore, the volume V(x) of the box is given by, _________________
V(x) = x(45 – 2x)(24 – 2x) = x(1080 – 90x – 48x + 4x²)
= 4x³ – 138x² +1080x
Hence,V’(x) = 12x² – 276x + 1080 = 12(x² – 23x + 90)
= 12(x – 18)(x – 5)
V'(x) = 24x – 276 = 12(2x – 23)
Now, V'(x) = 0 ⇒ x = 18, x = 5
It is not possible to cut off a square of side 18 cm from each comer of the rectangular sheet. :
Thus, x cannot be equal to 18.
When, x = 5
Then, V'(5) = 12[2 (5) – 23] = 12 (10 – 23) = 12(-13) = -156 < 0
By second derivative test, x = 5 is the point of local maxima.
Hence, the side of the square to be cut off to make the volume of the box maximum possible is 5 cm.

Question 6.
Show that of all the rectangles inscribed in a given fixed circle, the square has the maximum area.

Solution:
Let a rectangle of length 1 and breadth b be inscribed in the given circle of radius a .
Then, the diagonal passes through the centre and is of length 2a cm
Now, by applying the Pythagoras theorem, we have:

By the second derivative test, when l = \(\sqrt{2}\)a, then the area of the rectangle is the maximum. Since, l = b = \(\sqrt{2}\)a the rectangle is a square.
Hence, it has been proved that of all the rectangles inscribed in the given fixed circle, the square has the maximum area.

Question 7.
Show that the right circular cylinder of given surface and maximum volume is such that its height is equal to the diameter of the base.

Solution:
Let r and h be the radius and height of the cylinder respectively.
Then, the surface area (S) of the cylinder is given by, S = 2πr² + 2πrh
∴ h = \(\frac{S-2 \pi r^2}{2 \pi r}=\frac{S}{2 \pi}\left(\frac{1}{r}\right)-r\)
Let V be the volume of the cylinder
V = πr²h = πr² = \(\left[\frac{\mathrm{S}}{2 \pi}\left(\frac{1}{\mathrm{r}}\right)-\mathrm{r}\right]\) = \(\frac{\mathrm{Sr}}{2}\) -πr³
\(\frac{\mathrm{dV}}{\mathrm{dr}}=\frac{\mathrm{S}}{2}\) – 3πr² ⇒ \(\frac{d^2 V}{d r^2}\) = -6πr
Now, \(\frac{\mathrm{dV}}{\mathrm{dr}}\) = 0 ⇒ \(\frac{\mathrm{S}}{2}\) – 3πr² = 0
⇒ \(\frac{\mathrm{S}}{2}\) = 3πr²
⇒ r² = \(\frac{\mathrm{S}}{6 \pi}\)
When r² = \(\frac{\mathrm{S}}{6 \pi}\)
Then \(\frac{d^2 V}{d r^2}=-6 \pi\left(\sqrt{\frac{S}{6 \pi}}\right)<0\)
By second derivative test, the volume is the maximum when r² = \(\frac{\mathrm{S}}{6 \pi}\)
Now, when r² = \(\frac{\mathrm{S}}{6 \pi}\) Then, h = \(\frac{6 \pi \mathrm{r}^2}{2 \pi}\left(\frac{1}{\mathrm{r}}\right)\) – r = 3r – r = 2r
Hence, the volume is the maximum vyhen the height is twice the radius i.e., when the height is equal to the diameter.

Question 8.
Of all the closed cylindrical cans (right circular), of a given volume of 100 cubic centimetres, find the dimensions of the can which has the minimum surface area?

Solution:
Let r and h be the radius and height of the cylinder respectively.
Then, volume V of the cylinder is given by, V = πr² = 100 ⇒ h = \(\frac{100}{\pi \mathrm{r}^2}\)
Surface area is given by: S = 2πr² + 2πrh = 2πr² + \(\frac{200}{\mathrm{r}}\)
⇒ \(\frac{\mathrm{dS}}{\mathrm{~d} \mathrm{r}}\) = 4πr – \(\frac{200}{r^2}\) ⇒ \(\frac{\mathrm{d}^2 \mathrm{~S}}{\mathrm{dr}^2}\) = 4πr + \(\frac{400}{r^2}\)
Now, \(\frac{\mathrm{dS}}{\mathrm{dr}}\)= 0 ⇒ 4πr – \(\frac{200}{r^2}\) = 0
⇒ 4πr = \(\frac{200}{r^2}\)
⇒ r³ = \(\frac{200}{4 \pi}=\frac{50}{\pi}\)
⇒ r = \(\left(\frac{50}{\pi}\right)^{\frac{1}{3}}\)
When, r = \(\left(\frac{50}{\pi}\right)^{\frac{1}{3}}\) Then, \(\frac{\mathrm{d}^2 \mathrm{~S}}{\mathrm{dr}^2}\) > 0
By second derivative test, the surface area is the minimum when r = \(\left(\frac{50}{\pi}\right)^{\frac{1}{3}}\)
when r = \(\left(\frac{50}{\pi}\right)^{\frac{1}{3}}\) we have h = \(\frac{100}{\pi\left(\frac{50}{\pi}\right)^{\frac{2}{3}}}=\frac{2 \times 50}{(\pi)(50)^{\frac{2}{3}} \cdot \pi^{\frac{2}{3}}}=2\left(\frac{50}{\pi}\right)^{\frac{1}{3}}\)
Hence, the required dimensions of the can which has the minimum surface area is given by radius \(\left(\frac{50}{\pi}\right)^{\frac{1}{3}}\) cm and height \(2\left(\frac{50}{\pi}\right)^{\frac{1}{3}}\) cm.

Question 9.
A wire of length 28 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into a circle. What should be the length of the two pieces so that the combined area of the square and the circle is minimum?

Solution:
Let a piece of length l be cut from the given wire to make a square.
Then, the other piece of wire to be made into a circle is of length (28 – l).
Now, side of square is 1/4.
Let r be the radius of the circle.
Then, 2πr = 28 – l ⇒ r = \(\frac{1}{2 \pi}\)(28 – l)
The combined areas of the square and the circle A, is given by,

By second derivative test, the area (A) is the minimum when l = \(\frac{112}{\pi+4}\) cm.
Hence, the combined area is the minimum when the length of the wire in making the square is l = \(\frac{112}{\pi+4}\) cm while the length of the wire in making the circle is \(\left(28-\frac{112}{\pi+4}\right)=\frac{28 \pi}{\pi+4}\) cm.

Question 10.
Prove that the volume of the largest cone that can be inscribed in a sphere of radius R is 8/27 of the volume of the sphere.
Solution:
Let r and h be the radius and height of the cone respectively inscribed in a sphere of radius R.
Let V be the volume of the cone.Then V = \(\frac{1}{3}\)πr² h
Height of the cone is given by, h = R + AB = R + \(\sqrt{R^2-r^2}\) [ABC is a right angle]

When, r² = \(\frac{8}{9}\) R². Then \(\frac{d^2 \mathrm{~V}}{\mathrm{dr}^2}\) < 0
By second derivative test, the volume of the cone is the maximum, when r² = \(\frac{8}{9}\) R²
When, r² = \(\frac{8}{9}\) R².
Then, h = R + \(\sqrt{R^2-\frac{8}{9} R^2}=R+\sqrt{\frac{1}{9} R^2}=R+\frac{R}{3}=\frac{4}{3} R\)
∴ V = \(\frac{1}{3} \pi\left(\frac{8}{9} R^2\right)\left(\frac{4}{3} R\right)=\frac{8}{27}\left(\frac{4}{3} \pi R^3\right)=\frac{8}{27} \times(\text { Volume of sphere })\)
Hence, the volume of the largest cone that can be inscribed in the sphere is 8/27 the volume of the sphere.

Question 11.
Show that the right circular cone of least curved surface and given volume has an altitude equal to \(\sqrt{2}\) time the radius of the base.
Solution:
Let r and h be the radius and height of the cone, respectively.
Then, the volume (V) of the cone is given by, V = \(\frac{1}{3}\)πr² h ⇒ h = \(\frac{3 \mathrm{~V}}{\pi \mathrm{r}^2}\)
The surface area (S) of the cone is given by, S = πrl,
where l is the slant height

Thus, it can be easily verified that when r⁶ = \(\frac{9 V^2}{2 \pi^2}\), ⇒ \(\frac{\mathrm{d}^2 \mathrm{~S}}{\mathrm{~d} \mathrm{r}^2}\) > 0,
By second derivative test, the surface area of the cone is the least when r⁶ = \(\frac{9 V^2}{2 \pi^2}\)
Then, h = \(\frac{3 \mathrm{~V}}{\pi \mathrm{r}^2}=\frac{3 \mathrm{~V}}{\pi \mathrm{r}^2}\left(\frac{2 \pi^2 \mathrm{r}^6}{9}\right)^{\frac{1}{2}}=\frac{3}{\pi \mathrm{r}^2} \cdot \frac{\sqrt{2} \pi \mathrm{r}^3}{3}=\sqrt{2} \mathrm{r}\)
Hence, for a given volume, the right circular cone of the least curved surface has an altitude equal to \(\sqrt{2}\) times the radius of the base.

Question 12.
Show that the semi-vertical angle of the cone of the maximum volume and of given slant height is tan^-1\(\sqrt{2}\)

Solution:
Let θ be the semi-vertical angle of the cone.
It is clear that θ ∈ [o, \(\frac{\pi}{2}\)]
Let r, h and l be the radius, height, and the slant height of the cone respectively.
The slant height of the cone is given as constant.
Now, r = l sin θ and h = l cos θ
The volume V of the cone is given by, V = \(\frac{\pi l^3}{3}\)(sin²θ . cosθ)
∴ \(\frac{\mathrm{dV}}{\mathrm{~d} \theta}\) = \(\frac{\pi l^3}{3}\) (sin2θ(-sin θ) + cosθ2sinθ.cosθ)
= \(\frac{\pi l^3}{3}\)(-sin³θ + 2sinθcos²θ)
\(\frac{d^2 V}{d \theta^2}\) = \(\frac{\pi l^3}{3}\)[-3sin²θ cosθ + 2(sin θ . 2 cos θ(-sin θ)) + cos²θ(cos θ)]
= \(\frac{\pi l^3}{3}\) [-3sin²θcosθ – 4sin²θcosθ + 2cos³θ]
= \(\frac{\pi l^3}{3}\) [-7sin²θcosθ + 2cos³θ]
Now, \(\frac{\mathrm{dV}}{\mathrm{~d} \theta}\) = 0
⇒ \(\frac{\pi l^3}{3}\) [-sin³θ + 2sinθcos²θ] = 0
⇒ sin³θ = 2sinθcos²θ
⇒ tan² θ = \(\sqrt{2}\) since, sin θ ≠ 0
⇒ tan θ = \(\sqrt{2}\) since, θ ∈ [o, \(\frac{\pi}{2}\)]
We have, sin θ = \(\frac{\sqrt{2}}{\sqrt{3}}\), cosθ = \(\frac{1}{\sqrt{3}}\)
Now, \(\frac{\mathrm{d}^2 \mathrm{~V}}{\mathrm{~d} \theta^2}=\frac{\pi l^3}{3}\left[-7 \cdot \frac{2}{3} \cdot \frac{1}{\sqrt{3}}+2 \cdot \frac{1}{3 \sqrt{3}}\right]=\frac{\pi l^3}{3}\left[\frac{-12}{3 \sqrt{2}}\right]=-\frac{4 \pi l^3}{3 \sqrt{2}}\) < 0
By second derivative test, the volume V is the maximum when θ = tan^-1\(\sqrt{2}\)
Hence, for a given slant height, the semi-vertical angle of the cone of the maximum volume is tan^-1\(\sqrt{2}\)

Question 13.
Show that semi-vertical angle of right circular cone of given surface area and maximum volume is sin^-1\(\left(\frac{1}{3}\right)\)

Solution:
Let r be the radius, l be the slant height and h be the height of the cone of given surface area S.
Also, let α be the semi-vertical angle of the cone.
Then, S = πrl + πr² ⇒ l = \(\frac{\mathrm{S}-\pi \mathrm{r}^2}{\pi \mathrm{r}}\) …………….. (1)
Let V be the volume of the cone, Then V = \(\frac{1}{3}\)πr²h
V² = \(\frac{1}{9}\)π²r⁴h²
= \(\frac{1}{9}\)π²r⁴(l² – r²) [∵ l² = r² + h²]

Thus V is maximum when S = 4πr²
∴ S = πrl + πr² ⇒ 4πr² = πrl + πr²
⇒ 3πr² = πrl
⇒ l = 3r
Now, in ∆COB
sinα = \(\frac{O B}{B C}=\frac{r}{1}=\frac{r}{3 r}=\frac{1}{3}\)
∴ α = sin^-1\(\left(\frac{1}{3}\right)\)

Referring to the AP Inter 2nd Year Maths Study Material Chapter 9 Differential Equations Exercise 9c Solutions makes it easier to understand complex problems.

AP Inter 2nd Year Maths Differential Equations Solutions Exercise 9c

I.

Question 1.
Find the general solution of \(\frac{d y}{d x}=\frac{1-\cos x}{1+\cos x}\)
Solution:
Given D.E. is \(\frac{d y}{d x}=\frac{1-\cos x}{1+\cos x} \Rightarrow \frac{d y}{d x}=\frac{2 \sin ^2 \frac{x}{2}}{2 \cos ^2 \frac{x}{2}}=\tan ^2 \frac{x}{2}\)
⇒ \(\frac{d y}{d x}=\left(\sec ^2 \frac{x}{2}-1\right) \Rightarrow d y=\left(\sec ^2 \frac{x}{2}-1\right) d x\)
Integrating both sides, we get \(\int \mathrm{dy}=\int\left(\sec ^2 \frac{\mathrm{x}}{2}-1\right) \mathrm{dx}\)
⇒ y = \(\int \sec ^2 \frac{x}{2} d x-\int d x \Rightarrow y=2 \tan \frac{x}{2}-x+C\)

Question 2.
Find the general solution of \(\frac{d y}{d x}=\sqrt{4-y^2}\), (-2 < y < 2)
Solution:
Given D.E. is \(\int \frac{d y}{\sqrt{4-y^2}}=\int d x\)
Integrating both sides, we get \(\int \frac{d y}{\sqrt{4-y^2}}=\int d x\)
⇒ sin^-1\(\frac{y}{2}\) = x + C ⇒ \(\frac{y}{2}\) = sin(x + C) ⇒ y = 2 sin(x + C)

Question 3.
Find the general solution of \(\frac{d y}{d x}+y=1\), (y ≠ 1)
Solution:
Given D.E. is \(\frac{d y}{d x}+y=1 \Rightarrow \frac{d y}{d x}=1-y \Rightarrow \frac{d y}{1-y}=d x\)
Integrating both sides, we get
\(\int \frac{\mathrm{dy}}{1-\mathrm{y}}=\int \mathrm{dx}\) ⇒ -log (y – 1) = x + log C ⇒ – log C – log(y – 1) = x
⇒ -[log C + log(y – 1)] = x ⇒ log C(y – 1) = -x ⇒ C(y – 1) = e^-x
⇒ y = 1 + \(\frac{1}{C} e^{-x}\) ⇒ y = 1 + Ae^-x (Where A = \(\frac{1}{C}\))

Question 4.
Find the general solution of sec² x tan ydx + sec² ytan xdy = 0
Solution:
Given D.E. is sec² x tan ydx + sec² ytan xdy = 0
⇒ sec² xtan ydx = -(sec² y)(tan x) dy
⇒ \(\frac{\sec ^2 x}{\tan x} d x=-\frac{\sec ^2 y}{\tan y} d y\)
Integrating both sides, we get
\(\int \frac{\sec ^2 x}{\tan x} d x=-\int \frac{\sec ^2 y}{\tan y} d y\) ………….(1)
Let tan x = t \(\frac{d}{d x}\)(tan x) = \(\frac{d t}{d x}\) ⇒ sec² x = \(\frac{d t}{d x}\) ⇒ sec² xdx =dt
Now, \(\int \frac{\sec ^2 x}{\tan x} d x=\int \frac{1}{t} d t\) = log t = log(tan x) …………(2)
Similarly, \(\int \frac{\sec ^2 y}{\tan y} d y\) = log(tan y) ………….(3)
Using (1), (2), (3) we get log(tan x) = -log(tan y) + log C
log(tan x) = log\(\left(\frac{\mathrm{C}}{\tan \mathrm{y}}\right)\) ⇒ tan x = \(\frac{\mathrm{C}}{\tan \mathrm{y}}\) ⇒ tan x tan y = C

Question 5.
Find the general solution of (e^x + e^-x) dy – (e^x – e^-x)dx = 0
Solution:
Given D.E. is (e^x + e^-x) dy – (e^x – e^-x)dx = 0
⇒ (e^x + e^-x)dy = (e^x – e^-x)dx ⇒ dy = \(\left[\frac{e^x-e^{-x}}{e^x+e^{-x}}\right] d x\)
Integrating both sides, we get \(\int \mathrm{dy}=\int\left[\frac{\mathrm{e}^{\mathrm{x}}-\mathrm{e}^{-\mathrm{x}}}{\mathrm{e}^{\mathrm{x}}+\mathrm{e}^{-\mathrm{x}}}\right] \mathrm{dx}\) ……..(1)
Let (e^x + e^-x) = t ⇒ \(\frac{d}{d x}\)(e^x + e^-x) = \(\frac{d t}{d x}\) ⇒ (e^x + e^-x)dx = dt
Putting these values in equation (1), we get \(\int \mathrm{dy}=\int \frac{1}{\mathrm{t}} \mathrm{dt}+\mathrm{C}\)
⇒ y = log(t) + C ⇒ y = log(e^x + e^-x) + C

Question 6.
Find the general solution of \(\frac{d y}{d x}\) = (1 + x²)(1 + y²)
Solution:
Given D.E. \(\frac{d y}{d x}\) = (1 + x²)(1 + y²) ⇒ \(\frac{d y}{1+y^2}\) = (1 + x²)dx
Integratingboth sides, we get \(\int \frac{d y}{1+y^2}=\int\left(1+x^2\right) d x\)
⇒ tan^-1 y = \(\int \mathrm{dx}+\int \mathrm{x}^2 \mathrm{dx} \Rightarrow \tan ^{-1} \mathrm{y}=\mathrm{x}+\frac{\mathrm{x}^3}{\mathrm{x}}+\mathrm{C}\)

Question 7.
Find the general solution of y log ydx – xdy = 0
Solution:
Given D.E. is y log ydx – xdy = 0 ⇒ y log ydx = xdy ⇒ \(\frac{d y}{y \log y}=\frac{d x}{x}\)
Integrating both sides, we get \(\int \frac{d y}{y \log y}=\int \frac{d x}{x}\) ………….(1)
Let log y = ⇒ t \(\frac{d}{d y}(\log y)=\frac{d t}{d y} \Rightarrow \frac{1}{y}=\frac{d t}{d y} \Rightarrow \frac{1}{y} d y=d t\)
Putting these values in equation (1) we get \(\) ⇒ log t = log x + log C
⇒ log(log y) = log Cx ⇒ log y = Cx ⇒ y = e^Cx

Question 8.
Find the general solution of \(x^5 \frac{d y}{d x}\) = -y⁵
Solution:
Given D.E is \(x^5 \frac{d y}{d x}=-y^5 \Rightarrow \frac{d y}{y^5}=-\frac{d x}{x^5} \Rightarrow \frac{d x}{x^5}+\frac{d y}{y^5}=0\)
Integrating both sides, we get \(\int \frac{d x}{x^5}+\int \frac{d y}{y^5}\) = k
⇒ \(\int x^{-5} d x+\int y^{-5} d y=k \Rightarrow \frac{x^{-4}}{-4}+\frac{y^{-4}}{-4}\) = k
⇒ x^-4 + y^-4 = -4k ⇒ x^-4 + y^-4 = C (where C = -4k)

Question 9.
Find the general solution of \(\frac{d y}{d x}\) = sin^-1 x
Solution:
Given D.E. is \(\frac{d y}{d x}\) = sin^-1 x dy = sin^-1 xdx
Integrating both sides, we get \(\)
⇒ y = sin^-1 x.\(\text { (1) } \mathrm{dx}-\int\left[\left(\frac{\mathrm{d}}{\mathrm{dx}}\left(\sin ^{-1} \mathrm{x}\right)\right] \int(1) \mathrm{dx}\right] \mathrm{dx}\)
⇒ y = x sin^-1 x + \(\frac{-x}{\sqrt{1-x^2}} d x\) ………….(1)
Let 1 – x² = t ⇒ \(\frac{d}{d x}\left(1-x^2\right)=\frac{d t}{d x} \Rightarrow-2 x=\frac{d t}{d x} \Rightarrow x d x=-\frac{1}{2} d t\)
Puttingthese values in equation (1), we get
⇒ y = x sin^-1 x + \(\int \frac{1}{2 \sqrt{t}} d t \Rightarrow y=x \sin ^{-1} x+\frac{1}{2} \int(t)^{\frac{-1}{2}} d t\)
⇒ y = x sin^-1 x + \(\frac{1}{2} \cdot\left(\frac{\frac{1}{t^2}}{\frac{1}{2}}\right)+C \Rightarrow y=x \sin ^{-1} x+\sqrt{t}+C\)
y = x sin^-1 x + \(\sqrt{1-x^2}\) + C

Question 10.
Find the general solution of e^x tan ydx + (1 – e^x)sec² ydy = 0
Solution:
Given D.E. is e^x tan ydx + (1 – e^x)sec² ydy = 0
⇒ (1 – e^x) sec² y dy = -e^x tan y dx ⇒ \(\frac{\sec ^2 y}{\tan y} d y=\frac{-e^x}{1-e^x} d x\)
Integrating both sides, we get
\(\)
Let tan y = u ⇒ \(\frac{d}{d y}(\tan y)=\frac{d u}{d y} \Rightarrow \sec ^2 y=\frac{d u}{d y}\) ⇒ sec² ydy = du
Now, \(\int \frac{\sec ^2 y}{\tan y} d y=\int \frac{d u}{u}\) = log u = log(tan y) ……….(2)
Now, let(1 – e^x) = t ⇒ \(\frac{d}{d x}\)(1 – e^x) = \(\frac{d t}{d x}\)
⇒ -e^x = \(\frac{d t}{d x}\) ⇒ -e^x dx = dt
Now, \(\int \frac{-e^x}{1-e^x} d x=\int \frac{d t}{t}\) = log t = log(1 – e^x) ……………(3)
Sub (2) and (3) in (1) ⇒ log(tan y) = log(1 – e^x) + log C
⇒ log(tan y) = log[C(1 – e^x)] ⇒ tan y = C(1 – e^x)

Question 11.
Find a particular solution of \(\cos \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)\) = 1 (a ∈ R); y = 1 when x = 0
Solution:
Given D.E is \(\cos \left(\frac{d y}{d x}\right)\) = a \(\frac{d y}{d x}\) = cos^-1 a ⇒ dy = cos^-1 adx
Integrating both sides, we get
\(\int \mathrm{dy}=\cos ^{-1} \mathrm{a} \int \mathrm{dx}\) ⇒ y =cos^-1 a.x + C ⇒ y = cos^-1 a + C
Now, y = 1 when x = 0 ⇒ 1 = 0.cos^-1 a + C ⇒ C = 1
Thus, y = xcos^-1 a + 1 ⇒ \(\frac{y-1}{x}=\cos ^{-1} a \Rightarrow \cos \left(\frac{y-1}{x}\right)=a\), is a particular solution.

II.

Question 1.
Find a particular solution of \(\frac{d y}{d x}\) = y tan x; y = 1 when x = 0
Solution:
Given D.E. is \(\frac{d y}{d x}\) = y tan x ⇒ \(\frac{d y}{y}\) = tan xdx
Integrating both sides, we get \(\int \frac{d y}{y}=\int \tan x d x\) ⇒ log y = log(sec x) + log C
⇒ log y = log(sec xC) ⇒ y = C sec x
Now, y = 1, x = 0 ⇒ 1 = C × sec0 ⇒ 1 = C × 1 ⇒ C = 1, Thus, y = sec x is a particular solution.

Question 2.
Find the equation of a curve passing through the point (0, 0) and whose differential equation is y’ = e^x sin x.
Solution:
Given D.E. is y’ = e^x sin x ⇒ \(\frac{d y}{d x}\) = e^x sin x ⇒ dy = e^x sinx dx
Integrating both sides, we get \(\int d y=\int e^x \sin x d x \Rightarrow y=\int e^x \sin x d x\) ……………..(1)
Let I = \(\int e^x \sin x d x\)

⇒ 2y – 1 = e^x(sin x – cos x) which is the equation of the curve.

Question 3.
For the differential equation \(x y \frac{d y}{d x}\) = (x + 2)(y + 2). find the solution curve passing through the point (1, -1).
Solution:
Given D.E. is \(x y \frac{d y}{d x}\) = (x + 2)(y + 2) ⇒ \(\left(\frac{y}{y+2}\right) d y=\left(\frac{x+2}{2}\right) d x\)
Integrating both sides, we get
\(\int\left(1-\frac{2}{y+2}\right) d y=\int\left(1+\frac{2}{x}\right) d x \Rightarrow \int d y-2 \int \frac{1}{y+2} d y=\int d x+2 \int \frac{1}{x} d x\)
⇒ y – 2log(y + 2) = x + 2log x + C
⇒ y – x – C = log x² + log(y + 2)² ⇒ y – x – C = log[x²(y + 2)²]
Since, the curve passes through (1, -1) we have
⇒ -1 – 1 – C = log[(1)²(-1 + 2)²] ⇒ -2 – C = log 1 ⇒ 2 – C = 0 ⇒ C = -2
Thus, y – x + 2 = log[x²(y + 2)²] is the required solution of the curve.

Question 4.
Find the equation of a curve passing through the point (0, -2) given that at any point (x, y) on the curve, the product of the slope of its tangent and y coordinate of the point is equal to the x coordinate of the point.
Solution:
Let x and y be the x-coordinate and y-coordinate of the curve, respectively.
We know that the slope of a tangent to the curve in the coordinate axis is given by \(\frac{d y}{d x}\)
y.\(\frac{d y}{d x}\) = x ydy = xdx
Integrating both sides, \(\int y d y=\int x d x \Rightarrow \frac{y^2}{2}=\frac{x^2}{2}+C\) ⇒ y² – x² = 2C
Since, the curve passes through (0, -2) , we have (-2)² – 0² = 2C ⇒ 2C = 4
Thus, y² – x² = 4 is the required equation of the curve.

III.

Question 1.
Find a particular solution of (x³ + x² + x + 1)\(\frac{d y}{d x}\) = 2x² + x;y = 1 when x = 0
Solution:
Given D.E. is (x³ + x² + x + 1)\(\frac{d y}{d x}\) = 2x² + x;y = 1
⇒ \(\frac{d y}{d x}=\frac{2 x^2+x}{\left(x^3+x^2+x+1\right)} \Rightarrow d y=\frac{2 x^2+x}{\left(x^3+x^2+x+1\right)} d x\)
Integrating both sides, we get

⇒ A(x² + 1) + (Bx + C)(x + 1) = 2x² + x….(1)
Put x = -1 in (1) then A[(-1)² + 1] + 0 = 2(-1²) + 1 ⇒ 2A = 1 ⇒ A = 1/2
Put x = 0 in (1) then A + C = 0 => C = -A = -1/2
Equating the coefficients of x² we get A + B = 2 => B = 2 – A = 2 – \(\frac{1}{2}=\frac{3}{2}\)
∴ A = \(\frac{1}{2}\), B = \(\frac{3}{2}\) and C = \(\frac{-1}{2}\)
Substituting these values in (2) , we get

Thus, y = \(\frac{1}{4}\)[log(x + 1)²(x² + 1)³] – \(\frac{1}{2}\)tan^-1 x + 1, is a particular solution.

Question 2.
Find a particular solution of x(x² – 1)\(\frac{d y}{d x}\) = 1; y = 0 when x = 2
Solution:
Given D.E. is x(x² – 1)\(\frac{d y}{d x}\) = 1 ⇒ dy = \(\frac{d x}{x\left(x^2-1\right)} \Rightarrow d y=\frac{1}{x(x-1)(x+1)} d x\)
Integrating both sides, we get \(\int \mathrm{dy}=\int \frac{1}{\mathrm{x}(\mathrm{x}-1)(\mathrm{x}+1)} \mathrm{dx}\) ……..(1)
Let \(\frac{1}{(x-1)(x+1)}=\frac{A}{x}+\frac{B}{x-1}+\frac{C}{x+1}\) …….(2)
⇒ \(\frac{1}{x(x-1)(x+1)}=\frac{A(x-1)(x+1)+B x(x+1)+C x(x-1)}{x(x-1)(x+1)}\)
⇒ A(x – 1)(x + 1) + Bx(x + 1) + Cx(x – 1) = 1 ………..(3)
Put x =0 in (3) then A(0 – 1)(0 + 1) + B(0) + C(0) = 1 ⇒ -A = 1 ⇒ A = -1
Put x =1 in (3) then A(0) + B1(1 + 1) + C(0) = 1 ⇒ 2B = 1 ⇒ B = 1/2
Put x = -1 in (3) then A(0) + B(0) + C(-1)(-1 – 1) = 1 ⇒ 2C = 1 ⇒ C = 1/2
∴ A = -1, B = \(\frac{1}{2}\) and C = \(\frac{1}{2}\)
Substituting these values in (2), we get

Question 3.
At any point (x, y) of a curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point (- 4, -3). Find the equation of the curve given that it passes through (-2, 1).
Solution:
Given that (x, y) is point of contact of curve and tangent.
Slope of segment joining (x, y) and (-4, -3) is m₁ = \(\frac{y+3}{x+4}\)
We know that the slope of a tangent to the curve in the coordinate axis is \(\frac{d y}{d x}\)
Slope of tangent is m₂ = \(\frac{d y}{d x}\) But m₂ = 2m₁
⇒ \(\frac{d y}{d x}=2 \frac{(y+3)}{x+4} \Rightarrow \frac{d y}{y+3}=\frac{2 d x}{x+4}\)
Integrating both sides, we get \(\int \frac{d y}{y+3}=2 \int \frac{d x}{x+4}\) ⇒ log(y + 3) = 2log(x + 4) + log C
⇒ log (y + 3) = log C(x + 4)² ⇒ y + 3 = C(x + 4)²
Since, the curve passes through , (-2, 1) we have 1 + 3 = C(-2 + 4)² ⇒ 4 = 4C ⇒ C = 1
Thus, y + 3 = (x + 4)² is the required equation of the curve.

Question 4.
The volume of spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units and after 3 seconds it is 6 units. Find the radius of balloon after t seconds.
Solution:
Let the rate of change of volume of the balloon be k.
⇒ \(\frac{\mathrm{dV}}{\mathrm{dt}}=\mathrm{k} \Rightarrow \frac{\mathrm{~d}}{\mathrm{dt}}\left(\frac{4}{3} \pi \mathrm{r}^3\right)=\mathrm{k} \Rightarrow \frac{4}{3} \pi 3 \mathrm{r}^2 \frac{\mathrm{dr}}{\mathrm{dt}}=\mathrm{k} \Rightarrow 4 \pi \mathrm{r}^2 \mathrm{dr}=\mathrm{kdt}\)
Integrating both sides, we get \(\int 4 \pi \mathrm{r}^2 \mathrm{dr}=\int \mathrm{kdt} \Rightarrow 4 \pi \frac{\mathrm{r}^3}{3}=\mathrm{kt}+\mathrm{C} \Rightarrow 4 \pi \mathrm{r}^3=3(\mathrm{kt}+\mathrm{C})\)
At t = 0, we have r = 3
⇒ 4π × 27 = 3(k × 0 + C) ⇒108π = 3C ⇒ C = 36π
Now, at t = 3, we have r = 6 ⇒ 4π × 6³ = 3(k × 3 + C)
⇒ 864π = 3(3k + 36π) ⇒ 3k = 288π – 36π = 252π ⇒ k = 84π
Hence 4πr³ = 3[84πt + 36π] ⇒ 4πr³ = 4π(63t + 27)
⇒ r³ = 63t + 27 ⇒ r = (63t + 27)^{\(\frac{1}{3}\)}
Thus, the radius of the balloon after t seconds is (63t + 27)^{\(\frac{1}{3}\)} units.

Question 5.
In a bank, principal increases continuously at the rate of r%.per year. Find the value of r if Rs 100 double itself in 10 years (log_e2 = 0.6931).
Solution:
Let p, t and r represent the principle, time and rate of interest respectively.
The principle increases continuously at the rate of r% per year
⇒ \(\frac{d p}{d t}=\left(\frac{r}{100}\right) p \Rightarrow \frac{d p}{p}=\left(\frac{r}{100}\right) d t\)
Integrating both sides, we get \(\int \frac{\mathrm{dp}}{\mathrm{p}}=\frac{\mathrm{r}}{100} \int \mathrm{dt} \Rightarrow \log \mathrm{p}=\frac{\mathrm{rt}}{100}+\mathrm{k} \Rightarrow \mathrm{p}=\mathrm{e}^{\frac{\mathrm{rt}}{100}+\mathrm{k}}\)
It is given that p = 100 when t = 0 ⇒ 100 = e^k
Now, if t = 10 then p = 2 × 100 = 200
Hence, 200 = \(e^{\frac{r}{10}+k} \Rightarrow 200=e^{\frac{r}{10}} e^k \Rightarrow 200=e^{\frac{r}{10}} \cdot 100 \Rightarrow e^{\frac{r}{10}}=2 \Rightarrow \frac{r}{10}=\log _e 2\)
⇒ \(\frac{\mathrm{r}}{10}\) = 0.6931 ⇒ r = 6.931
Thus, the rate of interest, r = 6.931%

Question 6.
In a bank, principal increases continuously at the rate of 5% per year. An amount of Rs 1000 Is deposited with this bank, how much will it worth after 10 years (e^0.5 = 1.648).
Solution:
Let p and t be the principle and time, respectively.
The principle increases continuóusly at the rate of 5%per year.
\(\frac{d p}{d t}=5 \% \times p \Rightarrow \frac{d p}{d t}=\left(\frac{5}{100}\right) p \Rightarrow \frac{d p}{d t}=\frac{p}{20} \Rightarrow \frac{d p}{p}=\frac{d t}{20}\)
Integrating both sides, we get \(\int \frac{d p}{p}=\frac{1}{20} \int d t \Rightarrow \log p=\frac{t}{20}+C \Rightarrow p=e^{\frac{t}{20}+C}\)
Now, p = 1000 when t = 0 ∴ 1ooo = e^C
Now, at t= 10 and e^C = 1000 ⇒ p = \(e^{\frac{10}{20}+C} \Rightarrow p=e^{0.5} \times e^C\) ⇒ p = 1.648 × 1000 ⇒ p = 1648
∴ After 10 years Rs. 1000 becomes Rs 1648 at the rate of 5%.

Question 7.
In a culture, the bacteria count is 1,00,000. The number is increased by 10% in 2 hours. In how many hours will the count reach 2,00,000, if the rate of growth of bacteria is proportional to the number present?
Solution:
Let y be the number of bacteria at any instant t.
Rate of growth of the bacteria is proportional to the number present
⇒ \(\frac{d y}{d t} \propto y \Rightarrow \frac{d y}{d t}=k y \Rightarrow \frac{d y}{y}=k d t\)
Integrating both sides, we get \(\int \frac{d y}{y}=k \int d t \Rightarrow \log y=k t+C\)
Let y₀ be the number of bacteria at t = 0 ⇒ log y₀ = C ⇒ log y = kt + log y₀
⇒ log y – log y₀ = kt ⇒ log\(\left(\frac{\mathrm{y}}{\mathrm{y}_0}\right)\) = kt
Since, the number of bacteria increases by 10% in 2 hours. ⇒ \(\frac{y}{y_0}=\frac{110}{100} \Rightarrow \frac{y}{y_0}=\frac{11}{10}\)

The number of bacteria increases from 1,00,000 to 2,00,000 in 0.24 hr.

Referring to the AP Inter 2nd Year Maths Study Material Chapter 6 Application of Derivatives Exercise 6b Solutions makes it easier to understand complex problems.

AP Inter 2nd Year Maths Application of Derivatives Solutions Exercise 6b

I.

Question 1.
Show that the function given by f (x) = 3x + 17 is increasing on R.
Solution:
Let x₁ and x₂ be any two numbers in R.
Then x₁ < x₂ = (3x₁ + 17) < (3x₂ + 17) ⇒ f (x₁) < f (x₂)
Thus, f is strictly increasing on R.

Question 2.
Show that the function given by f (x) = e^2x is increasing on R.
Solution:
Let x₁ and x₂ be any two numbers in R.
Then x₁ < x₂ ⇒ 2x₁ < 2x₂
⇒ e^2x₁ < e^2x₂
⇒ f (x₁) < f (x₂)
Thus, f is strictly increasing on R.

Question 3.
Prove that the logarithmic function is increasing on (0, ∞).
Solution:
The given function is f(x) = log x ⇒ f (x) = \(\frac{1}{x}\)
For, x > 0, f'(x) = \(\frac{1}{x}\) > 0
Thus, the logarithmic function is strictly increasing in interval (0, ∞)

Question 4.
Prove that the function given by f(x) = x³ – 3x² + 3x – 100 is increasing in k.
Solution:
Given that f(x) = x³ – 3x² + 3x – 100
⇒ f'(x) = 3x² – 6x + 3 = 3(x² – 2x + 1) = 3(x – 1)²
For x ∈ R, (x – 1)² ≥ 0
So, f'(x) is always positive in R.
Thus, the function is increasing in R.

II.

Question 1.
Show that the function given by f (x) = sin x is
(a) increasing in (0, \(\frac{\pi}{2}\))
(b) decreasing in (\(\frac{\pi}{2}\), π)
(c) neither increasing nor decreasing in (0, π)
Solution:
Given that f(x) = sin x ⇒ f'(x) = cos x
(a) For x ∈ (0, \(\frac{\pi}{2}\)) ⇒ cos x > 0 ⇒ f(x) > 0. Thus, f is strictly increasing in (0, \(\frac{\pi}{2}\))
(b) For x ∈ (\(\frac{\pi}{2}\), π) ⇒ cos x < 0 ⇒ f(x) < 0. Thus, f is strictly decreasing in (\(\frac{\pi}{2}\), π)
From (a) & (b) it is clear that f is neither strictly increasing nor decreasing in (0, π)

Question 2.
Find the intervals in which the function f given by f(x) = 2x² – 3x is
(a)increasing (b) decreasing
Solution:
Given that f(x) = 2x² – 3x ⇒ f'(x) = 4x – 3
Now f'(x) = 0 ⇒ 4x – 3 = 0 ⇒ x = 3/4

In (\(\frac{3}{4}\), ∞), f'(x) = 4x – 3 > 0. Hence,f is strictly increasing in (\(\frac{3}{4}\), ∞)
In (-∞, \(\frac{3}{4}\)), f'(x) = 4x – 3 < 0. Here, f is strictly decreasing in (-∞, \(\frac{3}{4}\))

Question 3.
Find the intervals in which the function f given by f (x) = 2x³ – 3x² – 36x + 7 is
(a) increasing
(b) decreasing
Solution:
Given that f (x) = 2x³ – 3x² – 36x + 7

⇒ f'(x) = 6x² – 6x – 36 = 6(x² – x – 6) = 6(x + 2)(x – 3)
∴ f'(x) = 0 ⇒ x = -2, 3
In(-∞, 2)and(3, ∞),f'(x) > 0. In (-2, 3), f'(x) < 0
∴ f is strictly increasing in (-∞, -2), (3, ∞) and strictly decreasing in (-2, 3)

Question 4.
Find the intervals in which the function x² + 2x – 5 is strictly increasing or decreasing:
Solution:
Let f(x) = x² + 2x – 5 ⇒ f ’(x) = 2x + 2 = 2(x + 1)
(i) f(x) is increasing when f'(x) > 0 ⇒ x + 1 > 0 ⇒ x > – 1 ⇒ x ∈ (-1, ∞)
(ii) f(x) is decreasing when f'(x) <0 ⇒ X + 1<0 ⇒ x < -1 ⇒ x ∈ (-∞, -1).

Question 5.
Find the intervals in which the function 10 – 6x – 2x² is strictly increasing or decreasing:
Solution:
Let f(x) = 10 – 6x – 2x² ⇒ f'(x) = -6 – 4x = -2(2x+3)
x = –\(\frac{3}{2}\) divides the number line into two intervals (-∞, –\(\frac{3}{2}\)) and (-\(\frac{3}{2}\), ∞)
(i) In (-∞, –\(\frac{3}{2}\)), f'(x) = -2(2x + 3) < 0 ⇒ 2x + 3 > 0. Hence f is strictly increasing for x < \(\frac{-3}{2}\) (ii) In (-\(\frac{3}{2}\), ∞).f'(x) = -2(2x + 3) > 0 ⇒ 2x + 3 < 0 Hence f is strictly decreasing for x > \(\frac{-3}{2}\)

Question 6.
Find the intervals in which the function -2x³ – 9x² – 12x + 1 is strictly increasing or decreasing:
Solution:
Let f(x)= -2x³ – 9x² – 12x + 1
f'(x) = -6x² – 18x – 12 = -6(x² + 3x + 2) = -6(x + 1)(x + 2)
∴ f'(x) = 0 ⇒ x = -1; -2
x= -1 and x= -2 divide the number line into intervals (-∞, 2), (-2, -1) and (-1, ∞)

In (-2, -1), f (x) = -6(x + 1)(x + 2) > 0 Hence, f is strictly increasing in (-2, -1)
Hence, f is strictly increasing in (-2, -1)
In (-∞, -2) and (-1, ∞), f'(x) = -6(x + 1)(x + 2) < 0
Hence, f is strictly decreasing in (-∞, -2) ∪ (-1, ∞)

Question 7.
Find the intervals in which the function 6 – 9x – x² is strictly increasing or decreasing:
Solution:
Let f(x) = 6 – 9x – x² ⇒ f'(x) = -9 – 2x = -(2x + 9)
(i) f(x) is increasing when f'(x) > 0 ⇒ -(2x + 9) > 0 ⇒ 2x + 9 < 0
⇒ 2x < -9 ⇒ x < \(\frac{-9}{2}\) ⇒ x ∈ (-∞, \(\frac{-9}{2}\))

(ii) f(x) is decreasing when f'(x) < 0 ⇒ -(2x+9) < 0 ⇒ 2x + 9 > 0
⇒ 2x > -9 ⇒ x > \(\frac{-9}{2}\) ⇒ x ∈ (\(\frac{-9}{2}\), ∞)

Question 8.
Prove that the function f given by f (x) = x² – x + 1 is neither strictly increasing nor decreasing on (- 1, 1).
Solution:
Given that f(x) = x² – x + 1 ⇒ f'(x) = 2x – 1
Now, f'(x) = 0 ⇒ 2x – 1 = 0⇒ 2x = 1 ⇒ x = \(\frac{1}{2}\)

x = \(\frac{1}{2}\) divides the interval (-1, 1) into (-1, \(\frac{1}{2}\)) and (\(\frac{1}{2}\), 1)
In interval (-1, \(\frac{1}{2}\)), f'(x) = 2x – 1 < 0
Hence, f is strictly decreasing in (-1, \(\frac{1}{2}\))
In interval (\(\frac{1}{2}\), 1), f'(x) = 2x – 1 > 0
Hence, f is strictly increasing in (\(\frac{1}{2}\), 1).
Thus,f is strictly increasing nor strictly decreasing in interval (-1, 1)

Question 9.
For what values of a the function f given b f (x) = x² + a + 1 is incrcaing on [1, 2]?
Solution:
Given that f(x) = x² + ax + 1 = f'(x) = 2x + a
Now, the function f has to be strictly increasing on (1, 2]
Since, 2x + a is a linear function, its minimum occurs at x = 1
∴ minimum value = 2(1) + a = 2 + a
Since, the function is increasing
∴ 2 + a ≥ 0 ⇒ a ≥ -2

Question 10.
Prove that the function f given by f (x) = log sin x is increasing on (0, \(\frac{\pi}{2}\)) decreasing on (\(\frac{\pi}{2}\), π)
Solution:
Given that f (x) = log sin x ⇒ f (x) = \(\frac{1}{\sin x}\) (cos x) = cot x

In interval (0, \(\frac{\pi}{2}\)), f'(x) = cot x > 0
Hence, f is strictly increasing in (0, \(\frac{\pi}{2}\))
In interval (\(\frac{\pi}{2}\), π), f'(x) = cot x < 0. Hence, f is strictly decreasing in (\(\frac{\pi}{2}\), π)

Question 11.
Prove that the function f given by f (x) = log |cos x| is decreasing on (o, \(\frac{\pi}{2}\)) and increasing on (\(\frac{3\pi}{2}\), 2π)
Solution:
Given that f(x) = log |cos x| ⇒ f'(x) = \(\frac{1}{\cos x}\)(-sin x) = -tanx
In interval (0, \(\frac{\pi}{2}\)), tan x > 0 ⇒ – tan x < 0
Hence f(x) < 0
Thus, f is strictly decreasing on (0, \(\frac{\pi}{2}\))
In interval (\(\frac{3\pi}{2}\), 2π), tan x < 0 ⇒ -tan x > 0
Hence f'(x) > 0
Thus, f is strictly increasing on (\(\frac{3\pi}{2}\), 2π)

Question 12.
Find the values of x for which y = |x(x – 2)|² is an increasing function.
Solution:
Given that y = [x(x – 2)]² = [x² – 2x]²
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{\mathrm{d}}{\mathrm{dx}}\)[x² – 2x]² = 2(x² – 2x)(2x – 2) = 4x(x – 2)(x – 1)

Now \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 0 ⇒ 4x(x – 2)(x – 1) ⇒ x = 0, x = 2, x = 1
x = 0, x = 1 and x = 2 divide the number line intervals (-∞, 0), (0, 1),
In (1, 2) and (2, ∞), \(\frac{\mathrm{dy}}{\mathrm{dx}}\) > 0.
Hence y is strictly increasing in intervals (0, 1) and (2, ∞)
So, y is increasing for 0 < x < 1 and x > 2.

III.

Question 1.
Find the intervals in which the function (x + 1)³(x – 3)³ is strictly increasing or decreasing.
Solution:
f(x) = (x + 1)³(x – 3)³ ⇒ f'(x) = 3(x + 1)²(x – 3)³ + 3(x – 3)²(x + 1)³
= 3(x + 1)² (x – 3)² [x – 3 + x + 1] = 3(x + 1)² (x – 3)² (2x – 2)
= 6(x + 1)²(x – 3)²(x – 1)
∴ f'(x) = 0 = x = -1, 3, 1

x = -1 3,1 divides the number line into four intervals (-∞, -1), (-1, 1), (1, 3) and (3, ∞)
In(-∞, -1) and(-1, 1), f'(x) = -6(x + 1)²(x – 3)²(x – 1) < 0
Hence, f is strictly decreasing in (-∞, -1) and (-1, 1)
In(1, 3) and (3, ∞), f ‘(x) = -6(x + 1)² (x – 3)² (x – 1) > 0
Hence, f is strictly increasing in (1, 3) and (3, ∞)

Question 2.
Show that y = log(1 + x) – \(\frac{2 x}{2+x}\), x > -1, is an increasing function of x throughout its domain.
Solution:
Given that y = log(1 + x) – \(\frac{2 x}{2+x}\)
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{1+x}-\frac{(2+x)(2)-2 x(1)}{(2+x)^2}=\frac{1}{1+x}-\frac{4}{(2+x)^2}=\frac{x^2}{(1+x)(2+x)^2}\)
Now, \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 0 ⇒ \(\frac{x^2}{(2+x)^2}\) = 0 ⇒ x² = 0 ⇒ x = 0
Since, x > -1, x = 0 divides domain (-1, ∞) in two intervals -1 < x < 0 and x > 0
Case (i): -1 < x < 0 ⇒ x<0 and x > -1. We have x + 1 > 0, (2 + x)² >0, x² > 0
∴ \(\frac{x^2}{(1+x)(2+x)^2}\) > 0
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) > 0
Çasc (ii): x > 0 We have x + 1 > 0, (2 + x)² >0, x² > 0
∴ \(\frac{x^2}{(1+x)(2+x)^2}\) > 0
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) > 0
Hence, in both cases, we get \(\frac{\mathrm{dy}}{\mathrm{dx}}\) > 0
∴ f is an increasing function of x throughout its domain.

Question 3.
Prove that y = \(\frac{4 \sin \theta}{(2+\cos \theta)}\) – θ is an increasing function of θ in [0, \(\frac{\pi}{2}\)].
Solution:
Given that y = \(\frac{4 \sin \theta}{(2+\cos \theta)}\) – θ
⇒ \(\frac{d y}{d \theta}\) = \(\frac{(2+\cos \theta)(4 \cos \theta)-4 \sin \theta(-\sin \theta)}{(2+\cos \theta)^2}-1=\frac{8 \cos \theta+4 \cos ^2 \theta+4 \sin ^2 \theta}{(2+\cos \theta)^2}-1=\frac{8 \cos \theta+4}{(2+\cos \theta)^2}-1\)
Now \(\frac{d y}{d \theta}\) = 0 ⇒ \(\frac{8 \cos \theta+4}{(2+\cos \theta)^2}\) = 1 ⇒ 8 cos θ + 4 = 4 + cos²θ + 4 cos θ
⇒ cos²θ – 4cos θ = 0 ⇒ cos θ(cos θ – 4) = 0 ⇒ cos θ = 0 or cos θ = 4
∴ cos θ = 0 ⇒ θ = \(\frac{\pi}{2}\)
\(\frac{d y}{d \theta}=\frac{8 \cos \theta+4-\left(4+\cos ^2 \theta+4 \cos \theta\right)}{(2+\cos \theta)^2}=\frac{4 \cos \theta-\cos ^2 \theta}{(2+\cos \theta)^2}=\frac{\cos (4-\cos \theta)}{(2+\cos \theta)^2}\)
In interval [0, \(\frac{\pi}{2}\)], we have cos θ > 0
Also, 4 > cos θ ⇒ 4 – cos θ > 0
Hence, cos θ (4 – cos θ) > 0 and also (2 + cos θ)² > 0
∴ \(\frac{\cos \theta(4-\cos \theta)}{(2+\cos \theta)^2}\) > 0
Hence, \(\frac{d y}{d \theta}\) > 0
So, y is strictly increasing in (0, \(\frac{\pi}{2}\)) and the given function is continuous at x = 0 and x = π/2
Thus, y is increasing in interval [0, \(\frac{\pi}{2}\)]

Question 4.
Let I be any interval disjoint from |-1, 1|. Prove that the function f given by f(x) = x + \(\frac{1}{x}\) is increasing on I.
Solution:
Given I is any interval disjoint from [-1, 1] i.e., I ∩ [-1, 1] = Φ ⇒ I ∈ (-∞, -1) ∪ (1, ∞)
Now, f(x) = x + \(\frac{1}{x}\) ⇒ f (x) = 1 – \(\frac{1}{x^2}\)
Case (i): In (-∞, -1) clearly f'(x) = 1 – \(\frac{1}{x^2}\) > 0
Case (ii): In (1, ∞) clearly f'(x) = 1 – \(\frac{1}{x^2}\) > 0
Hence f'(x) > 0 in (-∞, -1) ∪ (1, ∞)
∴ f is strictly increasing function in the interval I disjoint from [-1, 1]

Referring to the AP Inter 2nd Year Maths Study Material Chapter 6 Application of Derivatives Exercise 6a Solutions makes it easier to understand complex problems.

AP Inter 2nd Year Maths Application of Derivatives Solutions Exercise 6a

I.

Question 1.
Find the rate of change of the area of a circle with respect to its radius r when (a) r = 3 cm (b) r = 4 cm
Solution:
(a) r = 3 cm
For the circle, we take radius = r and area = A
Given r = 3 cm, Area of circle A = πr²
Here, we have to find the rate of change of area A. w.r.t. ‘r’.
∴ A = πr² ⇒ \(\frac{\mathrm{dA}}{\mathrm{dr}}\) = π(2r) = 2πr
When r = 3,
\(\frac{\mathrm{dA}}{\mathrm{dr}}\) = 2π(3) = 6π
Thus, the area of the circle is changing at the rate of 6π cm²/s

(b) r = 4cm
For the circle, we take radius = r and area =A
Given r= 3 cm, Area of circle A = πr²
Here, we have to find the rate of change of area A w.r.t.’ r ‘.
∴ A = πr² ⇒ \(\frac{\mathrm{dA}}{\mathrm{dr}}\) = π(2r) = 2πr
When r = 5, \(\frac{\mathrm{dA}}{\mathrm{dr}}\) = 2π(4) = 8π.
Thus, the area of the circle is changing at the rate of 8π cm²/s.

Question 2.
The radius of a circle is increasing uniformly at the rate of 3 cm/s. Find the rate at which the area of the circle is increasing when the radius is 10 cm.
Solution:
For the circle, we take radius = r and area = A
We know that A = πr²
Given \(\frac{\mathrm{dr}}{\mathrm{dt}}\) = 3cm / s and r= 10 cm
Now \(\frac{\mathrm{dA}}{\mathrm{dr}}\) = \(\frac{\mathrm{d}}{\mathrm{dt}}\)(πr²) = 2πr \(\frac{\mathrm{dr}}{\mathrm{dt}}\)
= 2π(10)(3) = 60π cm²

Question 3.
An edge of a variable cube is increasing at the rate of 3 cm/s. How fast is the Volume of the cube increasing when the edge is 10 cm long?
Solution:
Let x be the length and V be the volume of the cube.
Given that \(\frac{\mathrm{dx}}{\mathrm{dt}}\) = 3cm/s ,x = 10cm
Hence, V = x³
Now, \(\frac{\mathrm{dV}}{\mathrm{dt}}\) = \(\frac{\mathrm{d}}{\mathrm{dt}}\)(x³) = 3x²\(\frac{\mathrm{dx}}{\mathrm{dt}}\) = 3(10)²(3) = 900cm³/s

Question 4.
A stone is dropped into a quiet lake and waves move in circles at the speed of 5 cm/s. At the instant when the radius of the circular wave is 8 cm, how fast is the enclosed area increasing?
Solution:
For the circle, we take radius = r and area = A
Given \(\frac{\mathrm{dr}}{\mathrm{dt}}\) = 5 and r = 8 Area of circle A = πr²
On diff. w.r.t ‘t’, we get \(\frac{\mathrm{dA}}{\mathrm{dt}}\) = (2πr)\(\frac{\mathrm{dr}}{\mathrm{dt}}\) = 2π(8)(5) = 80π cm²/s
Thus, the enclosed area is increasing at the rate of 80π cm²/s, when r = 8 cm.

Question 5.
The radius of a circle is increasing at the rate of 0.7 cm/s. What is the rate of increase of its circumference? ‘
Solution:
For the circle, we take radius = r and area = A
Given that \(\frac{\mathrm{dr}}{\mathrm{dt}}\) = 0.7 cm / s
We know that Circumference C = 2πr
Now, \(\frac{\mathrm{dC}}{\mathrm{dt}}\) = \(\frac{\mathrm{d}}{\mathrm{dt}}\)(2πr) = 2π\(\frac{\mathrm{dr}}{\mathrm{dt}}\) = 2π(0.7) = 1.4π cm / s

Question 6.
A balloon, which always remains spherical has a variable radius. Find the rate at which its volume is increasing with the radius when the later is 10 cm.
Solution:
For the sphere, we take radius = r and volume = V
Given that radius, r = 10 cm. We know that V = \(\frac{4}{3}\)πr³
∴ \(\frac{\mathrm{dV}}{\mathrm{dt}}\) = \(\frac{\mathrm{d}}{\mathrm{dr}}\)(\(\frac{4}{3}\)πr²) = \(\frac{4}{3}\)π(3r²) = 4πr² = 4π(10)² =400π
Thus, the volume of the balloon is increasing at the rate of 400π cm³/s.

Question 7.
The total cost C (x) in Rupees associated with the production of x units of an item is given by C (x) = 0.007x³ – 0.003x² + 15x + 4000. Find the marginal cost when 17 units are produced.
Solution:
We know that marginal cost is the rate of change of total cost with respect to the output.
∴ cost(MC) = \(\frac{\mathrm{dC}}{\mathrm{dx}}\) = 0.007(3x²) – 0.003(2x) + 15
= 0.021x² – 0.006x + 15 dx
When x = 17, MC = 0.021(17)² – 0.006(17) + 15
= 0.021(289) – 0.006(17) + 15
= 6069 – 0102 + 15
= 20.967
Hence, the required marginal cost is ₹ 20.967 (nearly).

Question 8.
The total revenue in Rupees received from the sale of x units of a product is given by R (x) = 13x² + 26x + 15. Find the marginal revenue when x = 7.
Solution:
Marginal revenue (MR) is die rate of change of the total revenue with respect to the number of units sold.
∴ MR = \(\frac{\mathrm{dR}}{\mathrm{dx}}\) = 13(2x)+26
= 26x + 26
When x = 7, MR = 26(7) + 26 = 182 + 26 = 208

II.

Question 1.
The volume of a cube is increasing at the rate of 8 cm³/s. How fast is the surface area increasing when the length of an edge is 12 cm?
Solution:
For the cube, we take length of the edge = x , Volume = V and Surface area = S
Given \(\frac{\mathrm{dV}}{\mathrm{dt}}\) = 8cm³ / s and x = 12 cm
Volume of the cube V = x³ On diff. w.r.t’t’, we get \(\frac{\mathrm{dV}}{\mathrm{dt}}\) = 3x²\(\frac{\mathrm{dx}}{\mathrm{dt}}\)
⇒ 8 = 3x²\(\frac{\mathrm{dx}}{\mathrm{dt}}\) ⇒ \(\frac{\mathrm{dx}}{\mathrm{dt}}\) = \(\frac{8}{3 x^2}\)
Surface area S = 6x²
On diff. w.r.t. ‘t’, we get \(\frac{\mathrm{dS}}{\mathrm{dt}}\) = 12x\(\frac{\mathrm{dx}}{\mathrm{dt}}\) = 12x\(\left(\frac{8}{3 x^2}\right)=\frac{32}{x}\)
So, when x = 12 cm ⇒ \(\frac{\mathrm{ds}}{\mathrm{dt}}\) = \(\frac{32}{12}\) = \(\frac{8}{3}\)cm² / s

Question 2.
The length x of a rectangle is decreasing at the rate of 5 cm/minute and the width y is increasing at the rate of 4 cm/minute. When x = 8cm and y = 6cm, find the rates of change of (a) the perimeter, and (b) the area of the rectangle.
Solution:
Given that \(\frac{\mathrm{dx}}{\mathrm{dt}}\) = -5cm / min , \(\frac{\mathrm{dy}}{\mathrm{dt}}\) = 4cm / min, x = 8cm and y = 6cm
(a) The perimeter of a rectangle is given by P = 2(x+y)
∴ \(\frac{\mathrm{dP}}{\mathrm{dt}}\) = 2\(\left(\frac{\mathrm{dx}}{\mathrm{dt}}+\frac{\mathrm{dy}}{\mathrm{dt}}\right)\) = 2(-5 + 4) = -2 cm / min

(b) The area of a rectangle is given by A = xy
⇒ \(\frac{\mathrm{dA}}{\mathrm{dt}}\) = \(\frac{\mathrm{dx}}{\mathrm{dt}}\)y + x\(\frac{\mathrm{dy}}{\mathrm{dt}}\) = -5y + 4x = (-5 × 6 + 4 × 8)cm² / min = 2cm² / min

Question 3.
A balloon, which always remains spherical on inflation, is being inflated by pumping in 900 cubic centimetres of gas per second. Find the rate at which the radius of the balloon increases when the radius is 15 cm.
Solution:
For the sphere, we take radius = r and volume = V
Given \(\frac{\mathrm{dV}}{\mathrm{dt}}\) = 900c.c/s, r = 15cm
Volume of the sphere V = \(\frac{4}{3}\)πr³
On diff w.r.t ‘t’, we get \(\frac{\mathrm{dV}}{\mathrm{dt}}\) = \(\frac{4}{3}\)π 3 r² \(\frac{\mathrm{dr}}{\mathrm{dt}}\)
⇒ 900 = 4π(15)²\(\frac{\mathrm{dr}}{\mathrm{dt}}\)
⇒ \(\frac{\mathrm{dr}}{\mathrm{dt}}\) = \(\frac{900}{4 \pi \times 15 \times 15}=\frac{900}{900 \pi}=\frac{1}{\pi}\) cm/s

Question 4.
A ladder 5 m long is leaning against a wall. The bottom of the ladder is pulled along the ground, away from the wall, at the rate of 2 cm/s. How fast is its height on the wall decreasing when the foot of the ladder is 4 m away from the wall ?
Solution:
Let the height of the wall at which the ladder is touching it be y
and the distance of its foot from the wall on the ground be x
Given that x = 4 cm, \(\frac{\mathrm{dx}}{\mathrm{dt}}\) = 2cm / s
Hence, x² + y² = 5² ⇒ y² = 25 – x² ⇒ y = \(\sqrt{25-x^2}\)
∴ \(\frac{d y}{d t}=\frac{d}{d t}\left(\sqrt{25-x^2}\right)=\frac{1}{2 \sqrt{25-x^2}}(-2 x) \frac{d x}{d t}=\frac{-x}{\left(\sqrt{25-x^2}\right)} \frac{d x}{d t}=\frac{-2 x}{\sqrt{25-x^2}}=\frac{-2 \times 4}{\sqrt{25-16}}=-\frac{8}{3}\) cm / s

Question 5.
A particle moves along the curve 6y = x³ + 2. Find the points on the curve at which the y-coordinate is changing 8 times as fast as the x-coordinate.
Solution:
From the question we have \(\frac{\mathrm{dy}}{\mathrm{dt}}\) = 8\(\frac{\mathrm{dx}}{\mathrm{dt}}\)
Given equation of the curve 6y = x³ + 2
Diff. w.r.t time we have 6\(\frac{\mathrm{dy}}{\mathrm{dt}}\) = 3x²\(\frac{\mathrm{dx}}{\mathrm{dt}}\) ⇒ 2\(\frac{\mathrm{dy}}{\mathrm{dt}}\) = x²\(\frac{\mathrm{dy}}{\mathrm{dt}}\)
⇒ 2(8\(\frac{\mathrm{dx}}{\mathrm{dt}}\)) = x²\(\frac{\mathrm{dx}}{\mathrm{dt}}\) ⇒ 16\(\frac{\mathrm{dx}}{\mathrm{dt}}\) = x²\(\frac{\mathrm{dx}}{\mathrm{dt}}\) ⇒ (x² – 16)\(\frac{\mathrm{dx}}{\mathrm{dt}}\) = 0
⇒ x² = 16 ⇒ x = ±4
When x = 4then y = \(\frac{4^3+2}{6}=\frac{66}{6}\) = 11
When x = -4 then y = \(\frac{\left(-4^3\right)+2}{6}=-\frac{62}{6}=-\frac{31}{3}\)
Thus, the points on the curve are (4, 11) and (-4, \(\frac{-31}{3}\))

Question 6.
The radius of an air bubble is increasing at the rate of \(\frac{1}{2}\) cm/s. At what rate is the volume of the bubble increasing when the radius is 1 cm?
Solution:
For the sphere, we take radius = r, Volume = V.
Given that \(\frac{\mathrm{dr}}{\mathrm{dt}}\) = \(\frac{1}{2}\)cm / s we have to find \(\frac{\mathrm{dV}}{\mathrm{dt}}\) at r = 1
Volume V = \(\frac{4}{3}\)πr³ ⇒ \(\frac{\mathrm{dV}}{\mathrm{dt}}\) = \(\frac{4}{3}\) π 3 r²\(\frac{\mathrm{dr}}{\mathrm{dt}}\) = 4π(1)²\(\frac{1}{2}\) = 2π cm³/s

Question 7.
A balloon, which always remains spherical, has a variable diameter \(\frac{3}{2}\)(2x + 1). Find the rate of change of its ‘oIurne with respect to x.
Solution:
Given that diameter d = \(\frac{3}{2}\)(2x +1). Hence, radius r = \(\frac{3}{4}\)(2x + 1)
We know that V = \(\frac{4}{3}\)πr³ = \(\frac{4}{3} \pi\left(\frac{3}{4}\right)^3\) (2x + 1)³ = \(\frac{9}{16}\) π(2x + 1)³
∴ \(\frac{\mathrm{dV}}{\mathrm{dx}}\) = \(\frac{9}{16}\) π\(\frac{\mathrm{d}}{\mathrm{dx}}\)(2x + 1)³= \(\frac{9 \pi}{16}\)3(2x + 1)² . 2
= \(\frac{27}{8}\)π(2x + 1)²

III.

Question 1.
Sand is pouring from a pipe at the rate of 12 cm³/s. The falling sand forms a cone on the ground in such a way that the height of the cone is always one-sixth of the radius of the base. How fast is the height of the sand cone increasing when the height is 4 cm?
Solution:
For the cone, we take radius = r, Volume = V, height = h
Given that h = \(\frac{1}{6}\)r ⇒ r = 6h, \(\frac{\mathrm{dV}}{\mathrm{dx}}\) = 12cm² / s and h = 4cm
We know that V = \(\frac{1}{3}\)πr²h.
∴ V = \(\frac{1}{3}\)π(6h)²h = 12πh³
\(\frac{\mathrm{dV}}{\mathrm{dx}}\) = 12π\(\frac{\mathrm{d}}{\mathrm{dh}}\)(h³) = 12π(3h²)\(\frac{\mathrm{dh}}{\mathrm{dt}}\) = 36πh² \(\frac{\mathrm{dh}}{\mathrm{dt}}\)
h = 4 cm, then 12 = 36π(4)² \(\frac{\mathrm{dh}}{\mathrm{dt}}\) ⇒ \(\frac{\mathrm{dh}}{\mathrm{dt}}\) = \(\frac{12}{36 \pi(16)}=\frac{1}{48 \pi}\) cm / s

Referring to the AP Inter 2nd Year Maths Study Material Chapter 9 Differential Equations Exercise 9b Solutions makes it easier to understand complex problems.

AP Inter 2nd Year Maths Differential Equations Solutions Exercise 9b

I.

Question 1.
Verify that y = e^x + 1 is a solution to y” – y’= 0
Solution:
Given function is y = e^x + 1; Differentiating both sides w.r.t x, we have
⇒ \(\frac{d y}{d x}=\frac{d}{d x}\left(e^x+1\right)\) ⇒ y’ = e^x ………..(1); Again differentiating both sides w.r.t x, we have
\(\frac{d}{d x}\left(y^{\prime}\right)=\frac{d}{d x}\left(e^x\right)\) ⇒ y” = e^x …………(2)
From (1) and (2), y” – y’= e^x – e^x = 0
Thus, the given function is the solution of corresponding differential equation. Hence verified.

Question 2.
Verify that y= x² + 2x + C is a solution to y’ – 2x – 2 = 0
Solution:
Given function is y = x² + 2x + c ⇒ y’ = \(\frac{d}{d x}\)(x² + 2x + c) ⇒ y’ = 2x + 2
∴ y’ – 2x – 2 = (2x + 2) – 2x – 2 = 0
Thus, the given function is the solution of corresponding differential equation. Hence verified.

Question 3.
Verify that y = cosx + C is a solution to y’ + sin x = 0
Solution:
Given function is y = cos x + C ⇒ y’ = \(\frac{d}{d x}\)(cos x + C) ⇒ y’ = -sin x
∴ y’ – sinx = -sinx + sinx = 0
Thus, the given function is the solution of corresponding differential equation. Hence verified.

Question 4.
Verify that y = \(\sqrt{1+x^2}\) is a solution to y’ = \(\frac{x y}{1+x^2}\)
Solution:
Given function is y = \(\sqrt{1+x^2} \Rightarrow y^{\prime}=\frac{d}{d x}\left(\sqrt{1+x^2}\right)=\frac{1}{2 \sqrt{1+x^2}} \cdot \frac{d}{d x}\left(1+x^2\right)\)
= \(\frac{2 x}{2 \sqrt{1+x^2}}=\frac{x}{2 \sqrt{1+x^2}}=\frac{x\left(\sqrt{1+x^2}\right)}{\left(1+x^2\right)}=\frac{x y}{1+x^2}\)
Thus, the given function is the solution of corresponding differential equation.Hence verified.

Question 5.
Verify that y = Ax is a solution to xy’ = y(x ≠ 0)
Solution:
Given function is y = Ax ⇒ y’ = \(\frac{d}{d x}\)(Ax) = A ∴ xy’ = xA = Ax = y
Thus, the given function is the solution of corresponding differential equation. Hence verified.

Question 6.
Verify that y = x sin x is a solution to xy’ = y + x\(\sqrt{x^2-y^2}\)(x ≠ 0 and x > y or x < -y)
Solution:
Given that y = x sinx ⇒ y’ = \(\frac{d}{d x}\)(x sinx) = sin x\(\frac{d}{d x}\)(x) + x\(\frac{d}{d x}\)(sinx) = sin x + x cosx
∴ xy’ = x(sin x + x cox x) = x sin x + x² cox x [∵ y = x sin x ⇒ \(\frac{\mathrm{y}}{\mathrm{x}}\) = sinx]
= y + x². \(\sqrt{1-\sin ^2 x}\) = y + x²\(\sqrt{1-\left(\frac{y}{x}\right)^2}\) = y + x\(\sqrt{x^2-y^2}\).
Hence verified.

Question 7.
Verify that xy = logy + C is a solution to y’ = \(\frac{y^2}{1-x y}\)(xy ≠ 1)
Solution:
Given that xy = log y + C ⇒ \(\frac{d}{d x}\)(xy + C) = \(\frac{d}{d x}\)(logy) ⇒ y\(\frac{d}{d x}\)(x) + x.\(\frac{d y}{d x}=\frac{1}{y} \frac{d y}{d x}\)
⇒ y + xy’ = \(\frac{1}{y}\).y’ ⇒ y² + xyy’ = y’ ⇒ (xy – 1)y’ = -y² ⇒ y ‘ = \(\frac{y^2}{1-x y}\). Hence verified

Question 8.
Verify that y – cos y = x is a solution to (y siny + cos y + x)y’ = y
Solution:
Given that y – cos y = x ⇒ \(\frac{d y}{d x}-\frac{d}{d x}\)(cos y) = \(\frac{d}{d x}\)(x)
⇒ y’ – (-sin y).y’ = 1 ⇒ y'(1 + sin y) = 1 ⇒ y’ = \(\frac{1}{1+\sin y}\) [∵ x = y – cos y]
∴ (y sin y + cos y + x)y’ = (y sin y + cos y + y – cos y) × \(\frac{1}{1+\sin y}\) = y(1 + sin y).\(\frac{1}{1+\sin y}\) = y
Hence verified.

Question 9.
Verify that x + y = tan^-1 y is a solution to y²y’ + y² + 1 = 0
Solution:
Given that x + y = tan^-1 y ⇒ \(\frac{d}{d x}\)(x + y) = \(\frac{d}{d x}\)(tan^-1y)
⇒ \(1+y^{\prime}=\left[\frac{1}{1+y^2}\right] y^{\prime} \Rightarrow y^{\prime}\left[\frac{1}{1+y^2}-1\right]=1\)
⇒ \(y^{\prime}\left[\frac{1-\left(1+y^2\right)}{1+y^2}\right]=1 \Rightarrow y^{\prime}\left[\frac{-y^2}{1+y^2}\right]=1 \Rightarrow y^{\prime}=\frac{-\left(1+y^2\right)}{y^2}\)
∴ y²y’ + y² + 1 = y²\(\left[\frac{-\left(1+y^2\right)}{y^2}\right]\) + y² + 1 = -1 – y² + y² + 1 = 0. Hence verified.

Question 10.
Verify that y = \(\sqrt{a^2-x^2}\); x (-a, a) is a solution to x + y\(\frac{d y}{d x}\) = 0(y ≠ 0)
Solution:
Given that y = \(\sqrt{a^2-x^2} \Rightarrow \frac{d y}{d x}=\frac{d}{d x}\left(\sqrt{a^2-x^2}\right) \Rightarrow \frac{d y}{d x}=\frac{1}{2 \sqrt{a^2-x^2}} \cdot \frac{d}{d x}\left(a^2-x^2\right)\)
⇒ \(\frac{d y}{d x}=\frac{1}{2 \sqrt{a^2-x^2}}(-2 x) \Rightarrow \frac{d y}{d x}=\frac{-x}{\sqrt{a^2-x^2}}\)
∴ x + y\(\frac{d y}{d x}=x+\sqrt{a^2-x^2} \times \frac{-x}{\sqrt{a^2-x^2}}\) = x – x = 0. Hence verified

Referring to the AP Inter 2nd Year Maths Study Material Chapter 9 Differential Equations Exercise 9a Solutions makes it easier to understand complex problems.

AP Inter 2nd Year Maths Differential Equations Solutions Exercise 9a

I.

Question 1.
Determine order and degree of \(\frac{d^4 y}{d x^4}\) + sin(y”’) = 0
Solution:
Given D.E is \(\frac{d^4 y}{d x^4}\) + sin(y”’) = 0 => y”” + sin(y””) = 0 dx
Highest order derivative in the differential equation is y””. Its order is four.
But the D.E is not a polynomial equation in its derivatives. So its degree is not defined.

Question 2.
Determine order and degree of y’+ 5y = 0
Solution:
Given D.E is y’ + 5y = 0
Highest order derivative in the D.E is y’. Its order is one.
It is a polynomial equation in y’. Highest power of y’ is 1. So degree of the D.E is one.

Question 3.
Determine order and degree \(\left(\frac{\mathrm{ds}}{\mathrm{dt}}\right)^4+3 \mathrm{~s} \frac{\mathrm{~d}^2 \mathrm{~s}}{\mathrm{dt}^2}\) = 0
Solution:
Highest order derivative in the given D.E is \(\frac{d^2 s}{d t^2}\). Its order is two.
It is a polynomial equation in \(\frac{d^2 s}{d t^2} \text { and } \frac{d s}{d t}\).
The power of \(\frac{d^2 s}{d t^2}\) is 1. So degree of the D.E is one.

Question 4.
Determine order and degree of \(\left(\frac{d^2 y}{d x^2}\right)^2+\cos \left(\frac{d y}{d x}\right)\) = 0
Solution:
Highest order derivative in the given D.E is \(\frac{d^2 y}{d x^2}\). Its order is 2.
Given differential equation is not a polynomial equation in its derivatives.
Degree of the D.E is not defined.

Question 5.
Determine order and degree of \(\frac{d^2 y}{d x^2}\) = cos3x + sin3x
Solution:
Highest order derivative in the given D.E is \(\frac{d^2 y}{d x^2}\). Its order is two.
It is a polynomial equation in \(\frac{d^2 y}{d x^2}\) and the power is 1. So degree of D.E is 1.

Question 6.
Determine order and degree of (y”’)² + (y”)³ + (y’)⁴ + y⁵ = 0
Solution:
Given D.E is (y”’)² + (y”)³ + (y’)⁴ + y⁵ = 0
Highest order derivative present in the D.E is y”‘. Its order is three.
Given D.E is a polynomial equation in y'”, y” and y’.
Highest powgr raised to y”‘ is 2. So degree of the D.E is 2.

Question 7.
Determine order and degree of y”’ + 2y” + y’= 0
Solution:
GivenD.E is y”’ + 2y’ + y’ = 0
Highest order derivative present in the differential equation is y'” . Its order is 3.
It is a polynomial equation in y'”, y” and y’. The highest power of y”’ is 1. Degree of the D.E is 1.

Question 8.
Determine order and degree of y’ + y = e^x
Solution:
Given D.E is y’ +y = e^x ⇒ y’ + y – e^x = 0
Highest order derivative present in the differential equation is y’. Its order is one.
Given D.E is a polynomial equation in y’ and the highest power of is one.
Degree of the D.E is 1.

Question 9.
Determine order and degree of y” + (y’)² + 2y = 0
Solution:
Given D.E is y” + (y’)² + 2y = 0
Highest order derivative present in the differential equation is y” . Its order is two.
Given D.E is a polynomial equation in y” and y’, the highest power of y” is one.
Degree of the D.E is 1.

Question 10.
Determine order and degree of y” + 2y’ + siny = 0
Solution:
Given D.E is y” + 2y’ + siny = 0
Highest order derivative present in the differential equation is y”. Its order is two.
This is a polynomial equation in y” and y’ the highest power of y” is one.
Degree of the D.E is 1

Referring to the AP Inter 2nd Year Maths Study Material Chapter 5 Continuity and Differentiability Exercise 5h Solutions makes it easier to understand complex problems.

AP Inter 2nd Year Maths Continuity and Differentiability Solutions Exercise 5h

I.

Question 1.
Differentiate (3x² – 9x + 5)⁹ w.r.t. x
Solution:
Let y = (3x² – 9x + 5)⁹
⇒ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 9(3x² – 9x + 5)⁸\(\frac{\mathrm{d}}{\mathrm{dx}}\)(3x² – 9x + 5) [∵ \(\frac{\mathrm{d}}{\mathrm{dx}}\) (f(x))ⁿ = n(f(x))^n-1\(\frac{\mathrm{d}}{\mathrm{dx}}\)f(x)]
= 9(3x² – 9x + 5)⁸[3(2x) – 9(1) + 0]
= 9(3x² – 9x + 5)⁸(6x – 9) = 27(3x² – 9x + 5)⁸(2x – 3).

Question 2
Differentiate sin³ x + cos⁶ x w.r.t. x
Solution:
Let y = sin³ x +cos⁶ x = (sin x)³ + (cos x)⁶
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = 3(sin x)²\(\frac{\mathrm{d}}{\mathrm{dx}}\) sin x + 6(cos x)⁵\(\frac{\mathrm{d}}{\mathrm{dx}}\)cos x [∵ \(\frac{\mathrm{d}}{\mathrm{dx}}\) (f(x))ⁿ = n(f(x))^n-1\(\frac{\mathrm{d}}{\mathrm{dx}}\)f(x)]
= 3 sin² x cos x – 6 cos⁵ x sin x
= 3 sin x cos x(sin x – 2 cos⁴ x)

II.

Question 1.
Differentiate (5x)^{3 cos 2x} w.r.t. x
Solution:
Let y = (5x)^{3 cos 2x} ………….. (i)
Taking logs of both sides of (1) we have
logy = log(5x)^{3 cos 2x} = 3 cos 2x log(5x)
Differentiating both sides w.r.t. x, we have

Question 2.
Differentiate sin^-1(x\(\sqrt{\mathrm{x}}\)), 0 ≤ x ≤ 1 w.r.t. x
Solution:
Let y = sin^-1(x\(\sqrt{\mathrm{x}}\)) = sin^-1 (x^3/2 [∵ x\(\sqrt{\mathrm{x}}\) = x¹ . x^1/2 = x^1+1/2 = x^3/2]
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{1}{\sqrt{1-\left(x^{3 / 2}\right)^2}} \frac{d}{d x} x^{3 / 2}\) [∵ \(\frac{\mathrm{d}}{\mathrm{dx}}\)sin^-1f(x) = \(\frac{1}{\sqrt{1-(f(x))^2}} \frac{d}{d x}\)f(x)]
= \(\frac{1}{\sqrt{1-x^3}} \frac{3}{2} x^{1 / 2}\)
= \(\frac{3 \sqrt{x}}{2 \sqrt{1-x^3}}\)
= \(\frac{3}{2} \sqrt{\frac{x}{1-x^3}}\)

Question 3.
Differentiate cos (a cos x + b sin x), for some constant a and b. w.r.t. x
Solution:
Let y = cos(a cos x + b sin x) for some constants a and b.
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = -sin(a cos x + b sin x)\(\frac{\mathrm{d}}{\mathrm{dx}}\)(a cos x + b sin x) [∵ \(\frac{\mathrm{d}}{\mathrm{dx}}\) cos f(x) = -sin f(x)\(\frac{\mathrm{d}}{\mathrm{dx}}\)f(x)]
= -sin(a cos x + b sin x)[-a sin x + b cos x]
= -(-a sin x + b cos x)sin(a cos x + b sin x)
= (a sin x – b cos x)sin(a cos x + b sin x).

Question 4.
Find \(\frac{\mathrm{dy}}{\mathrm{dx}}\) if y = 12(1 – cost), x = 10(t – sint), – \(\frac{\pi}{2}\) < t < \(\frac{\pi}{2}\)
Solution:
Given that y = 12(1 – cos t) and x = 10(t – sin t)
Differentiating both equations wr.t. t, we haye
\(\frac{\mathrm{dy}}{\mathrm{dt}}\) = 12\(\frac{\mathrm{d}}{\mathrm{dt}}\)(1 – cost) = 12(0 + sin t) = 12 sin t
\(\frac{\mathrm{dx}}{\mathrm{dt}}\) = 10\(\frac{\mathrm{d}}{\mathrm{dt}}\)(t – sin t) = 10(1 – cos t)
∴ \(\frac{d y}{d x}=\frac{d y / d t}{d x / d t}=\frac{12 \sin t}{10(1-\cos t)}=\frac{6}{5} \cdot \frac{2 \sin \frac{1}{2} \cos \frac{1}{2}}{2 \sin ^2 \frac{t}{2}}=\frac{6}{5} \frac{\cos \frac{1}{2}}{\sin \frac{t}{2}}=\frac{6}{5} \cot \frac{t}{2} .\)

Question 5.
Using the fact that sin (A + B) = sin A cos B + cos A sin B and the differentiation, obtain the sum formula for cosines.
Solution:
Given that sin (A + B) = sinA cosB + cosA sinB
Assuming A and B are functions of x and differentiating both sides w.r.t x, we have
cos(A+B)\(\frac{\mathrm{d}}{\mathrm{dx}}\)(A+B) = sin A \(\frac{\mathrm{d}}{\mathrm{dx}}\)(cos B) + cos B\(\frac{\mathrm{d}}{\mathrm{dx}}\)(sin A) + cos A\(\frac{\mathrm{d}}{\mathrm{dx}}\)(sin B) + sin B\(\frac{\mathrm{d}}{\mathrm{dx}}\)(cosA)
⇒ cos(A+B)\(\left(\frac{\mathrm{dA}}{\mathrm{dx}}+\frac{\mathrm{dB}}{\mathrm{dx}}\right)\) = -sin A sin B\(\frac{\mathrm{dA}}{\mathrm{dx}}\) + cos B cos A\(\frac{\mathrm{dA}}{\mathrm{dx}}\) + cos A cos B\(\frac{\mathrm{dB}}{\mathrm{dx}}\) – sin B sin A\(\frac{\mathrm{dA}}{\mathrm{dx}}\)
=(cos A cos B – sin A sin B) \(\frac{\mathrm{dB}}{\mathrm{dx}}\) +(cos A cos B – sin A sin B)\(\frac{\mathrm{dA}}{\mathrm{dx}}\)
= (cos A cos B – sin A sin B)\(\left(\frac{\mathrm{dB}}{\mathrm{dx}}+\frac{\mathrm{dA}}{\mathrm{dx}}\right)\)
Cancelling \(\left(\frac{\mathrm{dB}}{\mathrm{dx}}+\frac{\mathrm{dA}}{\mathrm{dx}}\right)\) both sides, we have cos(A + B)= cos A cos B – sin A sin B

Question 6.
If y = \(\left|\begin{array}{ccc}
f(x) & g(x) & h(x) \\
l & m & n \\
a & b & c
\end{array}\right|\), prove that \(\frac{d y}{d x}=\left|\begin{array}{ccc}
f^{\prime}(x) & g^{\prime}(x) & h^{\prime}(x) \\
l & m & n \\
a & b & c
\end{array}\right|\)
Solution:
Given that y = \(\left|\begin{array}{ccc}
f(x) & g(x) & h(x) \\
l & m & n \\
a & b & c
\end{array}\right|\)
Expanding the determinant along the first row,
y = f(x)(mc – nb) – g(x)(lc – na) + h(x)(lb – ma)
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = (mc – nb)\(\frac{\mathrm{d}}{\mathrm{dx}}\)f(x) – (lc – na)\(\frac{\mathrm{d}}{\mathrm{dx}}\)g(x) + (lb – ma)\(\frac{\mathrm{d}}{\mathrm{dx}}\)h(x)
= (mc – nb)f'(x) – (lc – na)g'(x) + (lb – ma)h'(x) ………… (i)
R.H.S = \(\frac{d y}{d x}=\left|\begin{array}{ccc}
f^{\prime}(x) & g^{\prime}(x) & h^{\prime}(x) \\
l & m & n \\
a & b & c
\end{array}\right|\)
= f'(x)(mc – nb) – g'(x)(lc – na) + h'(x)(lb – ma)
= (mc – nb)f'(x) – (lc – na)g'(x) + (lb – ma)h'(x) ………….. (ii)
From (i)and (ii), we have L.H.S. = RHS.

Question 7.
Differentiate the function (log x)^{cos x} w.r.t. x.
Solution:
Let y = (log x)^{cos x} ………………. (i)
⇒ log y = log(log x)^{cos x} = cos x log(logx) [∵ log mⁿ = n log m]
∴ \(\frac{\mathrm{d}}{\mathrm{dx}}\) log y = \(\frac{\mathrm{d}}{\mathrm{dx}}\)[cos x log(log x)]
⇒ \(\frac{1}{y}\frac{\mathrm{d}}{\mathrm{dx}}\) = cos x \(\frac{\mathrm{d}}{\mathrm{dx}}\) log (log x) + log(log x)\(\frac{\mathrm{d}}{\mathrm{dx}}\) cos x [By Product rule]
= cos x \(\frac{1}{\log x} \frac{d}{d x}\)log x + log(log x)(- sin x) = \(\frac{\cos x}{\log x} \frac{1}{x}\) -sin x log(log x)
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = y[\(\frac{\cos x}{x \log x}\) – sin x log(log x)]
Putting the value of y from (i), \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = (log x)^{c0s x}[\(\frac{\cos x}{x \log x}\) – sin x log(log x)]

Question 8.
Find \(\frac{\mathrm{dy}}{\mathrm{dx}}\) of the function xy = e^{(x – y)}
Solution:
Given that xy = = e^{(x – y)} ⇒ log(xy) = log e^{(x – y)}
⇒ log x + log y = (x – y) log e ⇒ log x + logy = x – y (∵ log e = 1)
Differentiating both sides w.r.t. x, we have \(\frac{\mathrm{d}}{\mathrm{dx}}\)log x + \(\frac{\mathrm{d}}{\mathrm{dx}}\) log y = \(\frac{\mathrm{d}}{\mathrm{dx}}\)x – \(\frac{\mathrm{d}}{\mathrm{dx}}\)y

Question 9.
Differentiate the function x (log x)^{log x}, x > 1 w.r.t x.
Solution:
Let y = (log x)^{log x}, x >1 …………… (i)
Taking log of both sides of (i), we have
log y = log(log x)^{log x} = log x log(log x)
Differentiating both sides w.r.t x,we have \(\frac{\mathrm{d}}{\mathrm{dx}}\)(logy) = \(\frac{\mathrm{d}}{\mathrm{dx}}\) (log x log(log x))

III.

Question 1.
Differentiate cot^-1\(\left[\frac{\sqrt{1+\sin \mathrm{x}}+\sqrt{1-\sin \mathrm{x}}}{\sqrt{1+\sin \mathrm{x}}-\sqrt{1-\sin \mathrm{x}}}\right]\), 0 < x < \(\frac{\pi}{2}\) w.r.t. x
Solution:

Question 2.
Differentiate (sin x – cos x)^{(sin x – cos x)}, \(\frac{\pi}{4}\) < x < \(\frac{3\pi}{4}\) w.r.t. x
Solution:
Let y = (sin x – cos x)^{(sin x – cos x)} …………… (i)
⇒ logy =log(sin x – cos x)^{(sin x – cos x)} = (sin x – cos x)log(sin x – cos x)
Differentiating both sides w.r.t x,we have
\(\frac{\mathrm{d}}{\mathrm{dx}}\)log y = \(\frac{\mathrm{d}}{\mathrm{dx}}\) (sin x – cos x)log(sin x – cos x)
\(\frac{\mathrm{d}}{\mathrm{dx}}\) log y = (sin x – cos x)\(\frac{\mathrm{d}}{\mathrm{dx}}\)log(sin x – cos x) + log(sin x – cos x)\(\frac{\mathrm{d}}{\mathrm{dx}}\)(sin x – cos x)
⇒ \(\frac{1}{y} \frac{d y}{d x}\) = (sin x – cos x)\(\frac{1}{(\sin x-\cos x)} \frac{d}{d x}\) (sin x – cos x) + log(sin x – cos x)(cos x + sin x)
=(cos x + sin x) + (cos x + sin x)log(sin x – cos x)
⇒ \(\frac{1}{y} \frac{d y}{d x}\) = (cos x + sin x)[1 + log(sin x – cos x)]
⇒ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = y(cos x + sin x)[1 + log(sin x – cosx)]
Putting the value of y from (i),
⇒ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = (sin x – cos x)^{(sin x – cos x)} (cos x + sin x)[1 + log(sin x – cos x)]

Question 3.
Differentiate x^x + x^a + a^x + a^a, for some fixed a > 0 and x > 0 w.r.t. x
Solution:
Let y = x^x + x^a + a^x + a^a
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{\mathrm{d}}{\mathrm{dx}}\) x^x + \(\frac{\mathrm{d}}{\mathrm{dx}}\)x^a + \(\frac{\mathrm{d}}{\mathrm{dx}}\)a^x + \(\frac{\mathrm{d}}{\mathrm{dx}}\)a^a
= \(\frac{\mathrm{d}}{\mathrm{dx}}\)x^x + ax^a-1 + a^x log a + 0 [∵ a^a is constant as 3³ = 27 is constant]
= \(\frac{\mathrm{d}}{\mathrm{dx}}\)x^x + ax^a-1 + a^x log a ………….. (i)
To find \(\frac{\mathrm{d}}{\mathrm{dx}}\) (x^x) We take u = x^x ……………… (ii)
Taking log on both sides of eqn (ii),we have log u = log x^x = x log x
∴ \(\frac{\mathrm{d}}{\mathrm{dx}}\) log u = \(\frac{\mathrm{d}}{\mathrm{dx}}\) (x log x) ⇒ \(\frac{1}{u} \frac{d u}{d x}\) = x \(\frac{\mathrm{d}}{\mathrm{dx}}\)(log x) + log x\(\frac{\mathrm{d}}{\mathrm{dx}}\)x (Product Rule)
= x\(\frac{1}{\mathrm{x}}\) + log x.1 = 1 + log x ⇒ \(\frac{\mathrm{du}}{\mathrm{dx}}\) = u(1 + log x)
\(\frac{\mathrm{d}}{\mathrm{dx}}\) x^x = x^x(1 + log x) [By putting the value of u from (ii)]
Putting this value in eqn. (i), \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = x^x(1 + log x) + ax^a-1 + a^x log a.

Question 4.
Find \(\frac{\mathrm{dy}}{\mathrm{dx}}\), if y = sin^-1x + sin^-1\(\sqrt{1-x^2}\), 0 < x < 1
Solution:

Question 5.
If x\(\sqrt{1+\mathrm{y}}\) + y\(\sqrt{1+\mathrm{x}}\) = 0, for -1 < x < 1, prove that \(\frac{d y}{d x}=-\frac{1}{(1+x)^2}\)
Solution:
Given that x\(\sqrt{1+\mathrm{y}}\) + y\(\sqrt{1+\mathrm{x}}\) = 0  …………. (i)
We shall first find y in terms of x
From eqn. (i), xx\(\sqrt{1+\mathrm{y}}\) = -y\(\sqrt{1+\mathrm{x}}\)
Squaring on both sides,
x²(1 + y) = y²(1 + x)
⇒ x² + x²y = y² + y²x or x² – y² = -x²y + y²x
⇒ (x – y)(x + y) = -xy(x – y)
Dividing both sides by (x – y) ≠ 0 (∵ x ≠ y)
x + y = -xy ⇒ y + xy = -x ⇒ y(1 + x) = -x
⇒ y = –\(\frac{x}{1+x}\)
Now differentiating both sides wrt.x, we have
\(\frac{d y}{d x}=-\frac{(1+x) \frac{d}{d x}(x)-x \frac{d}{d x}(1+x)}{(1+x)^2}=-\frac{(1+x) \cdot 1-x \cdot 1}{(1+x)^2}=-\frac{1}{(1+x)^2} .\)

Question 6.
If (x – a)² + (y – b)² = c², for some c > 0, prove that \(\frac{\left[1+{\frac{d y}{d x}^2}\right]^3}{d^2 y}\) is a constant independent of a and b.
Solution:
Given that (x – a)² + (y – b)² = c² …………. (i)
Differentiating both sides of eqn. (i) w.r.t. x,

Putting (x – a)² + (y – b)² = c² from (i)
= \(\frac{\left(c^2\right)^{3 / 2}}{-c^2}=\frac{-c^3}{c^2}\) = -c which is a constant and is independent of a and b

Question 7.
If cos y = x cos (a + y), with cos a ≠ ± , prove that \(\frac{d y}{d x}=\frac{\cos ^2(a+y)}{\sin a}\)
Solution:
Given that cos y = x cos (a + y)

Question 8.
If x = a(cos t + t sin t) and y = a(sin t – t cos t). find \(\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}\).
Solution:
Given x = a(cos t + t sin t) and y = a(sin t – t cos t),
Differentiating both eqns. w.r.t. t,we have
\(\frac{\mathrm{dx}}{\mathrm{dt}}\) = a(-sin t + t\(\frac{\mathrm{d}}{\mathrm{dt}}\)sin t + sin t\(\frac{\mathrm{d}}{\mathrm{dt}}\)t = a(-sin t + t cos t + sin t) = at cos t
\(\frac{\mathrm{dy}}{\mathrm{dt}}\) = a(cost – \(\frac{\mathrm{d}}{\mathrm{dt}}\)(t cos t)) = a(cos t – t\(\frac{\mathrm{d}}{\mathrm{dt}}\)cos t – cos t \(\frac{\mathrm{d}}{\mathrm{dt}}\)t) a(cos t + t sin t – cos t) = at sin t
∴ y = u + v
∴ \(\frac{d y}{d x}=\frac{d y / d t}{d x / d t}=\frac{\text { at } \sin t}{\text { at } \cos t}=\frac{\sin t}{\cos t}\) = tan t
Now differentiating both sides w.r.t.. x,we have \(\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}\) = \(\frac{\mathrm{d}}{\mathrm{dx}}\) (tan t) sec² t\(\frac{\mathrm{d}}{\mathrm{dt}}\)(t)
= sec² t\(\frac{\mathrm{d}}{\mathrm{dt}}\) = sec² t\(\left(\frac{1}{a t \cos t}\right)\) ………..(By(i))
= sec² t\(\left(\frac{\sec t}{\mathrm{at}}\right)=\frac{\sec ^3 t}{\mathrm{at}}\)

Question 9.
If y = e^{a cos-1x}, -1 ≤ x ≤ 1, show that (1 – x²)\(\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}\) – x\(\frac{\mathrm{dy}}{\mathrm{dx}}\) – a²y = 0
Solution:
Given that y = e^{a cos-1x}

Question 10.
Find the derivative of the function \(\frac{\cos ^{-1} \frac{x}{2}}{\sqrt{2 x+7}}\), -2 < x < 2 with respect to x.
Solution:
Let y = \(\frac{\cos ^{-1} \frac{x}{2}}{\sqrt{2 x+7}}\)
Applying the Quotient rule, we have

Question 11.
Find the derivative of the function x^x2-3 + (x – 3)^x2, for x > 3 with respect to x.
Solution:
Let y = x^x2-3 + (x – 3)^x2 for x > 3
Put u = x^x2-3 and v = (x – 3)^x2
∴ y = u + v
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{\mathrm{du}}{\mathrm{dx}}\) + \(\frac{\mathrm{dv}}{\mathrm{dx}}\) ………. (1)
Now u = x^(x2-3)
∴ Taking log of both sides, we have
log u = log x^(x2-3) = (x² – 3) log x.
Differentiating both sides w.r.t. x, we have
\(\frac{1}{\mathrm{u}}\frac{\mathrm{dy}}{\mathrm{dx}}\) = (x² – 3) \(\frac{\mathrm{d}}{\mathrm{dx}}\) log x + log x\(\frac{\mathrm{d}}{\mathrm{dx}}\) (x² – 3) = (x² – 3)\(\frac{1}{\mathrm{x}}\) + log x(2x – 0)
⇒ \(\frac{1}{\mathrm{u}}\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{x^2-3}{x}\) + 2x log x
∴ \(\frac{\mathrm{du}}{\mathrm{dx}}\) = u[\(\frac{x^2-3}{x}\) – 2 x log x]
\(\frac{\mathrm{du}}{\mathrm{dx}}\) = x^{(x2 – 3)} \(\left(\frac{x^2-3}{x}+2 x \log x\right)\) [By putting u = x^x2-3]
Now consider v = (x – 3)^x2 ⇒ log v = log (x – 3)^x2 = x² log(x – 3)
∴ \(\frac{\mathrm{d}}{\mathrm{dx}}\)log v = \(\frac{\mathrm{d}}{\mathrm{dx}}\)(x² log(x – 3))
⇒ \(\frac{1}{\mathrm{v}}\frac{\mathrm{dv}}{\mathrm{dx}}\) = x²\(\frac{\mathrm{d}}{\mathrm{dx}}\) log(x – 3) + log(x – 3)\(\frac{\mathrm{d}}{\mathrm{dx}}\)x²
= x²\(\frac{1}{x-3} \frac{d}{d x}\)(x – 3) + log(x – 3) . 2x
⇒ \(\frac{1}{\mathrm{v}}\frac{\mathrm{dv}}{\mathrm{dx}}\) = \(\frac{x^2}{x-3}\) + 2x log(x – 3)
⇒ \(\frac{\mathrm{dv}}{\mathrm{dx}}\) = v[\(\frac{x^2}{x-3}\) + 2x log(x – 3)]
= (x – 3)^x2[\(\frac{x^2}{x-3}\) + 2x log(x – 3)] ……………. (iii) [By putting v = (x – 3)^x2]
Putting values of \(\frac{\mathrm{du}}{\mathrm{dx}}\) and \(\frac{\mathrm{dv}}{\mathrm{dx}}\) from (ii) and (iii) in (i), we have
\(\frac{\mathrm{dy}}{\mathrm{dx}}\) = x^(x2-3)[\(\frac{x^2-3}{x}\) + 2x log x] + (x – 3)^x2 [\(\frac{x^2}{x-3}\) + 2x log(x – 3)]

Question 12.
If f (x) = |x|³ show that f”(x) exists for all real x and find it.
Solution:
Given f (x) = |x|³ = x³ if x ≥ 0 ………… (i) [∵ |x| = x if x ≥ 0]
and f(x) = |x|³ = (-x)³ = -x³ if x < 0 (ii) [∵ |x| = -x if x < 0]
f'(x) = 3x² if x >0 and f'(x) = -3x² if x < 0 ……………. (iii) (At x = 0, we can’t write the value of f(x) by usual rule of derivatives because x = 0 is a partitioning point of values of f(x) given by (i) and (ii)) ∴ f'(x) = 6x if x > 0 and f'(x) = -6x if x < 0 …………… (iv) ∴ From (iv), f”(x) exists for all x > 0 and for all x < 0 i.e., for all x ∈ R except at x = 0
(i) Let us discuss derivability of f(x) at x = 0
L f'(0) = \(\underset{\mathrm{x} \rightarrow 0-}{\mathrm{Lt}}\) \(\frac{f(x)-f(0)}{x-0}\) = \(\underset{\mathrm{x} \rightarrow 0-}{\mathrm{Lt}}\) \(\frac{-x^3-0}{x}\) [By (ii) and (i)]
= \(\underset{\mathrm{x} \rightarrow 0-}{\mathrm{Lt}}\) -x² = 0 (On putting x = 0)
R f'(0) = \(\underset{\mathrm{x} \rightarrow 0+}{\mathrm{Lt}}\) \(\frac{f(x)-f(0)}{x-0}\) = \(\underset{\mathrm{x} \rightarrow 0+}{\mathrm{Lt}}\) \(\frac{-x^3-0}{x-0}\) [By (i)]
= \(\underset{\mathrm{x} \rightarrow 0+}{\mathrm{Lt}}\) x² = 0 (On putting x = 0)
∴ Lf'(0) = Rf'(0) = 0
∴ f(x) is derivable at x = 0 and f'(0) = 0