TS 10th Class Maths Solutions Chapter 8 Similar Triangles Ex 8.1

Students can practice TS Class 10 Maths Solutions Chapter 8 Similar Triangles Ex 8.1 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 8 Similar Triangles Exercise 8.1

Question 1.
In ∆PQR, ST is a line such that \(\frac{\mathrm{PS}}{\mathrm{SQ}}\) = \(\frac{\mathrm{PT}}{\mathrm{TR}}\) and also ∠PST = ∠PRQ.
Prove that ∆PQR is an isosceles triangle. (AS2)
TS 10th Class Maths Solutions Chapter 8 Similar Triangles Ex 8.1 1
Solution:
In ∆PQR. ST divides PQ and PR in the same ratio.
\(\frac{\mathrm{PS}}{\mathrm{SQ}}\) = \(\frac{\mathrm{PT}}{\mathrm{TR}}\) (given)
∴ ST // QR
It is also given
that ∠PST = ∠PRQ …………… (1)
Since ST // QR and PQ is a transversal.
∠PST = ∠PQR (Corresponding angles) ………….. (2)
From (1) and (2) we have
∠PRQ = ∠PQR
In ∆PQR, ∠PRQ = ∠PQR
∴ PQ = PR
∴ ∆PQR is an isosceles triangle.

TS 10th Class Maths Solutions Chapter 8 Similar Triangles Ex 8.1

Question 2.
In the given figure, LM // CB and LN // CD, prove that \(\frac{\mathrm{AM}}{\mathrm{AB}}\) = \(\frac{\mathrm{AN}}{\mathrm{AD}}\). (AS2)
TS 10th Class Maths Solutions Chapter 8 Similar Triangles Ex 8.1 2
Solution:
In the figure, it is given that LM // CB and LN // CD.
In ∆ABC, ML // BC.
(By basic Proportionality Theorem,)
\(\frac{\mathrm{AM}}{\mathrm{AB}}\) = \(\frac{\mathrm{AL}}{\mathrm{AC}}\) ……….. (1)
In ∆ADC, LN // CD
(By Basic Proportionality Theorem,)
\(\frac{\mathrm{AN}}{\mathrm{AD}}\) = \(\frac{\mathrm{AL}}{\mathrm{AC}}\) ……….. (2)
From (1) and (2) we have
\(\frac{\mathrm{AM}}{\mathrm{AB}}\) = \(\frac{\mathrm{AN}}{\mathrm{AD}}\)
Alter :
In ∆ABC, ML // BC, AC is a transversal.
∠ALM = ∠ACB (Corresponding angles)
ML // BC, AB is a transversal.
∠AML = ∠ABC (Coressponding angles)
Now in ∆les AML and ABC,
∠ALM = ∠ACB;
∠AML = ∠ABC;
∠A is common.
∴ ∆AML ~ ∆ABC
∴ Their corresponding sides are proportional.
(i.e.,)\(\frac{\mathrm{AM}}{\mathrm{AB}}\) = \(\frac{\mathrm{AL}}{\mathrm{AC}}\) ……….. (1)
Similarity in ∆ADC, LN // CD.
So we can prove that
\(\frac{\mathrm{AN}}{\mathrm{AD}}\) = \(\frac{\mathrm{AL}}{\mathrm{AC}}\) ……….. (2)
Hence from (1) & (2), we get
\(\frac{\mathrm{AM}}{\mathrm{AB}}\) = \(\frac{\mathrm{AN}}{\mathrm{AD}}\)

TS 10th Class Maths Solutions Chapter 8 Similar Triangles Ex 8.1

Question 3.
In the given figure, DE // AC and DF // AE prove that \(\frac{\mathrm{BF}}{\mathrm{FE}}\) = \(\frac{\mathrm{BE}}{\mathrm{EC}}\). (AS2)
TS 10th Class Maths Solutions Chapter 8 Similar Triangles Ex 8.1 3
Solution:
In ∆ABC, DE // AC.
(∴ By Basic Proportionality Theorem,)
In ∆ABE, DF // AE
(∴ By Basic Proportionality Theorem,)
TS 10th Class Maths Solutions Chapter 8 Similar Triangles Ex 8.1 4
Hence proved.

Question 4.
Prove that a line drawn through the midpoint of one side of a triangle parallel to another side bisects the third side (Using basic proportionality theorem). (AS2)
Solution:
In ∆ABC, P is the mid-point of AB.
PQ is drawn parallel to BC, intersecting AC in Q.
By Basic Proportionality theorem,
TS 10th Class Maths Solutions Chapter 8 Similar Triangles Ex 8.1 5
Since P is the mid-point of AB, we have
AP = PB
∴ \(\frac{\mathrm{AQ}}{\mathrm{QC}}\) = \(\frac{\mathrm{AP}}{\mathrm{PB}}\)
⇒ \(\frac{\mathrm{AQ}}{\mathrm{QC}}\) \(\frac{\mathrm{AP}}{\mathrm{AP}}\) (∵ PB = AP)
⇒ \(\frac{\mathrm{AQ}}{\mathrm{QC}}\) = \(\frac{1}{1}\)
⇒ AQ = QC
Since AQ = QC, Q bisects AC.

TS 10th Class Maths Solutions Chapter 8 Similar Triangles Ex 8.1

Question 5.
Prove that a line joining the midpoints of any two sides of a triangle is parallel to the third side. (Using converse of basic proportionality theorem). (AS2)
Solution:
In ∆ABC, D and E are the mid-points of the sides AB and AC respectively.
TS 10th Class Maths Solutions Chapter 8 Similar Triangles Ex 8.1 6
In ∆ABC, DE divides the sides.
AB and AC in the same ratio.
∴ DE||BC (Basic Proportionality Theorem)

Question 6.
In the given figure, DE || OQ and DF || OR. Show that EF || QR. (AS2)
TS 10th Class Maths Solutions Chapter 8 Similar Triangles Ex 8.1 7
Solution:
In the given figure,
DE || OQ and DF || OR
In ∆PQO, DE || OQ
∴ \(\frac{\mathrm{PE}}{\mathrm{EQ}}\) = \(\frac{\mathrm{PD}}{\mathrm{DO}}\) (Basic Proportionality Theorem) ………….. (1)
Similarly in ∆PRO, DF || OR.
∴ \(\frac{\mathrm{PF}}{\mathrm{FR}}\) = \(\frac{\mathrm{PD}}{\mathrm{DO}}\) (Basic Proportionality Theorem) ………….. (2)
From (1) & (2), we have
\(\frac{\mathrm{PE}}{\mathrm{EQ}}\) = \(\frac{\mathrm{PF}}{\mathrm{FR}}\)
Now, in ∆PQR, we get that \(\frac{\mathrm{PE}}{\mathrm{EQ}}\) = \(\frac{\mathrm{PF}}{\mathrm{FR}}\)
∴ EF || QR (By converse of basic proportionality theorem).

TS 10th Class Maths Solutions Chapter 8 Similar Triangles Ex 8.1

Question 7.
In the adjacent figure, A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR. Show that BC || QR. (AS2)
TS 10th Class Maths Solutions Chapter 8 Similar Triangles Ex 8.1 8
Solution:
In ∆ OPQ, A and B are the points on
OP and OQ such that AB || PQ.
So, in ∆ OPQ, AB || PQ.
∴ \(\frac{\mathrm{OA}}{\mathrm{AP}}\) = \(\frac{\mathrm{OB}}{\mathrm{BQ}}\) (Basic proportionality theorem) …………… (1)
In ∆ OPR, A and C are the points on OP and OR such that AC || PR.
So, in ∆ OPR, AC || PR.
∴ \(\frac{\mathrm{OA}}{\mathrm{AP}}\) = \(\frac{\mathrm{OC}}{\mathrm{CR}}\) (Basic proportionality theorem) …………… (2)
From (1) & (2), we get \(\frac{\mathrm{OB}}{\mathrm{BQ}}\) = \(\frac{\mathrm{OC}}{\mathrm{CR}}\)
Now, in ∆ OQR, we get that \(\frac{\mathrm{OB}}{\mathrm{BQ}}\) = \(\frac{\mathrm{OC}}{\mathrm{CR}}\)
Therefore, BC || QR (Converse of the Basic proportionality theorem).

Question 8.
ABCD is a trapezium in which AB ||DC and its diagonals intersect each other at point ‘O’. Show that \(\frac{\mathrm{AO}}{\mathrm{BO}}\) = \(\frac{\mathrm{CO}}{\mathrm{DO}}\) (AS2)
Solution:
ABCD is a trapezium. AB || DC.
The diagonals AC and BD intersect in ‘O’.
Draw a line through ‘O’ parallel to AB, meeting AD in E and BC in F.
TS 10th Class Maths Solutions Chapter 8 Similar Triangles Ex 8.1 9

TS 10th Class Maths Solutions Chapter 8 Similar Triangles Ex 8.1

Question 9.
Draw a line segment of length 7.2 cm and divide it in the ratio 5:3. Measure the two parts. (AS3, AS5)
Solution:
TS 10th Class Maths Solutions Chapter 8 Similar Triangles Ex 8.1 10
Construction :
Draw a line segment AB = 7.2 cm. Make an angle BAX. Mark A1, A2, A3, A4, A5, A6, A7 on AX such that AA1 = A1A2 = A2A3 = A3A4 = A4A5 = A5A6 = A6A7 = A7A8. Mark D on AX such that AD = 5 equal parts and DC = 3 equal parts. Join BC. From D, draw DE || BC. E divides AB in the ratio 5:3. The parts are measured. AE = 4.2 cm on and EB = 2.7 cm.

TS Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(b)

Students must practice this TS Intermediate Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(b) to find a better approach to solving the problems.

TS Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(b)

I.
Question 1.
Find the set of vaines of x for which the binomial expansions of the following are valid.
i) (2 + 3x)\(\frac{-2}{3}\)
ii) (5 + x)\(\frac{3}{2}\)
iii) (7 + 3x)-5
iv) (4 – \(\frac{x}{3}\))\(\frac{-1}{2}\)
Solution:
We know that
(2 + 3x)\(\frac{-2}{3}\) = 2\(\frac{-2}{3}\) (1 + \(\frac{3}{2}\)x)\(\frac{-2}{3}\)
The expansion is valid only when |\(\frac{3}{2}\) x| < 1
⇒ |x| < \(\frac{2}{3}\)
⇒ x ∈ \(\left(\frac{-2}{3}, \frac{2}{3}\right)\).

ii) We know that (5 + x)3/2
= 53/2 (1 + \(\frac{x}{5}\))3/2
The expansion is valid only when
|\(\frac{x}{5}\)| < 1
⇒ |x| < 5
⇒ x ∈ (- 5, 5).

iii) We know that
(7 + 3x)-5 = 7-5 (1 + \(\frac{3 x}{7}\))-5
The expansion is valid only when |\(\frac{3 x}{7}\)| < 1
⇒ |x| < \(\frac{7}{3}\)
⇒ x ∈ \(\left(\frac{-7}{3}, \frac{7}{3}\right)\)

iii) We know that (4 – \(\frac{x}{3}\))\(\frac{-1}{2}\)
= \(4^{\frac{-1}{2}}\left(1-\frac{x}{12}\right)^{\frac{-1}{2}}\)
The expansion is valid only when |\(\frac{x}{12}\)| < 1
⇒ |x| < 12
⇒ x ∈ (- 12, 12).

TS Board Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(b)

Question 2.
i) 6th term of (1 + \(\frac{x}{2}\))-5
ii) 7th term of (1 – \(\frac{x^2}{3}\))-4
iii) 10th term of (3 – 4x)\(\frac{-2}{3}\)
iv) 5th term of (7 + \(\frac{8 y}{3}\))\(\frac{7}{4}\)
Solution:
i) General term in the expansion of (1 + x)-n is
Tr+1 = (- 1)r . n+r-1Cr . xr
∴ 6th term of (1 + x)-5 is
T6 = (- 1)5 . 9C5 (\(\frac{x}{2}\))-5 (∵ n = r = 5)
= \(\frac{-63}{16}\) x5.

ii) General term in the expansion of (1 + x)-n is
Tr+1 = (- 1)r . n+r-1Cr . xr
7th term in the eapansion of (1 – \(\frac{x^2}{3}\))-4 is
T7 = \({ }^9 C_6\left(\frac{x^2}{3}\right)^6\) (∵ n – 4; r = 6)
= \(\frac{28}{243}\) x12

iii) We know that
(3 – 4x)\(\frac{-2}{3}\) = \(3^{\frac{-2}{3}}\left(1-\frac{4 x}{3}\right)^{\frac{-2}{3}}\)

TS Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(b) 1

iv) We know that (7 + \(\frac{8 y}{3}\))\(\frac{7}{4}\)

TS Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(b) 2

TS Board Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(b)

Question 3.
Write down the first 3 terms In the expansion of
i) (3 +5x)\(\frac{-7}{3}\)
ii) (1 + 4x)-4
iii) (8 – 5x)\(\frac{2}{3}\)
iv) (2 – 7x)\(\frac{-3}{4}\)
Solution:
i) \((3+5 x)^{\frac{-7}{3}}=3^{\frac{-7}{3}}\left(1+\frac{5 x}{3}\right)^{\frac{-7}{3}}\)
We know that

TS Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(b) 3

ii) We know that
(1 + x)-n = 1 – nx + \(\frac{\mathrm{n}(\mathrm{n}+1)}{2 !}\) x2 – ………….
∴ (1 + 4x)-4 = 1 – 4 (4x) + \(\frac{4(5)}{2}\) (4x)2 – …………..
= 1 – 16x + 160x2 – ……………
∴ The first 3 terms are 1, – 16x, 160x2.

iii) (8 – 5x)\(\frac{2}{3}\)

TS Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(b) 4

iv) \((2-7 x)^{\frac{-3}{4}}=2^{\frac{-3}{4}}\left[1-\frac{7 x}{2}\right]^{\frac{-3}{4}}\)
We know that

TS Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(b) 5

TS Board Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(b)

Question 4.
Find the general term ((r + i)th term) in the expansion of
i) (4 + 5x)\(\frac{-3}{2}\)
ii) (1 – \(\frac{5 x}{3}\))-3
iii) (1 + \(\frac{4 x}{5}\))\(\frac{5}{2}\)
iv) (3 – \(\frac{5 x}{4}\))\(\frac{-1}{2}\)
Solution:
i) We know that

TS Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(b) 6

ii) General term in the expansion (1 – x)-n is

TS Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(b) 7

iii) General term in the expansion of (1 + X)p/q is

TS Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(b) 8

iv) We know that

TS Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(b) 9

TS Board Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(b)

II.
Question 1.
Find the coefficient of x10 in the expansion of \(\frac{1+2 x}{(1-2 x)^2}\).
Solution:
We know that
\(\frac{1+2 x}{(1-2 x)^2}\) = (1 + 2x) (1 – 2x)-2
= (1 + 2x) [1 + 2(2x) + 3 (2x)2 + 4 (2x)3 + …………. + 10 (2x)9 + 11 (2x)10 + …………]
∴ The coefficient of x10 in \(\frac{1+2 x}{(1-2 x)^2}\) is
= 11 (210) + 10 (2) (29)
= 210 (11 + 10)
= 21 × 210.

Question 2.
Find the coefficient of x4 in the expansion of (1 – 4x)-3/5.
Solution:
General term in the expansion of (1 – X)\(\frac{-p}{q}\) is

TS Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(b) 10

Question 3.
i) Find the coefficient of x5
ii) Find the coefficient of x8 in (1 nt)2
iii) Find the coefficient of x in (2+3×9
Solution:
i) We know that
\(\frac{(1-3 x)^2}{(3-x)^{\frac{3}{2}}}=(1-3 x)^2\left[3-\left.x\right|^{\frac{-3}{2}}\right.\)

TS Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(b) 11

TS Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(b) 12

TS Board Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(b)

ii) We know that \(\frac{(1+x)^2}{\left(1-\frac{2}{3} x\right)^3}=(1+x)^2\left[1-\frac{2}{3} x\right]^{-3}\)

TS Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(b) 13

iii) We know that
\(\frac{(2+3 x)^3}{(1-3 x)^4}\) = (2 + 3x)3 [1 – 3x]-4
= (8 + 36x + 54x2 + 27x3) [1 + 4C (3x) + 5C2 (3x)2 + 6C3 (3x)3 + ……….. + (r+3)Cr (3x)r + …………..]
Clearly the coefficient of xr in above expansion is
8 (r+3)Cr 3r + 36 (r+2)Cr-1 3r-1 + 54 (r+1)Cr-2 3r-2 + 27 rCr-3 3r-2
for coefficient of x7, put r = 7
∴The coefficient of x7 in \(\frac{(2 x+3)^3}{(1-3 x)^4}\) is
8 10C7 37 + 36 9C6 36 + 54 8C5 35 + 27 7C4 34 = 8 10C3 37 + 36 9C3 36 + 54 8C3 35 + 27 7C3 34.

TS Board Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(b)

Question 4.
Find the coefficient of x3 in the expansion of \(\frac{\left(1+3 x^2\right)^{\frac{3}{2}}}{(3+4 x)^{\frac{1}{3}}}\).
Solution:
\(\frac{\left(1+3 x^2\right)^{\frac{3}{2}}}{(3+4 x)^{\frac{1}{3}}}\) = (1 + 3x)3/2 (3 + 4x)-1/3

TS Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(b) 14

TS Board Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(b)

III.
Question 1.
Find the sum of the infinite series:
i) 1 + \(\frac{1}{3}+\frac{1 \cdot 3}{3 \cdot 6}+\frac{1 \cdot 3 \cdot 5}{3 \cdot 6 \cdot 9}\) + ………….
ii) 1 – \(\frac{4}{5}+\frac{4 \cdot 7}{5 \cdot 10}-\frac{4 \cdot 7 \cdot 10}{5 \cdot 10 \cdot 15}\) + ………….
iii) \(\frac{3}{4}+\frac{3 \cdot 5}{4 \cdot 8}+\frac{3 \cdot 5 \cdot 7}{4 \cdot 8 \cdot 12}\) + …………
iv) \(\frac{3}{4 \cdot 8}-\frac{3 \cdot 5}{4 \cdot 8 \cdot 12}+\frac{3 \cdot 5 \cdot 7}{4 \cdot 8 \cdot 12 \cdot 16}\) – …………..
Solution:
i) Let y = 1 + \(\frac{1}{3}+\frac{1 \cdot 3}{3 \cdot 6}+\frac{1 \cdot 3 \cdot 5}{3 \cdot 6 \cdot 9}\) + ………….

TS Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(b) 15

ii) Let y = – \(\frac{4}{5}+\frac{4 \cdot 7}{5 \cdot 10}-\frac{4 \cdot 7 \cdot 10}{5 \cdot 10 \cdot 15}\) + ………….

TS Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(b) 16

TS Board Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(b)

iii) \(\frac{3}{4}+\frac{3 \cdot 5}{4 \cdot 8}+\frac{3 \cdot 5 \cdot 7}{4 \cdot 8 \cdot 12}\) + …………

TS Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(b) 17

iv) Let y = \(\frac{3}{4 \cdot 8}-\frac{3 \cdot 5}{4 \cdot 8 \cdot 12}+\frac{3 \cdot 5 \cdot 7}{4 \cdot 8 \cdot 12 \cdot 16}\) – …………..

TS Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(b) 18

TS Board Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(b)

Question 2.
If t = \(\frac{4}{5}+\frac{4 \cdot 6}{5 \cdot 10}+\frac{4 \cdot 6 \cdot 8}{5 \cdot 10 \cdot 15}\) + ……….. ∞, then prove that 9t = 16.
Solution:
Given
t = \(\frac{4}{5}+\frac{4 \cdot 6}{5 \cdot 10}+\frac{4 \cdot 6 \cdot 8}{5 \cdot 10 \cdot 15}\) + ……….. ∞

TS Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(b) 19

Question 3.
If x = \(\frac{1 \cdot 3}{3 \cdot 6}+\frac{1 \cdot 3 \cdot 5}{3 \cdot 6 \cdot 9}+\frac{1 \cdot 3 \cdot 5 \cdot 7}{3 \cdot 6 \cdot 9 \cdot 12}+\ldots\) then prove that 9x2 + 24x = 11.
Solution:
Given x = \(\frac{1 \cdot 3}{3 \cdot 6}+\frac{1 \cdot 3 \cdot 5}{3 \cdot 6 \cdot 9}+\frac{1 \cdot 3 \cdot 5 \cdot 7}{3 \cdot 6 \cdot 9 \cdot 12}+\ldots\)

TS Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(b) 20

⇒ \(\frac{4}{3}\) + x = √3
⇒ 4 + 3x = 3√3
⇒ 16 + 9x2 + 24x = 27
⇒ 9x2 + 24x = 11.

TS Board Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(b)

Question 4.
If x = \(\frac{5}{(2 !) \cdot 3}+\frac{5 \cdot 7}{(3 !) \cdot 3^2}+\frac{5 \cdot 7 \cdot 9}{(4 !) \cdot 3^3}+\ldots \ldots\) then find the value of x2 + 4x.
Solution:
x = \(\frac{5}{(2 !) \cdot 3}+\frac{5 \cdot 7}{(3 !) \cdot 3^2}+\frac{5 \cdot 7 \cdot 9}{(4 !) \cdot 3^3}+\ldots \ldots\)

TS Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(b) 21

∴ (1) ⇒ 2 + x = \(\left(1-\frac{2}{3}\right)^{-\frac{3}{2}}\)
⇒ 2 + x = \(\left(\frac{1}{3}\right)^{\frac{-3}{2}}\)
⇒ 2 + x = √27
⇒ x2 + 4x = 23.

Question 5.
Find the sum to infinite terms of the series \(\frac{7}{5}\left(1+\frac{1}{10^2}+\frac{1 \cdot 3}{1 \cdot 2} \cdot \frac{1}{10^4}+\frac{1 \cdot 3 \cdot 5}{1 \cdot 2 \cdot 3} \cdot \frac{1}{10^6}+\ldots\right)\)
Solution:
Let y = \(\frac{7}{5}\left(1+\frac{1}{10^2}+\frac{1 \cdot 3}{1 \cdot 2} \cdot \frac{1}{10^4}+\frac{1 \cdot 3 \cdot 5}{1 \cdot 2 \cdot 3} \cdot \frac{1}{10^6}+\ldots\right)\)

TS Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(b) 22

TS Board Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(b)

Question 6.
Show that for any non-zero rational number x.
\(1+\frac{x}{2}+\frac{x(x-1)}{2 \cdot 4}+\frac{x(x-1)(x-2)}{2 \cdot 4 \cdot 6}+\ldots .\) = \(1+\frac{x}{3}+\frac{x(x+1)}{3 \cdot 6}+\frac{x(x+1)(x+2)}{3 \cdot 6 \cdot 9}+\ldots .\)
Solution:
L.H.S = \(1+\frac{x}{2}+\frac{x(x-1)}{2 \cdot 4}+\frac{x(x-1)(x-2)}{2 \cdot 4 \cdot 6}+\ldots .\)

TS Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(b) 23

TS 10th Class Physical Science Solutions Chapter 2 Chemical Equations

Telangana SCERT TS 10th Class Physical Science Study Material Pdf 2nd Lesson Chemical Equations Textbook Questions and Answers.

TS 10th Class Physical Science 2nd Lesson Questions and Answers Chemical Equations

Improve Your Learning
I. Reflections on concepts

Question 1.
What information do you get from a balanced chemical equation?
Answer:

  1. A chemical equation gives information about the reactants and products by means of their symbols and formulae.
  2. It gives the ratio of molecules of reactants and products.
  3. It gives the relative masses of reactants and products.
  4. If the masses are expressed in grams, then the equation also gives the molar ratios of reactants and products.
  5. We can calculate the volumes of gases liberated at given condition of temperature and pressure using molar mass and molar volume relationship.
  6. Using molar mass and Avagadro’s number we can calculate the number of molecules and atoms of different substances.

Question 2.
Why should we balance a chemical equation?
Answer:
Chemical reactions obeys law of conservation of mass. So, the total number of atoms of each element in the reactants must be equal to the total number of atoms of each element in the products. So we should have to balance chemical equation.

TS 10th Class Physical Science Solutions Chapter 2 Chemical Equations

Question 3.
Balance the following chemical equations.
(a) NaOH + H2SO4 → Na2SO4 + H2O
Answer:
Step 1: Write unbalanced equation
NaOH + H2SO4 → Na2SO4 + H2O

Step 2: Compare number of atoms of each element on both sides. Add the suitable coefficients to balance equation.
2NaOH + H2SO4 → Na2SO4 + 2H2O

Step 3 : Make sure the coefficients are reduced to their smallest whole number values.
2NaOH + H2SO4 → Na2SO4 + 2H2O

Step 4: Check the answer
TS 10th Class Physical Science Solutions Chapter 2 Chemical Equations 1

(b) KClO3 → KCl + O2
Answer:
Step 1: Write unbalanced equation.
KClO3 → KCl +O2

Step 2: Add suitable coefficients to balance equation on both sides.
2KClO3 → 2KCl +3O2

Step 3: Make sure the coefficients are reduced to their smallest whole number values.
2KClO3 → 2KCl +3O2

Step 4: Check the answer.
TS 10th Class Physical Science Solutions Chapter 2 Chemical Equations 2
(c) Hg (NO3)2 + Kl → Hg I2 + KNO3
Answer:
Step 1: Write unbalanced equation
Hg(NO3)2+KI → HgI2 +KNO3

Step 2: Add suitable coefficients to balance equation on both sides
Hg(NO3)2 +2KI → HgI2 +2KNO3

Step 3: Make sure the coefficients are reduced to their smallest whoíe number values.
Hg(NO3 )2 +2 Kl → Hg I2 + 2KNO3

Step 4: Check the answer
TS 10th Class Physical Science Solutions Chapter 2 Chemical Equations 3

Question 4.
Mention the physical states of the reactants and products of the following chemical reactions and balance the equations.
(a) C6H12O6 → C2H5OH + CO2
Answer:
C6H12O6(aq) → 2C2H5OH(aq)+2CO2(g)

(b) NH3 + Cl2 → N2 + NH4Cl(s)
Answer:
8NH3(g) +3Cl2(g) → N2(g) +6NH4Cl

(c) Na+H2O → NaOH+H2
Answer:
2Na(s)+2H2O (l) → 2NaOH (aq)+H2(g)

TS 10th Class Physical Science Solutions Chapter 2 Chemical Equations

II. Application Of Concepts

Question 1.
Balance the following chemical equations after writing the symbolic representation.
(a) Calcium hydroxide (s) + Nitric acid (aq) → Water(l) + Calcium nitrate(aq)
(b) Magnesium (s) + Iodine(s) → Magnesium Iodide (s)
Answer:
(a) Calcium hydroxide (s) + Nitric acid (aq) → Water(l) + Calcium nitrate(aq)
Answer:
Ca(OH)2(sol) + 2HNO3(so1) → 2H2O(l) + Ca(NO3)2(aq) Double displacement reaction

(b) Magnesium (s) + Iodine(s) → Magnesium Iodide (s)
Answer:
Mg(s) + I2(s) → MgI2(s)
Chemical combination reaction.

Question 2.
Write the following chemical reactions including the physical states of the substances and balance chemical equations
(a) Sodium hydroxide reacts with hydrochloric acid to form sodium chloride and water.
Answer:
(a) Sodium hydroxide reacts with hydrochloric acid to form sodium chloride and water.
Answer:
NaOH (aq) + HCl (aq) ) → NaCl (aq) +H2O (l)

(b) Barium chloride reacts with liquid sodium sulphate to leave Barium sulphate as a precipitate and also form liquid sodium chloride.
Answer:
BaCl2(aq)+ Na2SO4(aq) → BaSO4 ↓ +2 NaCl(aq )

Question 3.
Potassium nitrate and Sodium nitrate reacts separately with copper sulphate solution. Write balanced chemical equations for the above reactions.
Answer:

  1. Potassium nitrate reacts with copper sulphate to form potassium sulphate and copper nitrate.
    2KNO3 + CuSO4 → K2SO4 + Cu(NO3)2
    2. Sodium nitrate reacts with copper sulphate to form sodium sulphate and copper nitrate.
    2NaNO3 + CuSO4 → Na2SO4 + Cu(NO3)2

Higher Order Thinking Questions

Question 1.
2 moles of zinc reacts with a cupric chloride solution containing 6.023 × 1022 formula units of CuC12. Calculate the moles of copper obtained.
Zn(s) + CuCl2(aq) → ZnCl2(aq) + Cu (s)
Answer:
Given equation i.s
Zn(s) + CuCl2(aq) → ZnCl2(aq) + Cu(s)
1 mol + 1 mole → 1 mole + mole
From the above equation it is clear that 1 mole of zinc react with 1 mole of CuCl2 solution to give 1 mole of copper.
2 moIes of Zn required 2 moles (12.046 × 1022 formula units) of CuCl2, But only 1 mole (6.023 × 1022 formula units) of CuCl2 is available.
So, the No. of moles of copper obtained depends on amount of CuCl2 present.
∴6.023 × 1022 formula units (1 mole) of CuCl2 products 1 mole of copper.

Question 2.
1 mole of propane (C3H8) on combustion at STP gives A’ kilo Joules of heat energy. Calculate the heat liberated when 2.4 ltrs of propane on combustion at STP.
Answer:
The chemical equation for combination of propane is
C3H8 + 5O2 → 3CO2 + 4H2O + A (heat energy)
1 mole of propane gives ‘A kilojoules of heat energy.
(1 mole of any gas occupies 22.4 litres at SW)
i.e., 22.4 ltrs. of propane gives ‘A’ kilojoules of heat energy
⇒ 2.25 ltrs. of propane gives \(\frac{2.24}{22.4} \) x A = \(\frac{1}{10}\) A (0.1 A) heat energy.

TS 10th Class Physical Science Solutions Chapter 2 Chemical Equations

Question 3.
Calculate the mass and volume of Oxygen required at STP to convert 2.4 kg of graphite Into carbon dioxide.
Answer:
The chemical equation is
C + O2 → CO2 (∴ graphite is also carbon)
12 gm + 32 gm → 44 gm
1 mole + 1 mole → 1 mole
From the above equation
12 gm of Graphite requires 32 gm of oxygen
TS 10th Class Physical Science Solutions Chapter 2 Chemical Equations 4
1 mole of oxygen occupies 22.4 litres
200 moles Is oxygen occupIes 22.4 x 200 4480 litres.

TS 10th Class Physical Science Chemical Equations Intext Questions

Page 20

Question 1.
How do we know a chemical reaction has taken place?
Answer:
a. The original substances lose their characteristic properties. Hence these may be products with different physical states and colours permanently.

b. Chemical changes may be exothermic or endothermic i.e. they may involve liberation of heat energy or absorption of heat energy.

c. They may form an insoluble substance known as precipitate.

d. There may be gas liberation in a chemical change.

Page 21
Activity 1.
Take about 1 g of quick lime (calcium oxide) in a beaker. Add 10 ml of water to this.
Touch the beaker with your finger.

Question 2.
What do you notice?
Answer:
We notice that the beaker is hot when we touch it. The reason is that the calcium oxide (quick lime) reacts with water and in the process heat energy Is released. Calcium oxide dissolves in water producing colourless solution of Ca(OH)2.

TS 10th Class Physical Science Solutions Chapter 2 Chemical Equations

Question 3.
What is the nature of the solution?
Answer:
This solution Is a basic solution because a red litmus paper turns blue when dipped in the above solution

Activity 2
Take about 100 ml of water in a beaker and dissolve a small quantity of sodium sulphate (Na2SO4).
Take about 100ml of water In another beaker and dissolve a small quantity of barium chloride (BaCl2) observe the colours of the solutions obtained.

Question 4.
What are the colours of the above solutions?
Answer:
Colourless solutions

Question 5.
Can you name the solutions obtained?
Answer:
Sodium sulphate solution and barium chloride solution

Question 6.
Add Na2SO4 solution to BaCl2 solution and observe. Do you observe any change on mixing these solutions?
Answer:
A white precipitate of barium sulphate is formed

Activity 3
Take a few zinc granules in a conical flask. Add about 5 ml of dilute hydrochloric acid to the conical flask.

Question 7.
What changes do you notice?
Answer:
Effervescence Is observed

Question 8.
Keep a burning match stick near the mouth of the conical flask. with happens to burning match stick?
Answer:
It is put out. A pop sound is also heard.

Question 9.
Touch the bottom of the conical flask with your fingers. What do you notice?
Answer:
It Is hot

Question 10.
Is there any change in temperature?
Answer:
Temperature increases.

TS 10th Class Physical Science Solutions Chapter 2 Chemical Equations

Page 22

Question 11.
Can you write a chemical reaction in any other shorter way other than the way we disused above?
Answer:
The reaction of calcium oxide with water can be written using symbols of elements.
CaO + H2O → Ca(OH)2

Page 23
Question 12.
Is the number of atoms of each element are equal on both sides?
Answer:
The number of atoms of each element on both sides is equal.

Question 13.
Sodium sulphate reacts with barium chloride to give white precipitate, barium sulphate.
Na2SO4 + BaCl2 → BaSO4 ↓ + NaCl

Question 14.
Do the atoms of each element on left side equal to the atoms of the elements on the right side of the equation?
Answer:
No, the sodium and chlorine atoms are not balanced.

Page 25

Question 15.
Is it a balanced equation as per rules?
Answer:
Yes. But the coefficients are not the smallest numbers.

Question 16.
How do you say?
Answer:
Though the equation is balanced, the coefficients are not the smallest whole numbers. It would be necessary to divide all coefficients of equation by 2 to reach the final equation.
C3H8 + 5O2 → 3CO2 + 4H2O

Think And Discuss
You have brushed the wall with an aqueous suspension of Ca(OH)2. After two days the wall turned to white colour.

Question 1.
What are the steps involved in whitewashing of walls? (P.No. 23)
Answer:
A solution of slaked lime [Ca(OH)2] is prepared by adding water to quick lime [CaoO]. When Ca(OH)2 is applied to the wall ¡t reacts with carbon dioxide in air to form a thin layer of calcium carbonate giving a shiny finish to the walls.

TS 10th Class Physical Science Solutions Chapter 2 Chemical Equations

Question 2.
Write the balanced chemical reactions using the appropriate symbols. (P.No. 23)
Answer:
CaO(s) + H2O(l) → Ca(OH)2(Aq) + Q(heat energy)
TS 10th Class Physical Science Solutions Chapter 2 Chemical Equations 5

TS 10th Class Physical Science Chemical Equations Activities

Activity 1

Question 1.
How can you demonstrate action of quick lime with water? What is the nature of the product? (2 Marks)
Answer:

  1. Take about 1 g of quick lime (Calcium oxide) in a beaker. Add 10 ml of water to this. Touch the beaker with your fingers
  2. We will notice that the beaker is hot when we touch it. Hence that’s an exothermic reaction.
  3. The reason is that the Calcium oxide reacts with water and in that process heat energy is released.
  4. Calcium oxide dissolves In water producing colourless solution, This solution turns red litmus to blue. Hence the product is a base.
    CaO + H2O → Ca(OH)2 + Q

Activity 2

Question 2.
Explain the reaction between Sodium sulphate and Barium chloride.
Answer:

  1. luke about 100 ml of water in a beaker and dissolve a small quantity of sodium sulphate (Na2SO4)
  2. Take about 100 ml of water in another beaker and dissolve a small quantity of Barium chloride (BaCl2).
  3. These two solutions are colourless.
  4. Add Na2SO4 solution to BaCl2 solution and observe.
  5. Sodium sulphate solution on mixing with Barium chloride solution alarms a precipitate of Barium sulphate and also soluble Sodium chloride.
    Na2SO4(aq) + BaCl(aq) → BaSO4(s)↓ + 2NaCl(aq)
  6. This is a double displacement reaction.

TS 10th Class Physical Science Solutions Chapter 2 Chemical Equations 6

TS 10th Class Physical Science Solutions Chapter 2 Chemical Equations

Question 3.
Explain the reaction of Zinc with HCl and write a balanced equation.
(OR)
Write the required material and experimental procedure for the experiment, “Hydrochloric acid reacts with ‘Zn’pieces and liberates H2”.
Answer:
Material required :
(i) Conical flask
(ii) Zinc granules
(iii) HCl
(iv) Matchbox.

  1. Take a few zinc granules in a conical flask.
  2. Add about 5 ml of dilute Hydrochloric acid to the conical flask.
  3. We observe a gas evolving out.
  4. Now keep a burning match stick near the mouth of the conical flask.
  5. The match stick is put out with a ‘pop’ sound.
  6. I Touch the bottom of the conical flask. We feel hot. Hence it is an exothermic reaction.

TS 10th Class Physical Science Solutions Chapter 2 Chemical Equations 7
Zn(s) + 2HCl(l) → ZnCl2(s) + H2(g)

TS Inter 1st Year Maths 1A Hyperbolic Functions Important Questions

Students must practice these TS Inter 1st Year Maths 1A Important Questions Chapter 9 Hyperbolic Functions to help strengthen their preparations for exams.

TS Inter 1st Year Maths 1A Hyperbolic Functions Important Questions

I.

Question 1.
Show that \(t \tanh ^{-1}\left(\frac{1}{2}\right)=\frac{1}{2} \log _e^3\)
Solution:
TS Inter 1st Year Maths 1A Hyperbolic Functions Important Questions 1

Question 2.
If cosh \(x=\frac{5}{2}\) find the values of
(i) cosh (2x)
(ii) sinh (2x)
Solution:
TS Inter 1st Year Maths 1A Hyperbolic Functions Important Questions 2

TS Inter 1st Year Maths 1A Hyperbolic Functions Important Questions width=

Question 3.
Prove that for any x ∈ R sinh (3x) =3 sinh (3x) + 3sinh x +4sin3h3 x.
Solution:
LHS : sinh(3x) = sinh (2x + x)
= sinh (2x) cosh x + cosh (2x) sinh x
= (2sinh x cosh x) cosh x + (1 + 2sinh2 x) slnh x
= 2 sinh x cosh2 x + sinh x + 2sinh3 x
= 2sinh x (1 + sinh2 x) + sinh x + 2sinh3 x
= 3sinh x + 4sinh3 x = R.H.S.

Question 4.
Prove that for any x∈R
\(\tanh 3 x=\frac{3 \tanh x+\tanh ^3 x}{1+3 \tanh ^2 x}\)
Solution:
TS Inter 1st Year Maths 1A Hyperbolic Functions Important Questions 3

TS Inter 1st Year Maths 1A Hyperbolic Functions Important Questions width=

Question 5.
If cosh x = secθ then prove that \(\tanh ^2\left(\frac{x}{2}\right)=\tan ^2\left(\frac{\theta}{2}\right)\)
Solution:
TS Inter 1st Year Maths 1A Hyperbolic Functions Important Questions 4
Question 6.
sinhx = 5 then show that \(x=\log _e(5+\sqrt{26})\)
Solution:
TS Inter 1st Year Maths 1A Hyperbolic Functions Important Questions 5

TS Inter 1st Year Maths 1A Hyperbolic Functions Important Questions width=

II.

Question 1.
\(\theta \in\left(-\frac{\pi}{4}, \frac{\pi}{4}\right)\) and \(x=\log _{\mathrm{e}}\left(\cot \left(\frac{\pi}{4}+\theta\right)\right)\) that prove that
(i) cosh x = sec2θ
(ii) sin hx= – tan2θ.
Solution:
TS Inter 1st Year Maths 1A Hyperbolic Functions Important Questions 6
TS Inter 1st Year Maths 1A Hyperbolic Functions Important Questions 7

TS Inter 1st Year Maths 1A Hyperbolic Functions Important Questions width=

Question 2.
Draw the graph of y = sinh x.
Solution:
TS Inter 1st Year Maths 1A Hyperbolic Functions Important Questions 8

Question 3.
Draw the graph of y = cosh x.
Solution:
TS Inter 1st Year Maths 1A Hyperbolic Functions Important Questions 9

TS Inter 1st Year Maths 1A Trigonometric Equations Important Questions

Students must practice these TS Inter 1st Year Maths 1A Important Questions Chapter 7 Trigonometric Equations to help strengthen their preparations for exams.

TS Inter 1st Year Maths 1A Trigonometric Equations Important Questions

Question 1.
Solve 2cos2θ – \(\sqrt{3}\) sin θ+1 = 0
Solution:
TS Inter 1st Year Maths 1A Trigonometric Equations Important Questions 1

TS Inter 1st Year Maths 1A Trigonometric Equations Important Questions

Question 2.
Solve 1 +sin2θ = 3 sinθ cosθ
Solution:
Dividing by cos2 θ we get
⇒ sec2 θ + tan2 θ = 3 tanθ
⇒ 1 + 2 tan2 θ = 3 tanθ
⇒ 2tan2 θ – 3tan θ +1 = θ
⇒ 2 tan2 θ –  2tan θ –  tanθ + 1 = θ
⇒ (tanθ –  1)(2tan θ – 1) = θ
⇒ tanθ = 1 or tanθ = \(\frac{1}{2}\)
If tan θ = 1 then principal solution is α =\( \frac{\pi}{4}\)
General solution is θ = nπ + \(\frac{\pi}{4}\)  n ∈ Z
Let a be the principal solution of tan θ = \(\frac{1}{2}\)
Then the general solution is
θ = nπ+α,n∈ Z

Question 3.
Solve \(\sqrt{2}(\sin x+\cos x)=\sqrt{3}\)
Solution:
TS Inter 1st Year Maths 1A Trigonometric Equations Important Questions 3

Question 4.
Solve tanθ+3 cotθ = 5 secθ
Solution:
The given equation is
tanθ + 3 cotθ = 5 secθ
⇒ \(\frac{\sin \theta}{\cos \theta}+\frac{3 \cos \theta}{\sin \theta}=\frac{5}{\cos \theta}\)
⇒  sin2 θ + 3cos2θ= 5 sinθ
⇒  sin2 θ + 3(1 – sin2 θ) = 5 sinθ
⇒ sin2 θ + 3 – 3 sin2 θ = 5 sinθ
⇒ -2 sin θ – 5 sinθ + 3 = 0
⇒ 2 sin θ+ 5 sinθ – 3 = θ
⇒ 2 sin θ + 6 sinθ – sinθ – 3 = θ
⇒ 2 sin θ(sin θ + 3) –  1(sinθ + 3) = θ
⇒ (2sinθ – 1)(sinθ + 3)= θ
⇒ sin θ = \(\frac{1}{2}\) or sinθ = – 3.
sinθ =-3 is not admissible but for sinθ = \(\frac{1}{2}\) general solution is
θ=nπ+(-n)n ,n∈Z
Since the principal solution is α = \(\frac{1}{2}\)

TS Inter 1st Year Maths 1A Trigonometric Equations Important Questions

Question 5.
If acos2θ + bsin2θ =c has θ12 as its solutions then show that tan θ1 + tan θ1 = \(\frac{2 b}{c+a}\)
Solution:
Given equation is acos2θ + bsin2θ = c
TS Inter 1st Year Maths 1A Trigonometric Equations Important Questions 4
This is a quadratic equation in an θ.
Given θ12  are the solutions of θ.
tan θ1, tanθ2 are the roots of equation (1)
∴ Sum of the roots tan θ1 + tan θ2 = \(\frac{2 b}{a+c}\) and product of the roots
TS Inter 1st Year Maths 1A Trigonometric Equations Important Questions 5

Question 6.
Find all values of x in (-π, π) satisfying the equation g1+cosx+cos2 x +……………….. = 43
Solution:
TS Inter 1st Year Maths 1A Trigonometric Equations Important Questions 6

Question 7.
Solve sin x = \(\frac{1}{\sqrt{2}}\)
Solution:
Principal solution is α = \(\frac{\pi}{4}\)
General solution is
\(\mathrm{x}=\mathrm{n} \pi+(-1)^{\mathrm{n}} \frac{\pi}{4}, \mathrm{n} \in \mathrm{Z}\)

TS Inter 1st Year Maths 1A Trigonometric Equations Important Questions

Question 8.
Solve sin 2θ \(=\frac{\sqrt{5}-1}{4}\)
Solution:
The principal solution is α = 18°
∴ General solution is
TS Inter 1st Year Maths 1A Trigonometric Equations Important Questions 7
Question 9.
Solve tan2 θ = 3.
Solution:
TS Inter 1st Year Maths 1A Trigonometric Equations Important Questions 8
Question 10.
Solve 3cosec x = 4sinx.
Solution:
Given 3 cosec x = 4 sin x
TS Inter 1st Year Maths 1A Trigonometric Equations Important Questions 9

Question 11.
If x is acute and sin(x+10°) = cos(3x – 68°) find x in degrees.
Solution:
TS Inter 1st Year Maths 1A Trigonometric Equations Important Questions 10
When k = 0 we get x = 37°
If we take k = 1, 2 the value of x is not acute.
Hence the value of x is 37°.

TS Inter 1st Year Maths 1A Trigonometric Equations Important Questions

Question 12.
Solve cos 3θ = sin 2θ.
Solution:
TS Inter 1st Year Maths 1A Trigonometric Equations Important Questions 11

Question 13.
Solve 7 sin2θ+3cos2 θ= 4.
Solution:
Given 7 sin2θ+3cos2 θ= 4.
⇒ 7 sin 2θ + 3(1 -sin2 θ) = 4
⇒  4 sin 2θ = 1
⇒ sinθ = ± \(\frac{1}{2}\)
Principal solutions are \(\alpha=\pm \frac{\pi}{6}\) and general solution is given by general solution is given by
θ = nπ ± ,\(\alpha=\pm \frac{\pi}{6}\),n∈Z

Question 14.
Find general solution of θ which satisfies both the equations sinθ = \(-\frac{1}{2}\) and cosθ = \(-\frac{\sqrt{3}}{2}\)
Solution:
TS Inter 1st Year Maths 1A Trigonometric Equations Important Questions 12

Question 15.
Solve 4 sin x. sin 2x sin 4x = sin 3x
Solution:
Given sin 3x = 4 sinx sin 2x sin 4x
= 2sinx (2sin2x – sin4x)
⇒ 2 sin x (cos 2x – cos 6x)
⇒ sin3x = 2cos 2x sinx – 2 cos 6x sin x
⇒ sin 3x = sin 3x – sin x – 2 cos 6xsinx
⇒ 2cos 6x sinx + sinx= 0
⇒ sinx(2 cos 6x + 1)=0
⇒ sinx = 0 or cos 6x \(-\frac{1}{2}\)

Case (i): sin x = 0,  x= nπ, n∈Z is the general solution.
Case (ii) : cos 6x = \(-\frac{1}{2}\)
Principal solution is α \( =\frac{2 \pi}{3}\)
∴ General solution is 6x = 2πr ± \(\frac{2 \pi}{3}\)
\(x=\frac{n \pi}{3} \pm \frac{\pi}{9}, n \in Z\)

TS Inter 1st Year Maths 1A Trigonometric Equations Important Questions

Question 16.
If 0<θ<π solve cosθ cos2θ cos3θ = \(\frac{1}{4}\)
Solution:
TS Inter 1st Year Maths 1A Trigonometric Equations Important Questions 12
TS Inter 1st Year Maths 1A Trigonometric Equations Important Questions 13

Question 18.
Solve sin2x – cos2x = sinx – cosx
Solution:
The given equation can be written as
sin 2x – sin x = cos 2x – cos x
TS Inter 1st Year Maths 1A Trigonometric Equations Important Questions 14

TS Inter 1st Year Maths 1A Trigonometric Equations Important Questions

Question 19.
Solve cos 3x + cos 2x = sin + sin = \(\sin \frac{3 x}{2}+\sin \frac{x}{2} 0 \leq x \leq 2 \pi\)
Solution:
Given cos 3x + cos 2x = sin + sin = \(\sin \frac{3 x}{2}+\sin \frac{x}{2} 0 \leq x \leq 2 \pi\)
TS Inter 1st Year Maths 1A Trigonometric Equations Important Questions 15
TS Inter 1st Year Maths 1A Trigonometric Equations Important Questions 16

TS Inter 1st Year Maths 1A Inverse Trigonometric Functions Important Questions

Students must practice these TS Inter 1st Year Maths 1A Important Questions Chapter 8 Inverse Trigonometric Functions to help strengthen their preparations for exams.

TS Inter 1st Year Maths 1A Inverse Trigonometric Functions Important Questions

Question 1.
Prove that
TS Inter 1st Year Maths 1A Inverse Trigonometric Functions Important Questions 14
Solution:
TS Inter 1st Year Maths 1A Inverse Trigonometric Functions Important Questions 1

TS Inter 1st Year Maths 1A Inverse Trigonometric Functions Important Questions width=

Question 2.
Find the values of the following.

(i) \(\sin ^{-1}\left(-\frac{1}{2}\right)\)
Solution:
\(\sin ^{-1}\left(-\frac{1}{2}\right)=-\sin ^{-1}\left(\frac{1}{2}\right)=-\frac{\pi}{6}\)

(ii) \(\cos ^{-1}\left(-\frac{\sqrt{3}}{2}\right)\)
Solution:
TS Inter 1st Year Maths 1A Inverse Trigonometric Functions Important Questions 2

(iii) \(\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)\)
Solution:
\(\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)=\tan ^{-1}\left(\tan \frac{\pi}{6}\right)=\frac{\pi}{6}\)

TS Inter 1st Year Maths 1A Inverse Trigonometric Functions Important Questions width=

(iv) cot-1 (-1)
Solution:
TS Inter 1st Year Maths 1A Inverse Trigonometric Functions Important Questions 3

(v) sec -1  \((-\sqrt{2})\)
Solution:
TS Inter 1st Year Maths 1A Inverse Trigonometric Functions Important Questions 4

(vi) Cosec -1  \(\left(\frac{2}{\sqrt{3}}\right)\)
Solution:
TS Inter 1st Year Maths 1A Inverse Trigonometric Functions Important Questions 15

Question 3.
Find the values of the following.

(i) sin-1 \(\left(\sin \frac{4 \pi}{3}\right)\)
Solution:
TS Inter 1st Year Maths 1A Inverse Trigonometric Functions Important Questions 6

(ii) \(\tan ^{-1}\left(\tan \frac{4 \pi}{3}\right)\)
Solution:
TS Inter 1st Year Maths 1A Inverse Trigonometric Functions Important Questions 7

TS Inter 1st Year Maths 1A Inverse Trigonometric Functions Important Questions width=

Question 4.
Find the values of the following.

(i) \(\sin \left(\cos ^{-1} \frac{5}{13}\right)\)
Solution:
TS Inter 1st Year Maths 1A Inverse Trigonometric Functions Important Questions 8

(ii) \(\tan \left(\sec ^{-1} \frac{25}{7}\right)\)
Solution:
TS Inter 1st Year Maths 1A Inverse Trigonometric Functions Important Questions 9

(iii) \(\cos \left(\tan ^{-1} \frac{24}{7}\right)\)
Solution:
TS Inter 1st Year Maths 1A Inverse Trigonometric Functions Important Questions 10

Question 5.
Find the values of the following.

(i) \(\sin ^2\left(\tan ^{-1} \frac{3}{4}\right)\)
Solution:
TS Inter 1st Year Maths 1A Inverse Trigonometric Functions Important Questions 11

(ii) \(\sin \left(\frac{\pi}{2}-\sin ^{-1}\left(-\frac{4}{5}\right)\right)\)
Solution:
TS Inter 1st Year Maths 1A Inverse Trigonometric Functions Important Questions 12

(iii) \(\cos \left(\cos ^{-1}\left(-\frac{2}{3}\right)-\sin ^{-1}\left(\frac{2}{3}\right)\right)\)
Solution:
TS Inter 1st Year Maths 1A Inverse Trigonometric Functions Important Questions 13

TS Inter 1st Year Maths 1A Inverse Trigonometric Functions Important Questions width=

(iv) sec2 (cot-1 3) + cosec2 (tan-1 2)
Solution:
TS Inter 1st Year Maths 1A Inverse Trigonometric Functions Important Questions 16

Question 6.
Find the value of \(\cot ^{-1}\left(\frac{1}{2}\right)+\cot ^{-1}\left(\frac{1}{3}\right)\)
Solution:
TS Inter 1st Year Maths 1A Inverse Trigonometric Functions Important Questions 17

Question 7.
Prove that
\(\sin ^{-1}\left(\frac{4}{5}\right)+\sin ^{-1}\left(\frac{7}{25}\right)=\sin ^{-1}\left(\frac{117}{125}\right)\)
Solution:
TS Inter 1st Year Maths 1A Inverse Trigonometric Functions Important Questions 18

TS Inter 1st Year Maths 1A Inverse Trigonometric Functions Important Questions width=

Question 8.
If x ∈(-1, 1) prove that 2 tan-1 x = \(\tan ^{-1}\left(\frac{2 x}{1-x^2}\right)\)
Solution:
Given x ∈ (-1,1) and it tan-1 x = a then tan α = x and
TS Inter 1st Year Maths 1A Inverse Trigonometric Functions Important Questions 19

Question 9.
Prove that \(\sin ^{-1}\left(\frac{4}{5}\right)+\sin ^{-1}\left(\frac{5}{13}\right) +\sin ^{-1}\left(\frac{16}{65}\right)=\frac{\pi}{2}\)
Solution:
TS Inter 1st Year Maths 1A Inverse Trigonometric Functions Important Questions 20

Question 10.
Prove that cot-1 9+ cosec-1 \( \frac{\sqrt{41}}{4}=\frac{\pi}{4}\)
Solution:
TS Inter 1st Year Maths 1A Inverse Trigonometric Functions Important Questions 21

TS Inter 1st Year Maths 1A Inverse Trigonometric Functions Important Questions width=

Question 11.
Show that cot \(\begin{aligned} \cot \left(\operatorname{Sin}^{-1} \sqrt{\frac{13}{17}}\right) \\ = \sin \left(\operatorname{Tan}^{-1}\left(\frac{2}{3}\right)\right) \end{aligned}\)
Solution:
TS Inter 1st Year Maths 1A Inverse Trigonometric Functions Important Questions 22

Question 12.
Find the value of \(tan \left[2 \operatorname{Tan}^{-1}\left(\frac{1}{5}\right)-\frac{\pi}{4}\right]\)
Solution:
TS Inter 1st Year Maths 1A Inverse Trigonometric Functions Important Questions 23

Question 13.
Prove that \(\operatorname{Sin}^{-1}\left(\frac{4}{5}\right)+2 \operatorname{Tan}^{-1}\left(\frac{1}{3}\right)=\frac{\pi}{2}\)
Solution:
TS Inter 1st Year Maths 1A Inverse Trigonometric Functions Important Questions 24

TS Inter 1st Year Maths 1A Inverse Trigonometric Functions Important Questions width=

Question 14.
If sin-1 x + sin-1 y + sin-1 z = π, then prove that x4 + y4 + z4 + 4x2y2z2 = 2 (x2y2 + y2z2 + z2x2)
Solution:
Let sin-1 x = A, sin-1 y = B and sin-1z = C
then A+B+C = π …………………..(1)
and sinA = x, sin B = y, sin C = z
Now A+B = π – c
TS Inter 1st Year Maths 1A Inverse Trigonometric Functions Important Questions 25

Question 15.
If \(\operatorname{Cos}^{-1}\left(\frac{p}{a}\right)+\operatorname{Cos}^{-1}\left(\frac{q}{b}\right)\) =α the prove that
TS Inter 1st Year Maths 1A Inverse Trigonometric Functions Important Questions 36
Solution:
TS Inter 1st Year Maths 1A Inverse Trigonometric Functions Important Questions 26
TS Inter 1st Year Maths 1A Inverse Trigonometric Functions Important Questions 27

TS Inter 1st Year Maths 1A Inverse Trigonometric Functions Important Questions width=

Question 16.
Solve \(\sin ^{-1}\left(\frac{5}{x}\right)+\sin ^{-1}\left(\frac{12}{x}\right)=\frac{\pi}{2},(x>0)\)
Solution:
TS Inter 1st Year Maths 1A Inverse Trigonometric Functions Important Questions 28

Question 17.
Solve
TS Inter 1st Year Maths 1A Inverse Trigonometric Functions Important Questions 35
Solution:
TS Inter 1st Year Maths 1A Inverse Trigonometric Functions Important Questions 30

Question 18.
Solve \(\operatorname{Sin}^{-1} x+\operatorname{Sin}^{-1} 2 x=\frac{\pi}{3}\)
Solution:
TS Inter 1st Year Maths 1A Inverse Trigonometric Functions Important Questions 31
when \(x=-\frac{\sqrt{3}}{2 \sqrt{7}}\) value is not admissible
Since sin-1 x and sin-1 2x are negative
Hence \(x=-\frac{\sqrt{3}}{2 \sqrt{7}}\)

TS Inter 1st Year Maths 1A Inverse Trigonometric Functions Important Questions width=

Question 19.
If sin [2 Cos-1 (cot (2 Tan-1x)}] = 0 find x.
Solution:
TS Inter 1st Year Maths 1A Inverse Trigonometric Functions Important Questions 32

Question 20.
Prove that
TS Inter 1st Year Maths 1A Inverse Trigonometric Functions Important Questions 34
Solution:
Let cot-1 x=θ then cot θ = x and θ <x<π
∴ sin (cot-1x) = sinθ = \(\frac{1}{\operatorname{cosec} \theta}\)
TS Inter 1st Year Maths 1A Inverse Trigonometric Functions Important Questions 33

TS Inter 1st Year Maths 1A Inverse Trigonometric Functions Important Questions width=

Question 21.
Show that sec2 (tan-1) + cosec2 (cot-1 2) = 10.
Solution:
[1 + tan2 (tan-1(2)] + [1+ cot2 (cot-1(2))]
= 1 + 4 + 1 + 4 = 10

TS 10th Class Maths Solutions Chapter 9 Tangents and Secants to a Circle InText Questions

Students can practice 10th Class Maths Study Material Telangana Chapter 9 Tangents and Secants to a Circle InText Questions to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 9 Tangents and Secants to a Circle InText Questions

Do This

TS 10th Class Maths Solutions Chapter 9 Tangents and Secants to a Circle InText Questions 1

Question 1.
Draw a circle with any radius. Draw four tangents at different points. How many tangents can you draw to this circle ? (AS3, AS5) (Page No. 226)
Solution:
TS 10th Class Maths Solutions Chapter 9 Tangents and Secants to a Circle InText Questions 2
Let ‘O’ be the centre of the circle with radius ‘OA’. l, m, n, p and q be the tangents to the circle at A, B, C, D and E. We can draw a tangent at each point on the circle, i.e., infintely many tangents can be drawn to a circle.

Question 2.
How many tangents you can draw to circle from a point away from it ? (AS3) (Page No. 226)
Solution:
TS 10th Class Maths Solutions Chapter 9 Tangents and Secants to a Circle InText Questions 3
We can draw only two tangents from an exterior point.

TS 10th Class Maths Solutions Chapter 9 Tangents and Secants to a Circle InText Questions

Question 3.
In the below figure which are tangents to the circles ? (AS3) (Page No. 226)
Solution:
TS 10th Class Maths Solutions Chapter 9 Tangents and Secants to a Circle InText Questions 4
P and M are the tangents to the circles.

Question 4.
Draw a circle and a secant PQ of the circle on a paper as shown below. Draw various lines parallel to the secant on both sides of it. What happens to the length of chord coming closer and closer to the center of the circle ?
What is the longest chord ? How many tangents can you draw to a circle, which are parallel to each other. (Page No. 227)
Solution:
TS 10th Class Maths Solutions Chapter 9 Tangents and Secants to a Circle InText Questions 5
The length of the chord increases as it comes closer to the centre of the circle.

i) What is the longest chord ?  (Page No. 227)
Solution:
Diameter is the longest of all chords.

TS 10th Class Maths Solutions Chapter 9 Tangents and Secants to a Circle InText Questions

ii) How many tangents can you draw to a circle which are parallel to each other ? (Page No. 227)
Solution:
Only one tangent can be drawn parallel to a given tangent. To a circle, we can draw infinetely pairs of parallel tangents.

Try This

Question 1.
How can you prove the converse of the above theorem. “If a line in the plane of a circle is perpendicular to the radius at its end point on the circle, then the line is tangent to the circle”. (Page No. 228)
Solution:
Given : Circle with centre ‘O’, a point ‘A’ on the circle and the line AT’ perpendicular to OA’.
RTP : ‘AT’ is a tangent to the circle at A.
Construction :
TS 10th Class Maths Solutions Chapter 9 Tangents and Secants to a Circle InText Questions 6
Suppose ‘AT’ is not a tangent then ‘AT’ produced either way if necessary, will meet the circle again. Let it do so at R join OR

Proof : Since OA = OP (radii)
∴ ∠OAP = ∠OPA
But ∠OPA = 90°
∴ Two angles of a triangle are right angles which is impossible.
∴ Our supposition is false, hence AT is a tangent.

We can find some more results using the above theorem.

  1. Since there can be only one perpendicular OP at the point R it follows that one and only are tangent can be drawn to a circle at a given point on the circumference.
  2. Since there can be only one perpendicular to XY at the point R it follows that the perpendicular to a tangent at its point of contact passes through the centre.

TS 10th Class Maths Solutions Chapter 9 Tangents and Secants to a Circle InText Questions

Question 2.
How can you draw the tangent to a circle at a given point when the centre of the circle is not known ? (Page No. 229) Hint : Draw equal angles ∠QPX and ∠PRQ. Explain the construction.
Solution:
TS 10th Class Maths Solutions Chapter 9 Tangents and Secants to a Circle InText Questions 7
Steps of construction :

  1. Take a point ‘P’ and draw a chord PR through P
  2. Construct ∠PRQ and measure it.
  3. Construct ∠QPX at P equal to ∠PRQ.
  4. Extend PX on otherside. XY is the required tangent at R

Note : Angle between a tangent and chord is equal to angle in the alternate segment.

Try This

Question 1.
Use pythagorus theorem and write proof for the statement “the lengths of tangents drawn from an external point to a circle are equal”. (Page No. 231)
Solution:
Given : Two tangents PA and PB to a circle with centre ‘O’, from an exterior point P.
R.T.P : PA = PB
TS 10th Class Maths Solutions Chapter 9 Tangents and Secants to a Circle InText Questions 8
Proof : In ∆OAP; ∠OAP = 90°
∴ AP2 = OP2 – OA2 [∵ square of the hypotenuse is equal to the sum of squares on the other 2 sides → Pythagoras theorem]
⇒ (AP)2 = (OP)2 – (OB)2
[∵ OA = OB, radii of the same circle]
⇒ AP2 = BP2
[∴ In a ∆OBP; OB2 + BP2 = OP2
⇒ BP2 = OP2 – OB2]
⇒ AP2 = BP2
∴ PA = PB (CPCT)
Hence proved.

TS 10th Class Maths Solutions Chapter 9 Tangents and Secants to a Circle InText Questions

Question 2.
Draw a pair of radii OA and OB such that ∠BOA = 120°. Draw the bisector of ∠BOA and draw lines perpendicular to OA and OB at A and B. These lines meet on the bisector of ∠BOA at a point which is the external point and the perpendicular lines are the required tangents. Construct and justify. (Page No. 235)
Solution:
Justification :
OA ⊥ PA
OB ⊥ PB
Also in ∆OAP, ∆OBP
OA = OB
∠OAP = ∠OBP
OP = OP
∴ ∆OAP ≅ ∆OBP
∴ PA = PB [Q.E.D]
TS 10th Class Maths Solutions Chapter 9 Tangents and Secants to a Circle InText Questions 9

Do This

Question 1.
Shankar made the following pictures also
TS 10th Class Maths Solutions Chapter 9 Tangents and Secants to a Circle InText Questions 10
What shapes can they be broken into that we can find area easily ? (Page No. 237)
Solution:
TS 10th Class Maths Solutions Chapter 9 Tangents and Secants to a Circle InText Questions 11
Into two rectangles.

TS 10th Class Maths Solutions Chapter 9 Tangents and Secants to a Circle InText Questions 12
A rectangle and a circle

TS 10th Class Maths Solutions Chapter 9 Tangents and Secants to a Circle InText Questions 13
A cone and a secant

TS 10th Class Maths Solutions Chapter 9 Tangents and Secants to a Circle InText Questions 14
A rectangle and 2 segments

Make some more pictures and think of the shapes they can be divided into different parts.
Solution:
TS 10th Class Maths Solutions Chapter 9 Tangents and Secants to a Circle InText Questions 15
A cone and segment

TS 10th Class Maths Solutions Chapter 9 Tangents and Secants to a Circle InText Questions 16
A rectangle and a segment

TS 10th Class Maths Solutions Chapter 9 Tangents and Secants to a Circle InText Questions 17
A square and four segments

A cone and segment
A rectangle and a segment A square and four segments.

TS 10th Class Maths Solutions Chapter 9 Tangents and Secants to a Circle InText Questions

Do This

Question 1.
Find the area of sector, whose radius is 7 cm with the given angle : (Page No. 239)
(i) 60°
(ii) 30°
(iii) 72°
(iv) 90°
(v) 120°
Solution:
TS 10th Class Maths Solutions Chapter 9 Tangents and Secants to a Circle InText Questions 18

Question 2.
The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 10 minutes. (Page No. 239)
Solution:
Angle made by minute hand in 360°
1 min = \(\frac{360^{\circ}}{60^{\circ}}\) = 6°
Angle made by minute hand in
10 min = 10 × 6° = 60°
The area swept by minute hand is in the shape of a sector with radius.
r = 14 cm and angle x° = 60°
TS 10th Class Maths Solutions Chapter 9 Tangents and Secants to a Circle InText Questions 19
Area (A) = \(\frac{x^{\circ}}{360^{\circ}}\) × πr2
= \(\frac{60^{\circ}}{360^{\circ}}\) × \(\frac{22}{7}\) × 14 × 14
= \(\frac{616}{6}\) ⇒ 102.66 cm2
∴ Area swept by the minute hand in 10 minutes = 102.66 cm2

TS 10th Class Maths Solutions Chapter 9 Tangents and Secants to a Circle InText Questions

Try This

Question 1.
How can you find the area of major seg¬ment using are a of minor segment ?
Solution:
TS 10th Class Maths Solutions Chapter 9 Tangents and Secants to a Circle InText Questions 20
Area of the major segment = (Area of the circle) – (Area of the minor segment)

TS 10th Class Maths Solutions Chapter 9 Tangents and Secants to a Circle Optional Exercise

Students can practice 10th Class Maths Study Material Telangana Chapter 9 Tangents and Secants to a Circle Optional Exercise to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 9 Tangents and Secants to a Circle Optional Exercise

Question 1.
Prove that the angle between the two tan-gents drawn from an external point to a circle is supplementary to the angle subtended by the line segment joining the points of contact at the centre.
Solution:
PA and PB are tangents to the circle with centre ‘0’ from and external point P.
Join OA, OB and OR The angle between the tangents is ∠APB.
We have to prove that ∠AOB + ∠APB = 180°
PA is a tangent to the circle with centre ‘O’.
TS 10th Class Maths Solutions Chapter 9 Tangents and Secants to a Circle Optional Exercise 1
A is the point of contact. AO is the radius drawn through the point of contact i.e., A
∴ ∠PAO = 90°
PB is a tangent to the circle with centre ‘O’.
B is the point of contact. BO is the radius drawn through the point of contact i.e., B
∴ ∠PBO = 90°
OAPB is a quadrilateral.
The sum of the angles of a quadrilateral is 360°
∠APB + ∠PBO + ∠AOB + ∠PAO = 360°
∠APB + ∠AOB + ∠PAO + ∠PBO = 360°
∠APB + ∠AOB + 90° + 90° = 360°
∴ ∠APB + ∠AOB = 360° – 90° – 90°
= 360° – 180° = 180°
∠APB and ∠AOB are supplementary.

TS 10th Class Maths Solutions Chapter 9 Tangents and Secants to a Circle Optional Exercise

Question 2.
PQ is a chord of length 8cm of a circle of radius 5cm. The tangents at P and Q intersect at a point T (See figure). Find the length of TP.
Solution:
PQ is a chord of length 8cm of a circle with centre ‘O’.
Radius of the circle = 5 cm.
⇒ OP = OQ = 5cm
The tangents at P and Q meet at T. Join OT.
OT is the perpendicular bisector of PQ.
TS 10th Class Maths Solutions Chapter 9 Tangents and Secants to a Circle Optional Exercise 2
In triangle OPT, ∠OPT = 90°
∴ PT2 = OT2 – OP2 …………….. (1)
In ∆ PRO, ∠R = 90°
∴ OP2 = OR2 + PR2
⇒ OR2 = OP2 – PR2
= 52 – 42
= 25 – 16 = 9
∴ OR = \(\sqrt{9}\) = 3 cm
In the right triangle OPT,
PR is perpendicular to the hypotenuse OT.
∴ PR2 = TR. RO
42 = TR × 3 ⇒ TR = \(\frac{4^2}{3}\) = \(\frac{16}{3}\)
OT = OR + TR = \(\frac{3}{1}\) + \(\frac{16}{3}\) = \(\frac{25}{3}\)
From (1), PT2 = \(\left(\frac{25}{3}\right)^2\) – (5)2
= \(\frac{625}{9}\) – \(\frac{25}{1}\)
= \(\frac{625-225}{9}\) = \(\frac{400}{9}\)
∴ PT = \(\sqrt{\frac{400}{9}}\) = \(\frac{20}{3}\) cm

TS 10th Class Maths Solutions Chapter 9 Tangents and Secants to a Circle Optional Exercise

Question 3.
Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
Solution:
ABCD is a quadrilateral, circumscribing a circle whose centre is O.
The opposite sides AB and CD are subtend angles ∠AOB and ∠COD at the centre.
TS 10th Class Maths Solutions Chapter 9 Tangents and Secants to a Circle Optional Exercise 3
Similarly, the other pair of opposite sides AD and BC subtend angles ∠AOD and ∠BOC at the centre.
We have to prove that
∠AOB + ∠COD = 180°
and ∠AOD + ∠BOC = 180°
Join OR OQ, OR and OS.
We know that the tangents drawn from an ex¬ternal point to a circle subtend equal angles at the centre.
AP and AS are the tangents to the circle from A.
∴ ∠AOP = ∠AOS ……………. (1)
BP and BQ are the tangents to the circle from B.
∴ ∠BOP = ∠BOQ …………….. (2)
CQ and CR are the tangents to the circle from C.
∴ ∠COQ = ∠COR ………………. (3)
DR and DS are the tangents to the circle from D.
∴ ∠DOR = ∠DOS …………………. (4)
Adding (1), (2), (3) and (4), we get
∠AOP + ∠BOP + ∠COQ + ∠DOR + ∠AOS + ∠BOQ + ∠COR + ∠DOS = 360°
(∵ Sum of the angles formed at’O’is equal to 360°)
(∠AOP + ∠BOP) + (∠COR + ∠DOR) + (∠BOQ + ∠COQ) + (∠AOS + ∠DOS) = 360° 2∠AOP + 2∠COR + 2 ∠BOQ + 2 ∠DOS = 360°
⇒ ∠AOP + ∠COR + ∠BOQ + ∠DOS = 180°
⇒∠AOP + ∠BOQ + ∠COR + ∠DOS = 180°
⇒ ∠AOP + ∠BOP + ∠COR + ∠DOR = 180°
(∵ ∠BOQ = ∠BOP and ∠DOS = ∠DOR
⇒ ∠AOB + ∠COD = 180°
In ∆les BOP & BOQ)
Similarly, we can prove that
∠BOC + ∠AOD = 180°

TS 10th Class Maths Solutions Chapter 9 Tangents and Secants to a Circle Optional Exercise

Question 4.
Draw a line segment AB of length 8cm. Taking A as centre, draw a circle of radius 4cm and taking B as centre, draw another circle of radius 3cm. Construct tangents to each circle from the centre of the other circle.
Solution:
TS 10th Class Maths Solutions Chapter 9 Tangents and Secants to a Circle Optional Exercise 4

  1. Draw a line segment AB = 8 cm.
  2. Take A as centre and draw a circle (C1) of radius 4cm.
  3. Take B as centre and draw a circle (C2) of radius 3 cm.
  4. Draw the perpendicular bisector PQ of AB.
  5. Draw a circle with AB as diameter (C3). This circle intersects the circle with A as centre in T1 and T2. Join BT1 and BT2. These are the tangents to the circle C1.
  6. The circle with AB as diameter (C3) intersects the circle with B as centre (C2) in R1 and R2. Join AR1 and AR2. These are the tangents to the circle C2.

Question 5.
Let ABC be a right angled triangle in which AB = 6cm, BC = 8cm and ∠B = 90°. BD is the perpendicular from B on AC. The circle through B, C, D is drawn. Construct the tangents from A to this circle.
Solution:
Construction :
TS 10th Class Maths Solutions Chapter 9 Tangents and Secants to a Circle Optional Exercise 5

  1. Draw a line segment AB = 6cm.
    Make ∠ABX = 90°.
    Mark a point C on \(\overrightarrow{\mathrm{BX}}\) such that BC = 8cm. Join AC.
  2. Draw BD ⊥ AC intersecting AC in D.
  3. Draw the perpendicular bisectors of BC and CD. Let them intersect at ‘O’.
  4. Take ‘O’ as centre and OB = OD = OC as radius, draw a circle passing through B, D and C
  5. Join AO. Find M the mid point of AO.
  6. Take M as centre and radius as (AM = MO) draw a circle passing through A and C, intersecting the first circle at T1 and T2.
  7. Join AT1 and AT2.
  8. AT1 and AT2 are the required tangents.

TS 10th Class Maths Solutions Chapter 9 Tangents and Secants to a Circle Optional Exercise

Question 6.
Find the area of the shaded region in the figure, given in which two circles with centres A and B touch each other at the point C. If AC = 8 cm and AB = 3 cm.
TS 10th Class Maths Solutions Chapter 9 Tangents and Secants to a Circle Optional Exercise 6
Solution:
AC denotes the radius of the bigger circle.
AC = 8cm. (Given)
AB denotes the distance between the centres of two circles.
AB = 3cm (Given)
∴ BC denotes the radius of the smaller circle
BC = AC – AB = 8 – 3 = 5 cm
The Area of the shaded region = Area of the bigger circle – Area of the smaller circle =
= π(8)2 – π(5)2
= π[82 – 52]
= π[64 – 25]
= \(\frac{22}{7}\) × 39
= \(\frac{858}{7}\)
= 122.57 cm2

TS 10th Class Maths Solutions Chapter 9 Tangents and Secants to a Circle Optional Exercise

Question 7.
ABCD is a rectangle with AB = 14 cm and BC = 7cm. Taking DC, BC and AD as diameters, three semi-circles are drawn as shown in the figure. Find the area of the shaded region.
TS 10th Class Maths Solutions Chapter 9 Tangents and Secants to a Circle Optional Exercise 7
Solution:
ABCD is a rectangle.
Length of the rectangle AB = CD = 14 cm
Breadth of the rectangle AD = BC = 7cm
Area of the rectangle ABCD
= length × breadth
= 14 × 7
= 98 cm2
Two semi-circles are drawn on AD and BC clearly, They are of the same diameters.
Hence, Area of two semi-circles becomes area of one complete circle with diameter 7 cm.
So, radius of the circle (r) = \(\frac{7}{2}\) cm
Area of the circle = πr2
= \(\frac{22}{7}\) × \(\frac{7}{2}\) × \(\frac{7}{2}\)
= 38.5 cm2 ……………….. (1)
Area a of the semi-circle whose diameter is 14 cm = \(\frac{1}{2}\)πr2 ; Here, r = 7cm
= \(\frac{1}{2}\) × \(\frac{22}{7}\) × 7 × 7
= 77 cm2
Area of the shaded region in the rectangle
= Area of the rectangle – Area of the semicircle
= 98 – 77
= 21 cm2 ……………… (2)
Area of the total shaded region
= Area of the circle (shaded) + Area of the shaded region in the rectangle.
= (38.5 + 21) cm2
= 59.5 cm2

TS Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(a)

Students must practice this TS Intermediate Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(a) to find a better approach to solving the problems.

TS Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(a)

I.
Question 1.
Expand the following using binomial theorem.
i) (4x + 5y)5
ii) \(\left(\frac{2}{3} x+\frac{7}{4} y\right)^5\)
iii) \(\left(\frac{2 p}{5}-\frac{3 q}{7}\right)^6\)
iv) (3x + x – x2)4
Solution:
i) we know that
(x + a)n = \({ }^{\mathrm{n}} \mathrm{C}_0\) xn + \({ }^n C_1\) xn-1 a + \({ }^n C_2\) xn-2 a2 + ……………. + \({ }^n C_r\) + \({ }^n C_n\)
∴ (4x + 5y)7 = \({ }^7 \mathrm{C}_0\) (4x)7 + \({ }^7 \mathrm{C}_1\) (4x)6 (5y) + \({ }^7 \mathrm{C}_2\) (4x)5 (5y)2 + \({ }^7 \mathrm{C}_3\) (4x)4 (5y)3 + \({ }^7 \mathrm{C}_4\) (4x)3 (5y)4 + \({ }^7 \mathrm{C}_5\) (4x)2 (5y)5 + \({ }^7 \mathrm{C}_6\) (4x) (5y)6 + \({ }^7 \mathrm{C}_7\) (5y)7
= \(\sum_{r=0}^7{ }^7 C_r\) . (4x)7-r . (5y)r.

ii) We know that
(x + a)n = \({ }^{\mathrm{n}} \mathrm{C}_0\) xn + \({ }^n C_1\) xn-1 a + \({ }^n C_2\) xn-2 a2 + ……………. + \({ }^n C_r\) + \({ }^n C_n\)

\(\left(\frac{2}{3} x+\frac{7}{4} y\right)^5\) = \({ }^5 C_0\left(\frac{2}{3} x\right)^5+{ }^5 C_1\left(\frac{2}{3} x\right)^4\left(\frac{7}{4} y\right)+{ }^5 C_2\left(\frac{2}{3} x\right)^3\left(\frac{7}{4} y\right)^2 +{ }^5 C_3\left(\frac{2}{3} x\right)^3\left(\frac{7}{4} y\right)^3\) + \({ }^5 C_4\left(\frac{2}{3} x\right)\left(\frac{7}{4} y\right)^4+{ }^5 C_5\left(\frac{7}{4} y\right)^5\)

= \(\sum_{r=0}^5{ }^5 C_r \cdot\left(\frac{2}{3} x\right)^{5-r} \cdot\left(\frac{7}{4} y\right)^r\)

TS Board Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(a)

iii) We know that
(x + a)n = \({ }^{\mathrm{n}} \mathrm{C}_0\) xn + \({ }^n C_1\) xn-1 a + \({ }^n C_2\) xn-2 a2 + ……………. + \({ }^n C_r\) + \({ }^n C_n\)

TS Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(a) 1

iv) We know that
(3 + x – x2)4 = [3 + x (1 – x)]4
(x + a)n = \({ }^{\mathrm{n}} \mathrm{C}_0\) xn + \({ }^n C_1\) xn-1 a + \({ }^n C_2\) xn-2 a2 + ……………. + \({ }^n C_r\) + \({ }^n C_n\)
∴ [3 + x (1 – x)]4 = \({ }^4 C_0\) 34 + \({ }^4 C_1\)33x (1 – x + \({ }^4 C_2\) 32x2 (1 – x)2 + \({ }^4 C_3\) 3x3 (1 – x)3 + \({ }^4 C_4\) x4 (1 – x)4
= 81 + 108x (1 – x) + 54x2 (1 – 2x + x2) + 12x3 (1 – 3x + 3x2 – x3) + x4 (1 – 4x + 6x2 – 4x3 + x5
= 81 + 108x – 54x2 – 96x3 + 19x4 + 32x5 – 6x6 – 4x7 + x8.

Question 2.
Write down and simplify
i) 6th term in \(\left(\frac{2 x}{3}+\frac{3 y}{2}\right)^y\)
ii) 7th term in (3x – 4y)10
iii) 10th term \(\left(\frac{3 p}{4}-5 q\right)^{14}\)
iv) rth term in \(\left(\frac{3 a}{5}+\frac{5 b}{7}\right)^8\) (1 ≤ r ≤ 9).
Solution:
i) r + 1th term n the expansion of (x + a)n is given by
Tr+1 = \({ }^n C_r\) . xn-r . ar
∴ 6th term in the expansion of \(\left(\frac{2 x}{3}+\frac{3 y}{2}\right)^9\) is given by
T6 = \({ }^9 C_5 \cdot\left(\frac{2 x}{3}\right)^4 \cdot\left(\frac{3 y}{2}\right)^5\)
= 189 . x4 . y5

ii) (r + 1)th term in the expansion of (x + a)n is given by
Tr+1 = \({ }^n C_r\) . xn-r . ar
7th term In the expansion of (3x – 4y)10 is given by
T7 = \({ }^{10} \mathrm{C}_6\) (3x)4 (- 4y)6
= 280 . 125 x4 y6.

iii) (r + 1)th term in the expansion of (x + a)n is given by
Tr+1 = \({ }^n C_r\) . xn-r . ar
10th term in the expansion of \(\left(\frac{3 p}{4}-5 q\right)^{14}\) is given by
T10 = \({ }^{14} \mathrm{C}_9\left(\frac{3 p}{4}\right)^5\) (- 5q)9
= \(\frac{-(2002) \cdot 3^5 \cdot 5^9}{4^5}\) . p5 . q9.

iv) (r + )th term in the expansion of (x + a)n is given by
Tr+1 = \({ }^n C_r\) . xn-r . ar
∴ rth term in the expansion of \(\left(\frac{3 a}{5}+\frac{5 b}{7}\right)^8\) is given by
Tr = \({ }^8 C_{r-1}\left(\frac{3 a}{5}\right)^{9-r}\left(\frac{5 b}{7}\right)^{r-1}\).

TS Board Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(a)

Question 3.
Find the number of terms in the expansion of
i) \(\left(\frac{3 a}{4}+\frac{b}{2}\right)^9\)
ii) (3p + 4q)14
iii) (2x + 3y + z)7
Solution:
1) The expansion of (x + a)n contains (n + 1) terms.
∴ Expansion of \(\left(\frac{3 a}{4}+\frac{b}{2}\right)^9\) contains 10 terms.

ii) The expansion of (x + a)n contains (n + 1) terms.
∴ Expansion of (3p + 4q)14 contains 15 terms.

iii) Number of terms in the expansion of
(a + b + c)n = \(\frac{(n+1)(n+2)}{2}\)
∴ Expansion oF (2x + + z)7 contains \(\frac{(7+1)(7+2)}{2}\) = 36 terms.

Question 4.
Find the numerically greatest term(s) in the expansion of coefficients in (4x – 7y)49 + (4x + 7y)49.
Solution:
We know that
(4x – 7y)49 = \({ }^{49} \mathrm{C}_0\) (4x)49 – \({ }^{49} \mathrm{C}_1\)(4x)48 (7y) + \({ }^{49} \mathrm{C}_2\) (4x)47 (7y)2 – \({ }^{49} \mathrm{C}_3\) (4x)46 (7y)3 + ……………. + \({ }^{49} \mathrm{C}_49\) (7y)49 ………….(1)

(4x + 7y)49 = \({ }^{49} \mathrm{C}_0\) (4x)49 + \({ }^{49} \mathrm{C}_1\)(4x)48 (7y) + \({ }^{49} \mathrm{C}_2\) (4x)47 (7y)2 + \({ }^{49} \mathrm{C}_3\) (4x)46 (7y)3 + ……………. + \({ }^{49} \mathrm{C}_49\) (7y)49 ………….(2)

(1) + (2)
(4x – 7y)49 + (4x + 7y)49 = 2[\({ }^{49} \mathrm{C}_0\) (4x)49 + \({ }^{49} \mathrm{C}_2\) (4x)47 (7y)2 + \({ }^{49} \mathrm{C}_4\) (4x)45 (7y)4 + ………….. + \({ }^{49} \mathrm{C}_48\) (4x) (7y)48]
which contains 25 non-zero coefficients.

TS Board Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(a)

Question 5.
Find the sum of last 20 coefficients in the expansion of (1 + x)39.
Solution:
The last 20 coeffIcients in the expansion of (1 + x)39 are \({ }^{39} \mathrm{C}_{20},{ }^{39} \mathrm{C}_{21}, \ldots \ldots \ldots,{ }^{39} \mathrm{C}_{39}\)
We know that

TS Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(a) 2

∴ The sum of last 20 coefficients in expansion of (1 + x)39 is 238.

Question 6.
If A and B are coefficients of xn in the expansion of (1 + x)2n and (1 + x)2n – 1 respectively, then find the value of \(\frac{A}{B}\).
Solution:
Coefficient of xn in the expansion of (1 + x)2n is \({ }^{2 n} C_n\).
Coefficient of xn in the expansion of (1 + x)2n – 1 is \({ }^{2 n – 1} C_n\)
∴ A = \({ }^{2 n} C_n\) and B = \({ }^{2 n – 1} C_n\).
∴ \(\frac{A}{B}=\frac{{ }^{2 n} C_n}{2 n-1}=\frac{\frac{2 n !}{n ! n !}}{\frac{(2 n-1) !}{(n-1) ! \cdot n !}}\)
\(\frac{2 n !}{(2 n-1) ! n !} \cdot(n-1) !=\frac{2 n}{n}\)
\(\frac{A}{B}\) = 2.

TS Board Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(a)

II.
Question 1.
Find the coefficient of
i) x-6 in \(\left(3 x-\frac{4}{x}\right)^{10}\)
ii) x11 in \(\left(2 x^2+\frac{3}{x^3}\right)^{13}\)
iii) x2 in \(\left(7 x^3-\frac{2}{x^2}\right)^9\)
iv) x-7 in \(\left(\frac{2 x^2}{3}-\frac{5}{4 x^5}\right)^7\)
Solution:
i) General term in the expansion of (x + a)n is
Tr+1 = \({ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}\) . xn-r . ar
(r + 1)th term in the expansion of \(\left(3 x-\frac{4}{x}\right)^{10}\) is
Tr+1 = \({ }^{10} \mathrm{C}_{\mathrm{r}}\) . (3x)10-r . (\(\frac{-4}{x}\))r
Tr+1 = \({ }^{10} \mathrm{C}_{\mathrm{r}}\) . (3)10-r . (- 4)r . x10-2r.
for coefficient of x-6, 10 – 2r = – 6
⇒ 2r = 16
⇒ r = 8.
∴ Coefficient of x-6 is \({ }^{10} \mathrm{C}_8\) . 310-8 . (- 4)8 = 405 × 48.

ii) General term in the expansion of (x + a)n is
Tr+1 = \({ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}\) . xn-r . ar
(r + 1)th term in the expansion of \(\left(2 x^2+\frac{3}{x^3}\right)^{13}\) is
Tr+1 = \({ }^{13} C_r \cdot\left(2 x^2\right)^{13-r} \cdot\left(\frac{3}{x^3}\right)^r\)
Tr+1 = \({ }^{13} C_r \) . 213-r . 3r . x26-5r
For coefficients of x11, 26 – 5r = 11
⇒ 5r = 15
⇒ r = 3.
Coefficient of x11 in \(\left(2 x^2+\frac{3}{x^3}\right)^{13}\) is \({ }^{13} C_3 \) . 213-3 . 33
= 286 . 210 . 33.

iii) General term in the expansion of (x + a)n is
Tr+1 = \({ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}\) . xn-r . ar
∴ General term in the expansion of \(\left(7 x^3-\frac{2}{x^2}\right)^9\) is
Tr+1 = \({ }^9 C_r \cdot\left(7 x^3\right)^{9-r}\left(\frac{-2}{x^2}\right)^r\)
= \({ }^9 C_r\) 79-r (- 2)r . x27-5r
for coefficient of x2, 27 – 5r = 2
⇒ 5r = 25
⇒ r = 5.
∴ Coefficient of x2 is \({ }^9 \mathrm{C}_5\) . 74 . (- 2)5
= – 126 . 74 . 25.

iv) General term in the expansion of (x + a)n is
Tr+1 = \({ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}\) . xn-r . ar
General term in the expansion of \(\left(\frac{2 x^2}{3}-\frac{5}{4 x^5}\right)\) is
Tr+1 = \(={ }^7 C_r \cdot\left(\frac{2}{3} x^2\right)^{7-r} \cdot\left(\frac{-5}{4 x^5}\right)^r\)
= \({ }^7 \mathrm{C}_{\mathrm{r}} \cdot\left(\frac{2}{3}\right)^{7-\mathrm{r}}\left(\frac{-5}{4}\right)^{\mathrm{r}} \mathrm{x}^{14-7 \mathrm{r}}\)
For coefficient of x-7, 14 – 7r = – 7
⇒ 7r = 21
⇒ r = 3
∴ Coefficient of x-7 is \({ }^7 C_3 \cdot\left(\frac{2}{3}\right)^4 \cdot \frac{(-5)^3}{4^3}\)
= \(-\frac{35 \times 16 \times 125}{81 \times 64}=\frac{-4375}{324}\).

TS Board Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(a)

Question 2.
Find the term independent of x in the expansion of
i) \(\left(\frac{\sqrt{x}}{3}-\frac{4}{x^2}\right)^{10}\)
ii) \(\left(\frac{3}{\sqrt[3]{x}}+5 \sqrt{x}\right)^{25}\)
iii) \(\left(4 x^3+\frac{7}{x^2}\right)^{14}\)
iv) \(\left(\frac{2 x^2}{5}+\frac{15}{4 x}\right)^9\)
Solution:
i) General term in the expansion of (x + a)n is
Tr+1 = \({ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}\) . xn-r . ar
General term in the expansion of \(\left(\frac{\sqrt{x}}{3}-\frac{4}{x^2}\right)^{10}\) is
Tr+1 = \({ }^{10} C_r \cdot\left(\frac{\sqrt{x}}{3}\right)^{10-r} \cdot\left(\frac{-4}{x^2}\right)^r\)
= \({ }^{10} C_r \cdot \frac{x^{\frac{10-r}{2}}}{3^{10-r}} \cdot \frac{(-4)^r}{x^{2 r}}\)
= \({ }^{10} C_r \cdot \frac{(-4)^r}{3^{10-r}} \cdot x^{\frac{10-5 r}{2}}\)
For the independent term (i.e., the coefficient of x0)
put \(\frac{10-5 r}{2}\) = 0
⇒ r = 2.
∴ Term independent of ‘x’ in the expansion is
T3 = \({ }^{10} C_2 \frac{(-4)^2}{3^8} \times x^0=\frac{80}{729}\).

ii) General term in the expansion of (x + a)n is
Tr+1 = \({ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}\) . xn-r . ar
General term in the expansion of \(\left(\frac{3}{\sqrt[3]{x}}+5 \sqrt{x}\right)^{25}\) is
Tr+1 = \({ }^{25} C_r\left(\frac{3}{\sqrt[3]{x}}\right)^{25-r} \cdot(5 \sqrt{x})^r\)
= \({ }^{25} C_r\) . 325-r . 5r . x(5r – 50)/6
For the independent term,
(i.e., coefficient of x0)
Put \(\frac{5 r-50}{6}\) = 0
⇒ r = 10.
∴ Term independent of ‘x’ in the expansion is T11 = \({ }^{25} \mathrm{C}_{10}\) . 315 . 510.

TS Board Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(a)

iii) General term in the expansion of (x + a)nis
Tr+1 = \({ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}\) . xn-r . ar
General term in the expansion of \(\left(4 x^3+\frac{7}{x^2}\right)^{14}\) is
Tr+1 = \({ }^{14} C_r \cdot\left(4 x^3\right)^{14-r} \cdot\left(\frac{7}{x^2}\right)^r\)
= \({ }^{14} C_r\) . 414-r . 7r . x42-5r.
For the independent term, (i.e., coefficient of x0)
Put 42 – 5r = 0
which is impossible as ‘r’ is integer.
∴ Term independent of ‘x in the expansion is ‘0’.

iv) General term in the expansion of (x + a)nis
Tr+1 = \({ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}\) . xn-r . ar
General term in the expansion of \(\left(\frac{2 x^2}{5}+\frac{15}{4 x}\right)^9\) is
Tr+1 = \({ }^9 C_r\left(\frac{2 x^2}{5}\right)^{9-r} \cdot\left(\frac{15}{4 x}\right)^r\)
= \({ }^9 \mathrm{C}_{\mathrm{r}}\left(\frac{2}{5}\right)^{9-\mathrm{r}} \cdot\left(\frac{15}{4}\right)^{\mathrm{r}} \cdot \mathrm{x}^{18-3 \mathrm{r}}\)
For the independent term, (i.e. coefficient of x0)
Put 18 – 3r = 0
⇒ r = 6
Term independent of ‘x’ in the expansion is
T7 = \({ }^9 \mathrm{C}_6 \cdot\left(\frac{2}{5}\right)^3 \cdot\left(\frac{15}{4}\right)^6\)
= \(\frac{3^7 \cdot 5^3 \cdot 7}{2^7}\).

TS Board Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(a)

Question 3.
Find the middle term(s) in the expansion of
i) \(\left(\frac{3 x}{7}-2 y\right)^{10}\)
ii) \(\left(4 a+\frac{3}{2} b\right)^{11}\)
iii) (4x2 + 5x3)17
iv) \(\left(\frac{3}{a^3}+5 a^4\right)^{20}\)
Solution:
i) General term in the expansion of (x + a)nis
Tr+1 = \({ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}\) . xn-r . ar
General term in the expansion of \(\left(\frac{3 x}{7}-2 y\right)^{10}\) is
Tr+1 = \({ }^{10} C_{\mathrm{r}} \cdot\left(\frac{3 \mathrm{x}}{7}\right)^{10-\mathrm{r}}(-2 \mathrm{y})^{\mathrm{r}}\)
The expansion has 11 (odd number) terms.
Hence T6 is the only middle term.
Thus r = 5
∴ Middle term in the expansion is
T6 = \(-10 C_5\left(\frac{6}{7}\right)^5\) . x5 . y5.

ii) General term in the expansion of (x + a)nis
Tr+1 = \({ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}\) . xn-r . ar
General term in the expansion \(\left(4 a+\frac{3}{2} b\right)^{11}\) is
Tr+1 = \(\)
The expansion has 12 (even numbers) terms.
Hence T6 and T7 are the middle terms.
Thus r = 5 and r = 6
when r = 5,
T6 = \({ }^{11} C_5 \cdot(4 a)^6 \cdot\left(\frac{3}{2} b\right)^5\)
= 77 × 28 . 36 . a6 . b5.
when r = 6,
T7 = \({ }^{11} C_6(4 a)^5 \cdot\left(\frac{3}{2} b\right)^6\)
= 77 × 25 . 37 . a5 . b6.

TS Board Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(a)

iii) General term in the expansion of (x + a)nis
Tr+1 = \({ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}\) . xn-r . ar
General term in the expansion of (4x2 + 5x3)17 is
Tr+1 = \({ }^{17} \mathrm{C}_r\) . (4x2)17-r . (5x3)r.
The expansion has 18 (even number) terms.
Hence T9 and T10 are the middle terms.
Thus r = 8 and r = 9
When r = 8.
T9 = \({ }^{17} \mathrm{C}_8\) . (4x2)9 . (5x3)8
= \({ }^{17} \mathrm{C}_8\) . 49 . 58 . x42
When r = 9,
T10 = \({ }^{17} \mathrm{C}_9\) (4x2)8 (3)9
= \({ }^{17} \mathrm{C}_9\) . 48 . 59 . x43

iv) General term in the expansion of (x + a)nis
Tr+1 = \({ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}\) . xn-r . ar
General term in the expansion of \(\left(\frac{3}{a^3}+5 a^4\right)^{20}\) is
Tr+1 = \({ }^{20} C_r \cdot\left(\frac{3}{a^3}\right)^{20-r} \cdot\left(5 a^4\right)^r\)
The expansion has 21 (odd number) terms. Hence, T11 is the only middle term
Thus r = 10
∴ Middle term in the expansion is
T11 = \({ }^{20} C_{10} \cdot\left(\frac{3}{a^3}\right)^{10} \cdot\left(5 a^4\right)^{10}\)
= \({ }^{20} C_{10}\) . 1510 . a10.

Question 4.
Find the numerically greatest term(s) in the expansion of
i) (4 + 3x)15 when x = \(\frac{7}{2}\).
ii) (3x + 5y)12 when x = \(\frac{1}{2}\), y = \(\frac{4}{3}\)
iii) (4a – 6b)13 when a = 3, b = 5.
iv) (3 + 7x)n when x = \(\frac{4}{5}\), n = 5.
Solution:
i) We know that (4 + 3x)15 = 415 \(\left(1+\frac{3 x}{4}\right)^{15}\)
First we find the numerically greatest term in the expansion of \(\left(1+\frac{3 x}{4}\right)^{15}\)
Let X = \(\frac{3 x}{4}\)
x = \(\frac{7}{2}\), |X| = \(\left|\frac{3}{4} \times \frac{7}{2}\right|=\frac{21}{8}\)
Now \(\frac{(n+1)|X|}{1+|X|}=\frac{16 \times \frac{21}{8}}{1+\frac{21}{8}}=\frac{336}{29}\)
not an integer.
Its integral part m = \(\left[\frac{336}{29}\right]\) = 11.
∴ Tm+1, i.e., T12 is the numerically greatest term in the binomial expansion of \(\left(1+\frac{3 x}{4}\right)^{15}\)
∴ T12 = \({ }^{15} C_{11}\left(\frac{3 x}{4}\right)^{11}\)
= \({ }^{15} \mathrm{C}_{11}\left(\frac{21}{8}\right)^{11}\)
∴ Numerically greatest term in the expansion 01(4 + 3x)15 = \({ }^{15} C_{11}\left(4^{15}\right) \frac{21^{11}}{8^{11}}\)
= \({ }^{15} C_{11} \cdot \frac{21^{11}}{2^3}\).

TS Board Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(a)

ii) We know that
(3x + 5y)12 = (3x)12 . (1 + \(\frac{5 y}{3 x}\))12
First we find the numerically greatest term in the expansion of (1 + \(\frac{5 y}{3 x}\))12

TS Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(a) 3

iii) We know that
(4a – 6b)13 = (4a)13 (1 – \(\frac{6 b}{4 a}\))13
First we find the numerically greatest term in the expansion of (1 – \(\frac{6 b}{4 a}\))13
Let X = – \(\frac{6 b}{4 a}\)
As a = 3 and b = 5, |X| = \(\frac{5}{2}\)

TS Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(a) 4

∴ The numerically greatest terms in the expansion of (4a – 6b)13 are T10 and T11.
T10 = – 1213 . 13C9 . \(\left(\frac{5}{2}\right)^9\)
= – 13C9 . (12)4 . (30)9
and T11 = 1210 . 13C10 . \(\left(\frac{5}{2}\right)^10\)
= 143 . 217 . 313 . 510.

iv) We know that (3 + 7x)n = 3n (1 + \(\frac{7}{3}\) x)n
First we find numerically greatest term in the expansion of (1 + \(\frac{7}{3}\) x)n
Let X = \(\frac{7}{3}\) x
As x = \(\frac{4}{5}\), |X| = \(\frac{28}{15}\) and
\(\frac{(n+1)|X|}{1+|X|}=\frac{56}{5}\) and n = 15.

Its integral part,
m = \(\left[\frac{(\mathrm{n}+1)|\mathrm{X}|}{1+|\mathrm{X}|}\right]\) = 11
∴ Tm+1 i.e., T12 is numerically greatest term in the expansion of (1 + \(\frac{7}{3}\) x)n
T12 = 15C11 . \(\left(\frac{28}{15}\right)^{11}\)
∴ The numerically greatest terms in the expansion of (3 + 7x)n is
T12 = 315 . 15C11 . \(\left(\frac{28}{15}\right)^{11}\)
= 15C11 . \(\left(\frac{28}{5}\right)^{11}\) . 34.

TS Board Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(a)

Question 5.
Prove the following:
i) 2 . C0 + 5 . C1 + 8. C2 + ……………. + (3n + 2) Cn = (3n + 4) 2n-1
ii) C0 – 4 C1 + 7 . C2 – 10 . C3 + ……………. = 0,
if n is an even positive Integer.
iii) \(\frac{C_1}{2}+\frac{C_3}{4}+\frac{C_5}{6}+\frac{C_7}{8}+\ldots \ldots=\frac{2^n-1}{n+1}\)
iv) \(C_0+\frac{3}{2} \cdot C_1+\frac{9}{3} \cdot C_2+\frac{27}{4} \cdot C_3+\ldots \ldots+\frac{3^n}{n+1} C_n=\frac{4^{n+1}-1}{3(n+1)}\)
v) C0 + 2 . C1 + 4 . C2 + 8 . C3 + ………….. + 2n . Cn = 3n
Solution:
i) We know that
C0 = Cn, C1 = Cn-1, ………… Cr = Cn-r
Let S = 2 . C0 + 5 . C1 + 8 . C2 + ………….. + (3n – 1) . Cn-1 + (3n + 2) Cn ………….(1)
∴ S = (3n + 2) C0 + (3n – 1)C1 + (3n – 4)C2 + …………. + 5 . Cn-1 + 2 . Cn ………….(2)
(1) + (2) ⇒ 2S = (3n + 4) . C0 + (3n + 4) . C1 + (3n + 4) . C2 + …………… (3n + 4) . Cn
= (3n + 4) (C0 + C1 + C2 + …………… + Cn)
⇒ 2S = (3n + 4) . 2n
⇒ S = (3n + 4) . 2n-1
∴ 2 . C0 + 5 . C1 + 8 . C2 + ………… + (3n + 2) . Cn = (3n + 4) . 2n-1

ii) We know that 1, 4, 7, 10, ………… are in A.P.
(n + 1)th term,
Tn+1 = 1 + (n)3 = 3n + 1
∴ C0 – 4 . C1 + 7 . C2 – 10 . C3 + ……….. (n + 1) terms.
= \(\sum_{r=0}^n(-1)^r(3 r+1) C_r\)
= \(3 \sum_{r=0}^n(-1)^r r \cdot C_r+\sum_{r=0}^n(-1)^r \cdot C_r\)
= 3 (0) + 0 = 0
∴ C0 – 4 . C1 + 7 . C2 – 10 . C3 + ……….. = 0.

TS Board Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(a)

iii) Given L.H.S

TS Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(a) 5

iv) Given L.H.S

TS Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(a) 6

v) L.HS = C0 + 2 . C1 + 4 . C2 + 8 . C3 + ……….. + 2n . Cn
= C0 + C1 . 2 + C2 . 22 + C3 . 23 + ………. + Cn 2n
= (1 + 2)n
= 3n = R.H.S
∴C0 + 2 . C1 + 4 . C2 + 8 . C3 + ……….. + 2n . Cn = 3n.

TS Board Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(a)

Question 6.
Find the sum of the following:
i) \(\frac{{ }^{15} C_1}{{ }^{15} C_0}+2 \frac{{ }^{15} C_2}{{ }^{15} C_1}+3 \frac{{ }^{15} C_3}{{ }^{15} C_2}+\ldots+15 \frac{{ }^{15} C_{15}}{{ }^{15} C_{14}}\)
ii) Cn . C3 + C1 .C4 + C2 . C5 + + …………… + Cn-3 . Cn
iii) 22 C0 + 32 C1 + 42 C2 + …………. + (n + 2)2 Cn
iv) 3C0 + 6C1 + 12C2 + ……………. + 3 . 2n Cn
Solution:
i) We know that

TS Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(a) 7

ii) (1 + x)n = C0 + C1x + C2x2 + …………. + Cnxn …………..(1)
(x + 1)n = C0xn + C1xn – 1 + C23xn-2 + …………… + Cn …………….. (2)
(1) x (2) = (1 + x)2n
= (C0 + C1x + C2x2 + ……….. + Cnxn) (C0x + C1xn-1 + C2xn-2 + C3xn-3 + ………….. + Cn]
Comparing coefficients of xn-3 on bothsides,
2nCn-3 = C0 . C3 + C1 . C4 + C2 . C5 + …………. + Cn-3 . Cn
i.e., C0. C3 + C1 . C4 + C2. C5 + …………… + Cn-3 . Cn = 2nCn-3 = 2nCn+3
[∵ nCr = nCn-r].

TS Board Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(a)

iii) 22 C0 + 32 C1 + 42 C2 + …………. + (n + 2)2 Cn

TS Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(a) 8

= n (n – 1) 2n-2 + 5 n 2n-1 + 4 (2n)
= 2n-2 [n (n – 1) + 10n + 16]
= (n2 + 9n . 16) 2n-2.

iv) 3 . C0 + 6 . C1 + 12 . C2 + ……………. + 3 . 2n Cn
= \(\sum_{r=0}^n 3 \cdot 2^r \cdot C_r\)
= \(3 \sum_{r=0}^n 2^r \cdot C_r\)
= 3 [1 + C1 (2) + C2 (22) + C3 (23) + …………. + Cn 2n]
= 3 [1 + 2]n
= 3 . 3n
= 3n+1.

TS Board Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(a)

Question 7.
Using binomial theorem prove that 50n – 49n – 1 is divisible by 492 for all positive integers n.
Solution:
We know that,
50n – 49n – 1
= (1 + 49)n – 49n – 1
= [1 + nC1 . 49 + nC2 . 492 + nC3 . 493 + ………… + 49n] – 49n – 1
= nC2 . 492 + nC3 . 493 + …………… + nCn . 49n
= 492 [nC2 + nC3 . 49 + ……….. + nCn . 49n-2]
= 492 (a positive integer)
Hence 50n 49n – 1 is divisible by 492 ∀n ∈ N.

Question 8.
Using binomial theorem prove that 54n + 52n – 1 is divisible by 676 for all positive integers n.
Solution:
We know that, 54n + 52n – 1
= (52)2n + 52n – 1
= (25)2n+ 52n – 1
= (26 – 1)2n+ 52n – 1
= 2nC0 (26)2n2nC1 (26)2n-1 + 2nC2 (26)2n-2 + …………. + 2nC2n-2(26)22nC2n-1 (26) + 2nC2n + 52n – 1
= 262n + 2nC1 (26)2n-1 + 2nC2 (26)2n-2 + …………….. + 2nC2n-2 (26)2
= 262 [262n-22nC1 262n-3 + 2nC2 262n-4 + ……… + 2nC2n-2]
= 676 (some integer)
∴ 54n + 52n – 1 is divisible by 676, ∀n ∈ N.

TS Board Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(a)

Question 9.
If (1 + x + x2)n = a0 + a1x + a2x2 + ………….. + a2nx2n, then prove that
i) a0 + a1 + a2 + ……….. + a2n = 3n
ii) a0 + a2 + a4 + ………… + a2n = \(\frac{3^n+1}{2}\)
iii) a1 + a3 + a5 + ……………. + a2n-1 = \(\frac{3^n-1}{2}\)
iv) a0 + a3 + a6 + a9 + ………… = 3n-1
Solution:
Given (1 + x + x2)n/sup> = a0 + a1x + a2x2 + ………….. + a2nx2n
i) Subtituting x = 1 in (I), we have
a0 + a1 + a2 + ……….. + a2n = 3n …………. (1)
Substituting x = – 1 in (I), we have
a0 – a1 + a2 + ……….. + a2n = 1 …………….(2)

ii) (1) + (2)
⇒ 2a0 + 2a2 + 2a4 + ……….. + 2a2n = 3n + 1
⇒ a0 + a2 + a4 + ……….. + a2n = \(\frac{3^n+1}{2}\)

iii) (1) – (2)
⇒ 2a1 + 2a3 + 2a5 + ……….. + 2a2n-1 = 3n – 1
⇒ a1 + a3 + a5 + ……….. + a2n-1 = \(\frac{3^n-1}{2}\)

iv) Substituting x = ω, in (1), we have
a0 + a1ω + a2ω2 + ……….. + a2nω2n = 0
(∵ 1 + ω + ω2 = 0)

Substituting x = ω2 in (1), we have
a0 + a1ω2 + a2ω4 + a3ω6 + ……….. + a2nω4n = 0 …………… (4)
(1) + (3) +(4):
3a0 + a1 (1 + ω + ω2) + a2 (1 + ω + ω2) + a3 (1 + ω3 + ω6) + ………… + a2n (1 + ω2n + ω4n) = 3n
⇒ 3a0 + 3a3 + 3a6 + 3a9 + …………. = 3n
(∵ 1 + ω + ω2 = 0 and ω3 = 1)
⇒ a0 + a3 + a6 + a9 + ……….. = 3n-1

TS Board Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(a)

Question 10.
If (1 + x + x2 + x3)7 = b0 + b1x + b2x2 + ……………+ b21 x21, then find the value of
i) b0 + b2 + b4 + ……………. + b20
ii) b1 + b3 + b5 + ……………. + b21
Solution:
Given
(1 + x + x2 + x3)7 = b0 + b1x + b2x2 + ……………+ b21 x21 ……………(1)
Substituting x = 1 in (1),
we get b0 + b1 + b2 + ……………. + b21 = 47 ……….(2)
Substituting x = – 1 in (1),
we get b0 – b1 + b2 + ……………. – b21 = 0 …………… (3)
i) (2) + (3)
⇒ 2b0 + 2b2 + 2b4 + ……………. + b20 = 47
⇒ b0 + b2 + b4 + ……………. + b20 = 213.

ii) (2) – (3)
⇒ 2b1 + 2b3 + 2b5 + ……………. + 2b21 = 47
⇒ b1 + b3 + b5 + ……………. + b21 = 213.

Question 11.
If the coefficients of x11 and x12 in the binomial expansion of \(\left(2+\frac{8 x}{3}\right)^n\) are equal, find n.
Solution:
We know that \(\left(2+\frac{8 x}{3}\right)^n=2^n\left(1+\frac{4 x}{3}\right)^n\)
Coefficient of x11 in the expansion of \(\left(2+\frac{8 x}{3}\right)^n\) is nC11 . 2n . (\(\frac{4}{3}\))11
Coefficient of x12 in the expansion of \(\left(2+\frac{8 x}{3}\right)^n\) is nC12 . 2n . (\(\frac{4}{3}\))12
Given coefficients of x11 and x12 are same
nC11 . 2n . (\(\frac{4}{3}\))11 = nC12 . 2n . (\(\frac{4}{3}\))12
⇒ \(\frac{n !}{(n-11) ! 11 !}=\frac{n !}{(n-12) ! 12 !}\left(\frac{4}{3}\right)\)
⇒ 12 = (n – 11) \(\frac{4}{3}\)
⇒ 9 = n – 11
⇒ n = 11 + 9 = 20.

TS Board Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(a)

Question 12.
Find the remainder when 22013 is divided by 17.
Solution:
We have 22013
= 2 (22012)
= 2 (24)203
= 2 (16)503
= 2 (17 – 1)203
= 2 [503C0 17503503C1 17502 + 503C2 17501 – ……….. + 503C502 17 – 503C503]
= 2 [503C0 17503503C1 17502 + 503C2 17501 – ……….. + 503C502 17 – 503C503] – 2
= 17m – 2 where ‘m is some integer
∴ 22013 = 17m – 2 (or) 17k + 15
∴ The remainder is – 2 or 15.

Question 13.
If the coefficient of (2r + 4)th term and (3r + 4)th term in the expansion of (1 + x)21 are equal, find r.
Solution:
We know that coefficient of (r + 1)th term of (1 + x)n is nCr.
∴ Coefficient of (2r + 4)th term 0f (1 + x)21 is 21C2r+3
Also coefficient of (3r + 4)th term of (1 +x)21 is 21C3r+3.
Given coefficient of (2r + 4)th term and (3r + 4)th terms in the expansion of (1 + x)21 are equal.
21C2r+3 = 21C3r+4
∴ Either 2r + 3 = 3r + 3 0r 2r + 3 + 3 = 21
If 2r + 3 = 3r + 3 then r = 0,
If 2r + 3 + 3r = 21 then r = 3.
∴ r = 0 or r = 3.

TS Board Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(a)

III.
Question 1.
If the coefficients of x9, x10, x11 in the expansion of (1 + x)n are in AP. then prove that n2 – 4m + 398 = 0.
Solution:
Coefficient of xr in the expansion of (1 + x)n is nCr.
Given coefficients of x9, x10, x11 in the expansion of (1 + x)n are in A.P., then
2(nC10) = nC9 + nC11
⇒ \(2 \frac{n !}{(n-10) ! 10 !}=\frac{n !}{(n-9) ! 9 !}+\frac{n !}{(n-11) !+11 !}\)
⇒ \(\frac{2}{10(n-10)}=\frac{1}{(n-9)(n-10)}+\frac{1}{11 \times 10}\)
⇒ \(\frac{2}{(n-10) 10}=\frac{110+(n-9)(n-10)}{110(n-9)(n-10)}\)
⇒ 22 (n – 9) = 110 + n2 – 19n + 90
⇒ n2 – 41n + 398 = 0.

Question 2.
If 36, 84, 126 are three successive binomial coefficients in the expansion of (1 + then find n.
Solution:
Let us consider nCr-1, nCr and nCr+1 as three successive binomial coefficients of (1 + xy)n.
i.e., nCr-1 = 36; nCr-1 = 84 and nCr-1 = 126
Consider \(\frac{{ }^{n_C} C_r}{{ }^n C_{r-1}}=\frac{84}{36}\)
⇒ \(\frac{n-r+1}{r}=\frac{7}{3}\)
⇒ 3n + 3 = 10r …………..(1)
Similarly,
\(\frac{{ }^n C_{r+1}}{{ }^n C_r}=\frac{126}{84}\)
⇒ \(\frac{\mathrm{n}-\mathrm{r}}{\mathrm{r}+1}=\frac{3}{2}\)
⇒ 2n = 5r + 3 ………….(2)
⇒ 2n = 5 \(\left(\frac{3 \mathrm{n}+3}{10}\right)\) + 3 (∵ from (1))
⇒ n = 9.

TS Board Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(a)

Question 3.
If the 2nd, 3rd and 4th terms in the expansion of (a + x)n are respectively 240, 720, 1080. find a, x, n.
Solution:
General term in the expansion of (a + x)n is
Tr+1 = \({ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}\) . xn-r . ar
∴ 2nd term,T2 = nC1 . an-1 . x …………..(1)
3rd term, T3 = nC2 . an-2 . x2 ………..(2)
4th term, T4 = nC3 . an-3 . x3 …………(3)
Given T2 = 240, T3 = 720, T3 = 1080
From (1) and (2),

TS Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(a) 9

Sub. n = 5 in (1), we get
5 . a4 . x = 240 ………..(6)
Substituting n = 5 in (4), we get x = \(\frac{3 a}{2}\) ………….(7)
Substituting x = \(\frac{3 a}{2}\) in (6),
we get 5 . a4 = 240
⇒ a5 = 32
⇒ a = 2
Substituting in (7), we get x = 3
∴ a = 2; x = 3 and n = 5.

TS Board Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(a)

Question 4.
If the coefficients of rth, (r + 1)th and (r + 2)nd terms in the expansion of (1 + x)n are in A.P., then show that n2 – (4r + 1)n + 4r2 – 2 = 0.
Solution:
We know that, the general term, (r + 1)th term in the expansion of (1 + x)n is
Tr+1 = \({ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}\) . xn-r
∴ The coefficients of rth, (r + 1)th and (r + 2)nd terms in the expansion of (1 + x)n are nCr-1, nCr and nCr+1
Given nCr-1, nCr & nCr+1 are in A.P.

TS Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(a) 10

⇒ 2(n – r + 1) (r + 1) = r (r + 1) + (n – r) (n – r + 1)
⇒ 2nr + 2n – 2r2 – 2r + 2r + 2 = r2 + r + n2 – nr + n – nr + r2 + r
⇒ n2 – n (4r + 1) + 4r2 – 2 = 0.

Question 5.
Find the sum of the coefficients of x32 and x-18 in the expansion of (2x3 – \(\frac{3}{x^2}\))14.
Solution:
The general term in (2x3 – \(\frac{3}{x^2}\))14 is
Tr+1 = 14Cr (2x3)14-r (\(\frac{-3}{x^2}\))r
Tr+1 = (- 1)r . 14Cr . 214-r . 3r . x42-5r ………….(1)
For coefficient of x32, put
42 – 5r = 32
⇒ r = 2
Substituting r = 2 in (1), we get
T3 = 14C2 . 212 . 32 . x32
∴ Coefficient of x32 is 14C2 . 212 . 32 ………(2)
For coefficient of x-18, put
42 – 5r = 18
⇒ r=12
Substituting r = 12 in (1), we get
T13 = 14C2 . 22 . 312 . x-18
∴ Coefficient of x-18 is 14C12 . 22 . 312 ………….(3)
Hence sum of the coefficients of x32 and x-18 is
14C2 . 212 . 32 + 14C12 . 22 . 312
= 14C2 . 22 . 32 (210 + 310)
= 91 × 36(210 + 310).

TS Board Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(a)

Question 6.
If P and Q are the sum of odd terms and the sum of even terms respectively in the expansion of (x + a)n then prove that
i) P2 – Q2 = (x2 – a2)n
ii) 4PQ (x + a)2n – (x – a)2n
Solution:
We know that,
(x + a)n = nC0 xn + nC1 xn-1 . a + nC2 xn-2 . a2 + ………….. + nCn-1 x an-1 + nCn an
= (nC0 xn + nC2 xn-2 a2 + nC4 xn-4 a4 + …………) + (nC1 xn-1 a + nC3 xn-3 a3 + nC5 xn-5 a5 + …………..)
∴ (x + a)n = P + Q
∴ Sum of odd terms,
P = nC0 xn + nC2 xn-2 a2 + nC4 xn-4 a4 + …………
Sum of even terms,
Q = nC1 xn-1 a + nC3 xn-3 a3 + nC5 xn-5 a5 + …………..
∴ (x + a)n = P + Q
We know that
(x – a)n = nC0 xn a – nC1 xn-1 a + nC2 xn-2 a12nC3 xn-3 a3 + ……………. + nCn (- 1)n an
= (nC0 xn + nC2 xn-2 a2 + nC4 xn-4 a4 + …………) – (nC1 xn-1 a + nC3 xn-3 a3 + nC5 xn-5 a5 + …………..)
⇒ (x – a)n = P – Q

i) P2 – Q2 = (P + Q) (P – Q)
= (x + a)n (x – a)n
⇒ P2 – Q2 = (x2 – a2)n

ii) 4PQ = (P + Q)2 – (P – Q)2
= [(x + a)n]2 – [(x – a)n]2
⇒ 4PQ = (x + a)2n – (x – a)2n

Question 7.
If the coefficients of 4 consecutive terms in the expansion of (1 + x)n are a1, a2, a3, a4 respectively, then show that \(\frac{a_1}{a_1+a_2}+\frac{a_3}{a_3+a_4}=\frac{2 a_2}{a_2+a_3}\)
Solution:
Given a1, a2, a3, a4 are the coefficients of 4 consecutive terms in (1 + x)n respectively.
Let a1 = nCr-1
a2 = nCr
a3 = nCr+1
a4 = nCr+2

TS Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(a) 11

TS Board Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(a)

Question 8.
Prove that (2nC0)2 – (2nC1)2 + (2nC2)2 – (2nC3)2 + ……….. + (2nC2n)2 = (- 1)n 2nCn.
Solution:
We know that
(x + 1)2n = 2nC0 x2n + 2nC1 x2n-1 + 2nC2 x2n-2 + …………… + 2nC2n ……………(1)
(1 – x)2n = 2nC02nC1 x + 2nC2 x2 – …………… + 2nC2n x2n ……………(2)
From (1) and (2),
(1 – x)2n (1 + x)2n = (2nC02nC1 x + 2nC2 x2 + ……………… + 2nC2n x2n)
⇒ (1 – x2)2n = [2nC0 x2n + 2nC1 x2n-1 + 2nC2 x2n-2 + …………… + 2nC2n] . [2nC02nC1 x + 2nC2 x2 + …………… + 2nC2n x2n]
Equating coefficient of x2n on both sides, we get
(- 1)n . 2nCn = (2nC0)2 + ((2nC1)2 + (2nC2)2 – (2nC3)2 + ………….. + (2nC2n)2

Question 9.
Prove that
(C0 + C1) (C1 + C2) (C2 + C3) ………….. (Cn-1 + Cn) = \(\frac{(\mathbf{n}+1)^{\mathbf{n}}}{\mathbf{n} !}\) . C0 . C1 . C2 ………….. Cn.
Solution:
Given L.H.S
(C0 + C1) (C1 + C2) (C2 + C3) ………….. (Cn-1 + Cn)

TS Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(a) 12

= R.H.S
(C0 + C1) (C1 + C2) (C2 + C3) ………….. (Cn-1 + Cn) = \(\frac{(\mathbf{n}+1)^{\mathbf{n}}}{\mathbf{n} !}\) . C0 . C1 . C2 ………….. Cn.

TS Board Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(a)

Question 10.
Find the term independent of x in (1 + 3)n (1 + \(\frac{1}{3 x}\))n.
Solution:
We know that (1 + 3)n (1 + \(\frac{1}{3 x}\))n = (\(\frac{1}{3 x}\))n (1 + 3x)2n
= \(\frac{1}{3^n \cdot x^n} \sum_{r=0}^{2 n}\left({ }^{2 n} C_r\right)(3 x)^r\)
For the term, independent of x put r = n.
∴ The term independent of x in (1 + 3)n (1 + \(\frac{1}{3 x}\))n is \(\frac{1}{3^n}\) 2nCn . 3n.

Question 11.
Show that the middle term In the expansion of (1 + x)2n is \(\frac{1 \cdot 3 \cdot 5 \ldots \ldots(2 n-1)}{n !}\) (2x)n.
Solution:
The expansion of (1 + x)2n contains 2n + 1 terms.
∴ Middle term is (n + 1)th term

TS Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(a) 13

TS Board Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(a)

Question 12.
If (1 + 3x – 2x2)10 = a0 + a1x + a2x2 …………… + a20x20, then prove that
i) a0 + a1 + a2 + …………… + a20 = 210
ii) a0 – a1 + a2 – a3 + …………. + a20 = 410
Solution:
Given
(1 + 3x – 2x2)10 = a0 + a1x + a2x2 …………… + a20x20 …………….(1)
i) Put x = 1, in (1), we get
(1 + 3 – 2)10 = a0 + a1 + a2 + …………… + a20
⇒ a0 + a1 + a2 + …………… + a20 = 210.

ii) Put x – 1 in (1), we get
(1 – 3 – 2)10 = a0 – a1 + a2 – a3 + …………. + a20
a0 – a1 + a2 – a3 + …………. + a20 = 410.

Question 13.
If (3√3 + 5)2n + 1 = x and f = x – [x] (where [x] the integral part of x), find the value of x.f.
Solution:
Given (3√3 + 5)2n + 1 = x
and f = x – [x]
∴ 0 < f < 1
Let us consider F = (3√3 + 5)2n + 1
We know that
5 < 3√3 < 6 = 0 < 3√3 – 5 < 1
⇒ 0 < (3√3 – 5)2n + 1 < 1
⇒ 0 < F < 1
⇒ – 1 < – F < 0
Let x = I + f (where E is an integer)
Now
i + f – F = (3√3 + 5)2n + 1 + (3√3 – 5)2n + 1
= [2n + 1C0 (3√3)2n+1 + 2n + 1C1 (3√3)2n . 5 + 2n + 1C2 (3√3) . 52 + …………..] – [2n + 1C (3√3)2n+12n + 1C1 (3√3)2n . 5 + 2n + 1C2 (3√3)2n-1 . 52 – …………]
= 2 [2n + 1C1 (3√3) 5 + 2n + 1C3 (3√3)2n-2 53 + …………. + (2n + 1)C(2n+1) (5)2n+1]
= 2k, where k is an integer
∴ I + f – F is an even integer
⇒ f – F is also integer.
∴ (1) + (2)
⇒ – 1 < f – F < 1
⇒ f – F = 0
⇒ f = F
∴ x . f = x . F
= (3√3 + 5)2n + 1 . (3√3 – 5)2n + 1
= (27 – 25)2n + 1
= 22n + 1.

TS Board Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(a)

Question 14.
If R, n are posilve integers, n is odd, 0< F < 1 and if (5√5 + 11) = R + F, then prove that
i) R is an even integer and
ii) (R + F). F = 4n
Solution:
Given R, n are positive integers, 0 < F < 1 and
(5√5 + 11)n = R + F …………….(1)
i) Consider (5√5 – 11)n = f …………….(2)
We know that 11 < 5√5 < 12
⇒ 0 < 5√5 – 11 < 1
⇒0 < (5√5 – 11)n < 1
⇒ 0 < f < 1
⇒ – 1 < – f < 0 ………….(3)
From (1) and (3)
– 1 < F -f < 1 ……………(4)
Also R + F – f
= (5√5 + 11)n – (5√5 – 11)n
= [nC0 (5√5)n + nC1 (5√5)n-1 (11) + nC2 (5√5)n-2 . 112 + nC3 (5√5)n-3 . 113 + ………… + nCn (11)n] – [nC0 (5√5)nnC1 (5√5)n-1 (11) + nC2 (5√5)n-2 . 112nC3 (5√5)n-3 . 113 + ………….. + (- 1)nCn (11)n]
= 2 [nC1 (5√5)n-1 (11) + nC3 (5√5)n-3 (11)3 + ……………]
= 2k, where k is an integer (∵ f is odd)
∴ R + F – f is an even Integer.
⇒ F – f is also an interger. (∵ R is integer)
From (4), we have F – f = 0
⇒ F = f
∴ R is an even Integer.

ii) Now (R + F) . F
= (5√5 + 11)n (5√5 + 11)n
= ((5√5)2 – 112)n = 4n
∴ (R + F) . F = 4n.

Question 15.
If I, n are positive Integers, 0 < f < 1 and if (7 + 4√3) = I + f, then show that
i) I is an odd integer and
ii) (I + f) (I – f) =
Solution:
Given I and n are positve integers.
0 < f < I and (7 + 4√3)n = I + f ………….(1)

i) Let us consider (7 – 4√3)n = F
We know that 6 < 4√3 < 7
⇒ – 7 < – 4√3 < – 6
⇒ 0 < 7 – 4√3 < 1
⇒ 0 < (7 – 4√3)n < 1
⇒ 0 < F < 1
we have 0 < F + f < 2 …………(2)
Now I + f + F = (7 + 4√3)n + (7 – 4√3)n
= [nC0 7n + nC1 7n-1 (4√3) + nC2 7n-2 (4√3)2 + ………….. + nCn (4√3)n] + [nC0 7nnC1 7n-1 (4√3) + nC2 7n-2 (4√3)2 – ………….. + nCn (- 1)n (4√3)n]]
= 2[nC0 7n + nC2 7n-2 (4√3)2 + nC4 7n-4 (4√3)4 + ……………]
= 2k, where k is an integer.
∴ I + f + F is an even integer.
∴ f + F is also an integer. (∵ I is an integer)
from (2) we have f + F = I ………..(3)
∴ I + I is an even integer
⇒ I is an odd integer.

ii) Also (I + f) (I – f) = (I + f) F
= (7 + 4√3)n (7 – 4√3)n
= (72 – (4√3)2)n = 1.

TS Board Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(a)

Question 16.
If n is a Positive integer, prove that \(\sum_{r=1}^n \mathbf{r}^3 \cdot\left(\frac{{ }^n C_r}{{ }^n C_{r-1}}\right)^2=\frac{n(n+1)^2(n+2)}{12}\).
Solution:
We know that \(\frac{{ }^{{ }^r} C_r}{{ }^n C_{r-1}}=\frac{n-r+1}{r}\)

TS Inter 2nd Year Maths 2A Solutions Chapter 6 Binomial Theorem Ex 6(a) 14

Question 17.
Find the number of irrational terms in the expansion of (51/6 + 21/8)100.
Solution:
Number of terms in the expansion of (51/6 + 21/8)100 are 101.
General term in the expansion of (x + y)n is
Tr+1 = nCr xn-r . yr
∴ General term in the expansion of (51/6 + 21/8)100
Tr+1 = 100Cr . \(\left(5^{\frac{1}{6}}\right)^{100-r} \cdot\left(2^{\frac{1}{8}}\right)^r\)
= 100Cr . \(\cdot 5^{\frac{100-r}{6}} \cdot 2^{\frac{r}{8}}\)
For Tr+1 to be a rational.
Clearly ‘r’ is a multiple of 8 and 100 – r is a multiple of 6
∴ r = 16, 40, 64. 88.
Number of rational terms are 4.
∴ Number of irrational terms are 101 – 4 = 97.

TS 10th Class Maths Solutions Chapter 9 Tangents and Secants to a Circle Ex 9.3

Students can practice 10th Class Maths Study Material Telangana Chapter 9 Tangents and Secants to a Circle Ex 9.3 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 9 Tangents and Secants to a Circle Exercise 9.3

Question 1.
A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding : (use π = 3.14)
(i) Minor segment
(ii) Major segment. (A.P. Mar. ’16, June ’15)
Solution:
TS 10th Class Maths Solutions Chapter 9 Tangents and Secants to a Circle Ex 9.3 1
Given :
Angle subtended by the chord = 90°
Radius of the circle = 10 cm
Area of the minor segment = Area of the sector POQ – Area of ∆POQ
Area of the sector = \(\frac{\mathrm{x}^{\circ}}{360^{\circ}}\) × πr2
= \(\frac{90}{360}\) × 3.14 × 10 × 10 = 78.5
Area of the triangle POQ
= \(\frac{1}{2}\) × Base × Height
= \(\frac{1}{2}\) × 10 × 10 = 50
∴ Area of the minor segement
⇒ 78.5 – 50 = 28.5 cm2
Area of the major segment = Area of the circle – Area of the minor segment
⇒ 3.14 × 10 × 10 – 28.5
⇒ 314 – 28.5 cm2
⇒ 285.5 cm2

Question 2.
A chord of a circle of radius 12 cm. subtends an angle of 120° at the centre. Find the area of the corresponding minor segment of the circle, (use π = 3.14 and \(\sqrt{3}\) = 1.732).
Solution:
Radius of the circle r = 12 cm
Area of the sector = \(\frac{\mathrm{x}^{\circ}}{360^{\circ}}\) πr2
Here x = 120°
TS 10th Class Maths Solutions Chapter 9 Tangents and Secants to a Circle Ex 9.3 2
= \(\frac{120^{\circ}}{360^{\circ}}\) × 3.14 × 12 × 12
= 150.72
Drop a perpendicular from ‘O’ to the chord ‘PQ’
∆OPM = ∆OQM [∵ OP = OQ, ∠P = ∠Q; angles opposite to equal sides OP & OQ, ∠OMP = ∠OMQ by A.A.S]
∴ ∆OPQ = ∆OPM + ∆OQM
= 2(∆OPM)
Area of ∆OPM = \(\frac{1}{2}\) × PM × OM
But cos 30° = \(\frac{\mathrm{PM}}{\mathrm{OP}}\)
[∴ In ∆OPQ ∠POQ = 120° ∠OPQ = ∠OQP = \(\frac{180-120^{\circ}}{2}\) = 120°]
∴ PM = \(\frac{12 \times \sqrt{3}}{2}\) = 6\(\sqrt{3}\)
Also sin 30° = \(\frac{\mathrm{OM}}{\mathrm{OP}}\)
⇒ \(\frac{1}{2}\) = \(\frac{\mathrm{OM}}{12}\) ⇒ OM = \(\frac{12}{2}\) = 6
∴ ∆OPM = \(\frac{1}{2}\) × 6\(\sqrt{3}\) × 6
= 18 × 1.732 = 31.176 cm2
∴ ∆OPQ = 2 × 31.176 = 62.352 cm2
Area of the minor segment PQ = (Area of the sector) – (Area of the ∆OPQ)
= 150.72 – 62.352 = 88.368 cm2

TS 10th Class Maths Solutions Chapter 9 Tangents and Secants to a Circle Ex 9.3

Question 3.
A car has two wipers which do not overlap. Each wiper has a blade of length 25cm sweeping through an angle of 115°. Find the total area cleaned at each sweep of the blades (use π = \(\frac{22}{7}\))
Solution:
TS 10th Class Maths Solutions Chapter 9 Tangents and Secants to a Circle Ex 9.3 3
Angle made by the each blade = 115°
It is evident that each wiper sweeps a sector of a circle of radius 25cm and sector angle 115°.
Area of a sector = \(\frac{\theta^{\circ}}{360^{\circ}}\) × πr2
Total area cleaned at each sweep of the blades
= 2 × \(\frac{\theta^{\circ}}{360^{\circ}}\) × πr2
= 2 × \(\frac{115^{\circ}}{360^{\circ}}\) × \(\frac{22}{7}\) × 25 × 25
= 2 × \(\frac{23}{72}\) × \(\frac{22}{7}\) × 25 × 25
= \(\frac{23}{36}\) × \(\frac{22}{7}\) × 25 × 25
= 1254.96 cm2

Question 4.
Find the area of the shaded region in the figure, where ABCD is a square of side 10cm and semicircles are drawn with each side of the square as diameter. (use π = 3.14).
TS 10th Class Maths Solutions Chapter 9 Tangents and Secants to a Circle Ex 9.3 4
Solution:
Let us mark the four unshaded regions as I, II, III and IV.
TS 10th Class Maths Solutions Chapter 9 Tangents and Secants to a Circle Ex 9.3 5
Area of I + Area of III = Area of ABCD – Areas of 2 semicircles with radius 5 cm.
= 10 × 10 – 2 × \(\frac{1}{2}\) × 7π × 52
= 100 – 3.14 × 25
⇒ 100 – 78.5 ⇒ 21.5 cm2
Similarly, Area of II + Area of IV = 21.5 cm2
So, area of the shaded region
= are a of ABCD – Area of unshaded region
⇒ 100 – 2 × 21.5
= 100 – 43 = 57 cm2

TS 10th Class Maths Solutions Chapter 9 Tangents and Secants to a Circle Ex 9.3

Question 5.
Find the area of the shaded region in the figure, if ABCD is a square of side 7cm and APD and BPC are semi-circles. (use π = \(\frac{22}{7}\))
TS 10th Class Maths Solutions Chapter 9 Tangents and Secants to a Circle Ex 9.3 6
Solution:
Given,
ABCD is a square of side 7 cm
Area of the shaded region
= Area of ABCD – Area of 2 semi-circles with 7 radius (\(\frac{7}{2}\) = 3.5 cm)
APD and BPC are semi-circles 1 22
= 7 × 7 – 2 × \(\frac{1}{2}\) × \(\frac{22}{7}\) × 3.5 × 3.5
= 49 – 38.5 = 10.5 cm2
∴ Area of shaded region = 10.5 cm2

Question 6.
In figure, OACB is a quadrant of a circle with centre ‘O’ and radius 3.5 cm. If OD = 2 cm, find the area of the shaded region, (use π = \(\frac{22}{7}\)).
TS 10th Class Maths Solutions Chapter 9 Tangents and Secants to a Circle Ex 9.3 7
Solution:
Given OACB is a quadrant of a circle radius 3.5 cm OD = 2 cm
Area of the shaded region = Area of the sector – Area of ABOD
= \(\frac{\mathrm{x}^{\circ}}{360^{\circ}}\) × πr2 – \(\frac{1}{2}\) OB . OD
= \(\frac{99^{\circ}}{360^{\circ}}\) × \(\frac{22}{7}\) × 3.5 × 3.5 – \(\frac{1}{2}\) × 3.5 × 2
= \(\frac{1}{4}\) × \(\frac{22}{7}\) × 3.5 × 3.5 – 3.5
= \(\frac{1}{4}\) × \(\frac{22}{7}\) × 12.25 – 3.5
= 9.625 – 3.5 = 6.125 cm2
Area of shaded region = 6.125 cm2

TS 10th Class Maths Solutions Chapter 9 Tangents and Secants to a Circle Ex 9.3

Question 7.
AB and CD are respectively arcs of two concentric circles of radii 21cm and 7cm with centre ‘O’ (see figure) If ∠AOB = 30°, find the area of the shaded region. (Use π = \(\frac{22}{7}\))
TS 10th Class Maths Solutions Chapter 9 Tangents and Secants to a Circle Ex 9.3 8
Solution:
Area of the shaded region
= Area of sector OAB – Area of the sector COD
= \(\frac{30^{\circ}}{360^{\circ}} \) × \(\frac{22}{7}\) × 21 × 21 – \(\frac{30^{\circ}}{360^{\circ}} \) × \(\frac{22}{7}\) × 7 × 7
= \(\frac{30^{\circ}}{360^{\circ}} \) × \(\frac{22}{7}\) (21 × 21 – 7 × 7)
= \(\frac{1}{12}\) × \(\frac{22}{7}\) (441 – 49)
= \(\frac{1}{6}\) × \(\frac{11}{7}\) × 392
= \(\frac{1}{6}\) × 11 × 56
= \(\frac{11 \times 28}{3}\)
= 102.67 cm2

TS 10th Class Maths Solutions Chapter 9 Tangents and Secants to a Circle Ex 9.3

Question 8.
Calculate the area of the designed region in figure, common between the two quadrants of the circles of radius 10 cm. each, (use π = 3.14)
TS 10th Class Maths Solutions Chapter 9 Tangents and Secants to a Circle Ex 9.3 9
Solution:
Radius of the circle (r) = 10 cm
Area of the designed region
= 2 [Area of quadrant ABYD – Area of ∆ABD]
= 2 [\(\frac{1}{4}\) × πr2 – \(\frac{1}{2}\) × Base × Height]
= 2 [(\(\frac{1}{4}\) × 3.14 × 10 × 10) – (\(\frac{1}{2}\) × 10 × 10)]
= 2 [78.5 – 50]
⇒ 2 × 28.5
= 57 cm2

TS 10th Class Maths Solutions Chapter 10 Mensuration Optional Exercise

Students can practice 10th Class Maths Study Material Telangana Chapter 10 Mensuration Optional Exercise to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 10 Mensuration Optional Exercise

Question 1.
A golf ball has diameter equal to 4.1cm. Its surface has 150 dimples each of radius 2mm. Calculate total surface area which is exposed to the surroundings (Assume that the dimples are all hemispherical) [π = \(\frac{22}{7}\)]
Solution:
Area exposed = surface area of the ball – total area of 150 dimples with radius 2 mm
= 4πr2 – 150 × πr2
= 4 × \(\frac{22}{7}\) × \(\frac{4.1}{2}\) × \(\frac{4.1}{2}\) – 150 × \(\frac{22}{7}\) × \(\frac{2}{10}\) × \(\frac{2}{10}\)
[∵ 2 mm = \(\frac{2}{10}\) cm]
= 52.831 – 18.85 = 33.972 cm2

TS 10th Class Maths Solutions Chapter 10 Mensuration Optional Exercise

Question 2.
A cylinder of radius 12 cm. contains water to a depth of 20 cm. when a spherical iron ball is dropped in to the cylinder and thus the level of water is raised by 6.75 cm. Find the radius of the ball. [π = \(\frac{22}{7}\)]
Solution:
Rise in the water level is seen as a cylinder of radius ‘r’ = r1 = 12 cm
Height, h = 6.75 cm
Volume of the rise = Volume of the spherical iron ball dropped
πr12h = \(\frac{4}{3}\) πr23
r12h = \(\frac{4}{3}\) π23
12 × 12 × 6.75 cm3 = \(\frac{4}{3}\) × r23 cm3
r23 = \(\frac{3}{4}\) × 12 × 12 × 6.75
= 9 × 12 × 6.75
= 108 × 5.75
r23 = 729
r23 = 9 × 9 × 9
∴ 729 = (3 × 3) × (3 × 3) × (3 × 3)
∴ Radius of the ball r = r2 = 9 cm.
TS 10th Class Maths Solutions Chapter 10 Mensuration Optional Exercise 1

Question 3.
A solid toy is in the form of a right circular cylinder with a hemispherical shape at one end and a cone at the other end. Their common diameter is 4.2 cm. and height of the cylindrical and conical portion are 12 cm. and 7 cm. respectively. Find the volume of the solid toy. [π = \(\frac{22}{7}\)]
Solution:
TS 10th Class Maths Solutions Chapter 10 Mensuration Optional Exercise 2
Volume of the toy
= volume of the hemisphere + volume of the cylinder + volume of the cone.
= \(\frac{2}{3}\) πr3 + πr2h1 + \(\frac{1}{3}\) πr2h2
= πr2(\(\frac{2}{3}\)r + h1 + \(\frac{\mathrm{h}_2}{3}\))
= \(\frac{22}{7}\) × \(\frac{4.2}{2}\) × \(\frac{4.2}{2}\)[\(\frac{2}{3}\) × \(\frac{4.2}{2}\) + 12 + \(\frac{7}{3}\)]
= 11 × 0.6 × 2.1 [1.4 + 12 + \(\frac{7}{3}\)]
= 13.86[13.4 + \(\frac{7}{3}\)]
= 13.86 × \(\left[\frac{40.2+7}{3}\right]\)
= \(\frac{13.86 \times 47.2}{3}\) = 218.064 cm3

TS 10th Class Maths Solutions Chapter 10 Mensuration Optional Exercise

(or)

Hemisphere :
Radius = \(\frac{\text { diameter }}{2}\) = \(\frac{4.2}{2}\) = 2.1 cm
V = \(\frac{2}{3}\) πr3 = \(\frac{2}{3}\) × \(\frac{22}{7}\) × 2.1 × 2.1 × 2.1
= 19.404 cm3

Cylinder :
Radius, r = \(\frac{\mathrm{d}}{2}\) = \(\frac{4.2}{2}\) = 2.1 cm
height, h = 12 cm
V = πr2h = \(\frac{22}{7}\) × 2.1 × 2.1 × 12
= 166.32 cm3

Cone :
Radius, r = \(\frac{\mathrm{d}}{2}\) = \(\frac{4.2}{2}\) = 2.1cm
Height, h = 7 cm
V = \(\frac{1}{3}\) πr2h = \(\frac{1}{3}\) × \(\frac{22}{7}\) × 2.1 × 2.1 × 7
= 32.34 cm3
∴ Total volume = 19.404 + 166.32 + 32.34
= 218.064 cm3.

Question 4.
Three metal cubes with edges 15 cm., 12 cm. and 9 cm. respectively are melted together and formed into a single cube. Find the diagonal of this cube.
Solution:
Edges l1 = 15 cm, l2 = 12 cm, l13 = 9 cm.
TS 10th Class Maths Solutions Chapter 10 Mensuration Optional Exercise 3
Volume of the resulting cube = Sum of the volumes of the three given cubes
L3 = l31h
L3 = l13 + l23 + l33
L3 = 153 + 123 + 93
L3 = 3375 + 1728 + 729
L3 = 5832 = 18 × 18 × 18
∴ Edge of the new cube l = 18 cm
Diagonal = \(\sqrt{3 l^2}\) = \(\sqrt{3 \times 18^2}\)
= \(\sqrt{3 \times 324}\) = \(\sqrt{972}\) = 31.176 cm.

TS 10th Class Maths Solutions Chapter 10 Mensuration Optional Exercise

Question 5.
A hemi-spherical bowl of internal diameter 36 cm. contains a liquid. This liquid is to be filled in cylindrical bottles of radius 3 cm. and height 6 cm. How many bottles are required to empty the bowl ?
Solution:
Let the number of bottles required = n.
Then total valume of a ’n’ bottles = volume of the hemispherical bowl.
n. πr12h = \(\frac{2}{3}\) πr22h
Bottle :
Radius, r1 = 3 cm
Height, h = 6 cm
Volume, V = πr12h
= \(\frac{22}{7}\) × 3 × 3 × 6 = \(\frac{1188}{7}\)
∴ Total volume of n bottles = n × \(\frac{1188}{7}\) cm3.
Bowl :
TS 10th Class Maths Solutions Chapter 10 Mensuration Optional Exercise 4
∴ 72 bottles are required to empty the bowl.