Mahesh – Page 23 – TS Board Solutions

TS 10th Class Maths Solutions Chapter 3 Polynomials Optional Exercise

February 21, 2024February 20, 2024 by Mahesh

Students can practice TS 10th Class Maths Solutions Chapter 3 Polynomials Optional Exercise to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 3 Polynomials Optional Exercise

Question 1.
Verify that the numbers given along side the cubic polynomials below are their zeroes. Also verify the relationship between the zeros and the coefficients in each case :
i) 2x³ + x² – 5x + 2;(\(\frac{1}{2}\), 1, -2)
ii) x³ + 4x² + 5x – 2; (2, 1, 1)
Solution:
i) Given polynomial is 2x³ + x² – 5x + 2
Comparing the given polynomial with ax³ + bx² + cx + d,
We get a = 2, b = 1, c = -5 and d = 2
p(\(\frac{1}{2}\)) = 2(\(\frac{1}{2}\))³ + (\(\frac{1}{2}\))³ – 5(\(\frac{1}{2}\)) + 2
= \(\frac{1}{4}\) + \(\frac{1}{4}\) – \(\frac{5}{2}\) + \(\frac{2}{1}\)
= \(\frac{1+1-10+8}{4}\) = \(\frac{0}{4}\) = 0
P(1) = 2(1)³ + (1)² – 5(1) + 2
= 2 + 1 – 5 + 2 = 0
p(-2) = 2(-2)³ + (-2)² – 5(-2) + 2
= 2(-8) + 4 + 10 + 2
= -16 + 16 = 0
∴ \(\frac{1}{2}\), 1, and -2 are the zeroes of 2x³ + x² – 5x + 2
So, α = \(\frac{1}{2}\), β = 1 and γ = -2
Therefore,
α + β + γ = \(\frac{1}{2}\) + 1 + (-2)
= \(\frac{1+2-4}{2}\) = \(\frac{-1}{2}\) = \(\frac{-b}{a}\)
αβ + βγ + γα = (\(\frac{1}{2}\))(1) + (1)(-2) + -2(\(\frac{1}{2}\))
= \(\frac{1}{2}\) – 2 – 1
= \(\frac{1-4-2}{2}\) = \(\frac{-5}{2}\) = \(\frac{c}{a}\)
And αβγ = \(\frac{1}{2}\) × 1 × (-2) = -1 = \(\frac{-2}{2}\) = \(\frac{-\mathrm{d}}{\mathrm{a}}\)

ii) Given polynomial is x³ – 4x² + 5x – 2
Comparing the given polynomial with ax³ + bx² + cx + d, We get a = 1, b = -4, c = 5 and d = -2
p(2) = (2)³ – 4(2)² + 5(2) – 2
= 8 – 16 + 10 – 2 = 0
p(1) = (1)³ – 4(1)² + 5(1) – 2
= 1 – 4 + 5 – 2 = 0
∴ 2, 1 and 1 are the zeroes of x³ – 4x² + 5x – 2.
So, α = 2, β = 1 and γ = 1
Therefore,
α + β + γ = 2 + 1 + 1
= 4 = \(\frac{-(-4)}{1}\) = \(\frac{-b}{a}\)
αβ + βγ + γα = 2(1) + (1)(1) + (1) (2)
= 2 + 1 + 2 = 5
= \(\frac{5}{1}\) = \(\frac{c}{a}\)
and
αβγ = (2) (1) (1) = 2 = \(\frac{-(-2)}{1}\) = \(\frac{-\mathrm{d}}{\mathrm{a}}\)

Question 2.
Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time and the product of its zeroes as 2, -7, -14 respectively.
Answer:
Let the cubic polynomial be
ax³ + bx² + cx + d and its zeroes be α, β and γ.
Then
α + β + γ = 2 = \(\frac{-(-2)}{1}\) = \(\frac{-b}{a}\)
αβ + βγ + γα = -7 = \(\frac{-7}{1}\) = \(\frac{c}{a}\)
and αβγ = -14 = \(\frac{-14}{1}\) = \(\frac{-d}{a}\)
∴ a = 1, b = -2, c = -7 and d = 14.
So, one cubic polynomial which satisfies the given conditions will be
x³ – 2x² – 7x + 14

Question 3.
If the zeroes of the polynomial x³ – 3x² + x + 1 are a – b, a, a + b, find a and b.
Answer:
Given polynomial is x³ – 3x² + x + 1
Since, (a – b), a, (a + b) are the zeroes of the polynomial x³ – 3x² + x + 1.
Therefore, sum of the zeroes = (a – b) + a + (a + b)
⇒ 3a = 3 ⇒ a = 1
= \(\frac{-(-3)}{1}\) = 3
∴ Sum of the products of its zeroes taken two at a time.
= a(a – b) +a(a + b) + (a + b) (a – b)
= \(\frac{1}{1}\) = 1
⇒ a² – ab + a² + ab + a² – b² = 1
⇒ 3a² – b² = 1
So, 3(1)² – b² = 1
⇒ 3 – b² = 1
⇒ b² = 2 ⇒ b = \(\sqrt{2}\) = ±\(\sqrt{2}\)
Here, a = 1 and b = ±\(\sqrt{2}\)

Question 4.
If two zeroes of the polynomial x⁴ – 6x³ – 26x² + 138x – 35 are 2 ± \(\sqrt{3}\) , find other zeroes.
Answer:
Given polynomial is
x⁴ – 6x³ – 26x² + 138x – 35
We have, 2 ± \(\sqrt{3}\) are two zeroes of the polynomial
p(x) = x⁴ – 6x³ – 26x² + 138x – 35
Let x = 2 ± \(\sqrt{3}\)
So, x – 2 = ±\(\sqrt{3}\)
On squaring, we get x² – 4x + 4 = 3, i.e., x² – 4x + 1 = 0
Let us divide p(x) by x² – 4x + 1 to obtain other zeroes
TS 10th Class Maths Solutions Chapter 3 Polynomials Optional Exercise 9
p(x) = x⁴ – 6x³ – 26x² + 138x – 35
= (x² – 4x + 1) (x² – 2x – 35)
= (x² – 4x + 1) (x² – 7x + 5x – 35)
= (x² – 4x + 1) [x(x – 7) +5(x – 7)]
= (x² – 4x + 1) (x + 5) (x – 7)
So, (x + 5) and (x – 7) are other factors of p(x)
So, -5 and 7 are other zeroes of the given polynomial.

Question 5.
If the polynomial
x⁴ – 6x³ – 16x² + 25x + 10 is divided by another polynomial x² – 2x + k, the remainder comes out to be x + a, find k and a.
Answer:
Given polynomial is
x⁴ – 6x³ + 16x² – 25x + 10 and another polynomial is x² – 2x + k
Remainder is x + a
Let us divide
x⁴ – 6x³ + 16x² – 25x + 10 by x² – 2x + k.
TS 10th Class Maths Solutions Chapter 3 Polynomials Optional Exercise 10
∴ Remainder = (2k – 9)x – (8 – k)k + 10
But the remainder is given as x + a.
On comparing their coefficients we have
2k – 9 = 1 ⇒ 2k = 10
⇒ k = 5 and -(8 – k) k + 10 = a
So, a = -(8 – 5) 5 + 10 = -3 × 5 + 10
= -15 + 10 = -5
Hence, k = 5 and a = -5

TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b)

February 21, 2024February 20, 2024 by Mahesh

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

TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Exercise 6(b)

I. Evaluate the following integrals.

Question 1.
∫e^2x dx, x ∈ R
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) I Q1

Question 2.
∫sin 7x dx, x ∈ R
Solution:
Let 7x = t then 7 dx = dt
⇒ dx = \(\frac{1}{7}\) dt
∴ ∫sin 7x dx = \(\frac{1}{7}\) ∫sint dt
= \(-\frac{1}{7}\) cos t
= \(-\frac{1}{7}\) cos 7x + c

Question 3.
\(\int \frac{x}{1+x^2} d x\), x ∈ R
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) I Q3

Question 4.
∫2x sin(x² + 1) dx, x ∈ R
Solution:
Let x² + 1 = t then 2x dx = dt
∴ ∫2x sin(x² + 1) dx = ∫sin t dt
= -cos t + c
= -cos(x² + 1) + c

Question 5.
\(\int \frac{(\log x)^2}{x} d x\) on I ⊂ (0, ∞).
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) I Q5

Question 6.
\(\int \frac{e^{Tan^{-1} x}}{1+x^2} d x\) on I ⊂ (0, ∞).
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) I Q6

Question 7.
\(\int \frac{\sin \left({Tan}^{-1} x\right)}{1+x^2} d x\), x ∈ R
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) I Q7

Question 8.
\(\int \frac{1}{8+2 x^2} d x\) on R.
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) I Q8

Question 9.
\(\int \frac{3 x^2}{1+x^6} d x\) on R.
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) I Q9

Question 10.
\(\int \frac{2}{\sqrt{25+9 x^2}} d x\) on R.
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) I Q10

Question 11.
\(\int \frac{3}{\sqrt{9 x^2-1}} d x\) on (\(\frac{1}{3}\), ∞).
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) I Q11

Question 12.
∫sin mx cos nx dx on R, m ≠ n, m and n are positive integers.
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) I Q12

Question 13.
∫sin mx sin nx dx on R, m ≠ n, m and n are positive integers.
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) I Q13

Question 14.
∫cos mx cos nx dx on R, m ≠ n, m and n are positive integers.
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) I Q14

Question 15.
∫sin x sin 2x sin 3x dx on R.
Solution:
Consider sin x sin 2x sin 3x = \(\frac{1}{2}\) (2 sin x sin 2x sin 3x)
= \(\frac{1}{2}\) [cos(3x – 2x) – cos(3x + 2x)] sin x
= \(\frac{1}{2}\) [sin x cos x – sin x cos 5x]
= \(\frac{1}{4}\) [2 sin x cos x – 2 sin x cos 5x]
= \(\frac{1}{4}\) [sin 2x – [sin(5x + x) + sin(x – 5x)]
= \(\frac{1}{4}\) [sin 2x – [sin 6x – sin 4x]]
= \(\frac{1}{4}\) [sin 2x – sin 6x + sin 4x]
∴ ∫sin x sin 2x sin 3x dx = \(\frac{1}{4}\) ∫sin 2x dx – \(\frac{1}{4}\) ∫sin 6x dx + \(\frac{1}{4}\) ∫sin 4x dx
= \(-\frac{1}{8}\) cos 2x + \(\frac{1}{24}\) cos 6x – \(\frac{1}{16}\) cos 4x + c
= \(\frac{1}{4}\left[\frac{\cos 6 x}{6}-\frac{\cos 4 x}{4}-\frac{\cos 2 x}{2}\right]\) + c

Question 16.
\(\int \frac{\sin x}{\sin (a+x)} d x\) on I ⊂ R – {nπ – a : n ∈ Z}.
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) I Q16

II. Evaluate the following integrals.

Question 1.
\(\int(3 x-2)^{\frac{1}{2}} d x\) on (\(\frac{2}{3}\), ∞)
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) II Q1

Question 2.
\(\int \frac{1}{7 x+3} d x\) on I ⊂ R – {\(-\frac{3}{7}\)}
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) II Q2

Question 3.
\(\int \frac{\log (1+x)}{1+x} d x\) on (-1, ∞).
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) II Q3

Question 4.
∫(3x² – 4)x dx on R.
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) II Q4

Question 5.
\(\int \frac{d x}{\sqrt{1+5 x}} \text { on }\left(-\frac{1}{5}, \infty\right)\)
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) II Q5

Question 6.
∫(1 – 2x³) x² dx on R.
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) II Q6

Question 7.
\(\int \frac{\sec ^2 x}{(1+\tan x)^3} d x\) on I ⊂ R – {nπ – \(\frac{\pi}{4}\) : n ∈ Z}
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) II Q7

Question 8.
∫x³ sin(x⁴) dx on R.
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) II Q8

Question 9.
\(\int \frac{\cos x}{(1+\sin x)^2} d x\) on I ⊂ R – {2nπ + \(\frac{3 \pi}{2}\) : n ∈ Z}
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) II Q9

Question 10.
∫\(\sqrt[3]{\sin x}\) cos x dx on [2nπ, (2n + 1)π], (n ∈ Z).
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) II Q10

Question 11.
∫2x \(e^{x^2}\) dx on R.
Solution:
Let x² = t then 2x dx = dt
∴ ∫2x \(e^{x^2}\) dx = ∫e^t dt
= e^t + c
= \(e^{x^2}\) + c

Question 12.
\(\int \frac{e^{\log x}}{x} d x\) on (0, ∞).
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) II Q12

Question 13.
\(\int \frac{x^2}{\sqrt{1-x^6}} d x\) on I ∈ (-1, 1)
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) II Q13

Question 14.
\(\int \frac{2 x^3}{1+x^8} d x\) on R.
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) II Q14

Question 15.
\(\int \frac{x^8}{1+x^{18}} d x\) on R. (Mar. ’09)
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) II Q15

Question 16.
\(\int \frac{e^x(1+x)}{\cos ^2\left(x e^x\right)} d x\) on I ⊂ R – {x ∈ R : cos(xe^x) = 0}. (Mar. ’10, ’04)
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) II Q16

Question 17.
\(\int \frac{{cosec}^2 x}{(a+b \cot x)^5} d x\) on I ⊂ R – {x ∈ R : a + b cot x = 0}, where a, b ∈ R, b ≠ 0.
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) II Q17

Question 18.
∫e^x sin e^x dx on R.
Solution:
Let t = e^x then dt = e^x dx
∴ ∫e^x sin e^x dx = ∫sin t dt
= -cos t + c
= -cos(e^x) + c

Question 19.
\(\int \frac{\sin (\log x)}{x} d x\) on (0, ∞).
Solution:
Let log x = t then \(\frac{1}{x}\) dx = dt
∴ \(\int \frac{\sin (\log x)}{x} d x\) = ∫sin t dt
= -cos t + c
= -cos(log x) + c

Question 20.
\(\int \frac{1}{x \log x} d x\) on (1, ∞).
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) II Q20

Question 21.
\(\int \frac{(1+\log x)^n}{x} d x\) on (e^-1, ∞), n ≠ -1.
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) II Q21

Question 22.
\(\int \frac{\cos (\log x)}{x} d x\) on (0, ∞).
Solution:
Let log x = t then \(\frac{1}{x}\) dx = dt
∴ \(\int \frac{\cos (\log x)}{x} d x\) = ∫cos t dt
= sin t + c
= sin(log x) + c

Question 23.
\(\int \frac{\cos \sqrt{x}}{\sqrt{x}} d x\) on (0, ∞).
Solution:
Let √x = t then \(\frac{1}{2 \sqrt{x}}\) dx = dt
∴ \(\int \frac{\cos \sqrt{x}}{\sqrt{x}} d x\) = 2 ∫cos t dt
= 2 sin t + c
= 2 sin(√x) + c

Question 24.
\(\int \frac{2 x+1}{x^2+x+1} d x\) on R.
Solution:
Let x² + x + 1 = t then (2x + 1) dx = dt
∴ \(\int \frac{2 x+1}{x^2+x+1} d x=\int \frac{d t}{t}\)
= log|t| + c
= log|x² + x + 1| + c

Question 25.
\(\int \frac{a x^{n-1}}{b x^n+c} d x\) where n ∈ N, a, b, c are real numbers, b ≠ 0 and x ∈ I ⊂ {x ∈ R : xⁿ ≠ \(-\frac{c}{b}\)}
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) II Q25

Question 26.
\(\int \frac{1}{x \log x[\log (\log x)]} d x\) on (1, ∞).
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) II Q26

Question 27.
∫coth x dx on R.
Solution:
∫coth x dx = \(\int \frac{\cosh x}{\sinh x} d x\)
Let sinh x = t then cosh x dx = dt
∴ ∫coth x dx = \(\int \frac{\mathrm{dt}}{\mathrm{t}}\)
= log|t| + c
= log|sinh x| + c

Question 28.
\(\int \frac{1}{\sqrt{1-4 x^2}} d x \text { on }\left(-\frac{1}{2}, \frac{1}{2}\right)\).
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) II Q28

Question 29.
\(\int \frac{d x}{\sqrt{25+x^2}}\) on R.
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) II Q29

Question 30.
\(\int \frac{1}{(x+3) \sqrt{x+2}}\) on I ⊂ (-2, ∞). (New Model Paper & May ’12)
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) II Q30

Question 31.
\(\int \frac{1}{1+\sin 2 x} d x\) on I ⊂ R – {\(\frac{n \pi}{2}+(-1)^n \frac{\pi}{4}\) : n ∈ Z}.
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) II Q31

Question 32.
\(\int \frac{x^2+1}{x^4+1} d x\) on R.
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) II Q32

Question 33.
\(\int \frac{d x}{\cos ^2 x+\sin 2 x}\) on I ⊂ R / ({(2n + 1)\(\frac{\pi}{2}\) : n ∈ Z} ∪ {2nπ + \({tan}^{-1} \frac{1}{2}\) : n ∈ Z})
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) II Q33

Question 34.
\(\int \sqrt{1-\sin 2 x} d x\) on I ⊂ [2nπ – \(\frac{3 \pi}{4}\), 2nπ + \(\frac{\pi}{4}\)], n ∈ Z.
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) II Q34

Question 35.
\(\int \sqrt{1+\cos 2 x} d x\) on I ⊂ [2nπ – \(\frac{\pi}{2}\), 2nπ + \(\frac{\pi}{2}\)], n ∈ Z.
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) II Q35

Question 36.
\(\int \frac{\cos x+\sin x}{\sqrt{1+\sin 2 x}} d x\) on I ⊂ [2nπ – \(\frac{\pi}{4}\), 2nπ + \(\frac{3 \pi}{4}\)], n ∈ Z.
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) II Q36

Question 37.
\(\int \frac{\sin 2 x}{(a+b \cos x)^2} d x\) on {R, if |a| > |b|, I ⊂ {x ∈ R : a + b cos x ≠ 0}, if |a| < |b|}.
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) II Q37

Question 38.
\(\int \frac{\sec x}{(\sec x+\tan x)^2} d x\) on I ⊂ R – {(2n + 1)\(\frac{\pi}{2}\), n ∈ Z}.
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) II Q38

Question 39.
\(\int \frac{d x}{a^2 \sin ^2 x+b^2 \cos ^2 x}\) on R, a ≠ 0, b ≠ 0.
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) II Q39

Question 40.
\(\int \frac{d x}{\sin (x-a) \sin (x-b)}\) on I ⊂ R – ({a + nπ : n ∈ Z} ∪ {b + nπ : n ∈ Z}).
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) II Q40

Question 41.
\(\int \frac{1}{\cos (x-a) \cos (x-b)} \mathbf{d x}\) on I ⊂ R – ({a + \(\frac{(2 n+1) \pi}{2}\) : n ∈ Z} ∪ {b + \(\frac{(2 n+1) \pi}{2}\) : n ∈ Z})
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) II Q41

III. Evaluate the following integrals.

Question 1.
\(\int \frac{\sin 2 x}{a \cos ^2 x+b \sin ^2 x} d x\) on I ⊂ R – {x ∈ R | a cos²x + b sin²x = 0}
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) III Q1

Question 2.
\(\int \frac{1-\tan x}{1+\tan x} d x\) for x ∈ I ⊂ R – {nπ – \(\frac{\pi}{4}\) : n ∈ Z}
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) III Q2

Question 3.
\(\int \frac{\cot (\log x)}{x} d x\), x ∈ I ⊂ (0, ∞) – {e^nπ : n ∈ Z}
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) III Q3

Question 4.
∫e^x cot e^x dx, x ∈ I ⊂ R – {log nπ : n ∈ Z}
Solution:
Let e^x = t then e^x dx = dt
∴ ∫e^x cot x dx = ∫cot t dt
= log|sin t| + c
= log|sin(e^x)| + c

Question 5.
∫sec(tan x) sec²x dx on I ⊂ {x ∈ E : tan x ≠ \(\frac{(2 k+1) \pi}{2}\) for any k ∈ Z), where E = R – {\(\frac{(2 n+1) \pi}{2}\), n ∈ Z}
Solution:
Let tan x = t, then sec²x dx = dt
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) III Q5

Question 6.
\(\int \sqrt{\sin x} \cos x d x\) on [2nπ, (2n + 1)π], (n ∈ Z).
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) III Q6

Question 7.
∫tan⁴x sec²x dx, x ∈ I ⊂ R – {\(\frac{(2 n+1) x}{2}\) : n ∈ Z}.
Solution:
Let tan x = t then sec²x dx = dt
∴ ∫tan⁴x sec²x dx = ∫t⁴dt
= \(\frac{t^5}{5}\) + c
= \(\frac{\tan ^5 x}{5}\) + c

Question 8.
\(\int \frac{2 x+3}{\sqrt{x^2+3 x-4}} d x\), x ∈ I ⊂ R – [-4, 1]
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) III Q8

Question 9.
\(\int {cosec}^2 x \sqrt{\cot x} d x\) on (0, \(\frac{\pi}{2}\)].
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) III Q9

Question 10.
∫sec x log(sec x + tan x) dx on (0, \(\frac{\pi}{2}\)).
Solution:
Let log(sec x + tan x) = t then \(\frac{1}{\sec x+\tan x}\) (sec x tan x + sec²x) dx = dt
⇒ \(\frac{\sec x(\sec x+\tan x) d x}{\sec x+\tan x}\) = dt
⇒ sec x dx = dt
∴ ∫sec x log(sec x + tan x) dx = ∫t dt + c
= \(\frac{\mathrm{t}^2}{2}\) + c
= \(\frac{1}{2}\) [log(sec x + tan x)]² + c

Question 11.
∫sin³x dx on R.
Solution:
sin 3x = 3 sin x – 4 sin³x
⇒ 4 sin³x = 3 sin x – sin 3x
⇒ sin³x = \(\frac{3 \sin x-\sin 3 x}{4}\)
∴ ∫sin³x dx = \(\frac{1}{2}\) ∫(3 sin x – sin 3x) dx
= \(\frac{3}{4}\)∫sin x dx – \(\frac{1}{4}\)∫sin 3x dx
= \(-\frac{3}{4}\) cos x + \(\frac{1}{12}\) cos 3x + c
= \(\frac{1}{12}\) (cos 3x – 9 cos x] + c

Question 12.
∫cos³x dx on R.
Solution:
cos 3x = 4 cos³x – 3 cos x
⇒ 4cos³x = cos 3x + 3 cos x
⇒ cos³x = \(\frac{1}{4}\)(cos 3x + 3 cos x)
∴ ∫cos3x dx = \(\frac{1}{4}\)∫(cos 3x + 3 cos x) dx
= \(\frac{1}{12}\) sin 3x + \(\frac{3}{4}\) sin x dx + c
= \(\frac{1}{12}\) [sin 3x + 9 sin x] + c

Question 13.
∫cos x cos 2x dx on R.
Solution:
We have cos A cos B = \(\frac{1}{2}\) [cos(A + B) + cos(A – B)]
⇒ cos x cos 2x = \(\frac{1}{2}\) [cos 3x + cos x]
∴ ∫cos x cos 2x dx = \(\frac{1}{2}\) [∫cos 3x + ∫cos x] dx
= \(\frac{1}{2}\left(\frac{1}{3}\right)\) sin 3x + \(\frac{1}{2}\) sin x + c
= \(\frac{1}{6}\) sin 3x + \(\frac{1}{2}\) sin x + c
= \(\frac{1}{6}\) [sin 3x + 3 sin x] + c

Question 14.
∫cos x cos 3x dx on R.
Solution:
cos x cos 3x = \(\frac{1}{2}\) [cos 4x + cos 2x]
∴ ∫cos x cos 3x dx = \(\frac{1}{2}\) ∫cos 4x dx + \(\frac{1}{2}\) ∫cos 2x dx
= \(\frac{1}{8}\) sin 4x + \(\frac{1}{4}\) sin 2x + c

Question 15.
∫cos⁴x dx on R.
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) III Q15

Question 16.
\(\int x \sqrt{4 x+3} d x\) on (\(-\frac{3}{4}\), ∞).
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) III Q16

Question 17.
\(\int \frac{d x}{\sqrt{a^2-(b+c x)^2}}\) on {x ∈ R : |b + cx| < a}, where a, b, c are real numbers c ≠ 0 and a > 0.
Solution:
Let b + cx = t then c dx = dt
⇒ dx = \(\frac{1}{c}\) dt
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) III Q17

Question 18.
\(\int \frac{d x}{a^2+(b+c x)^2}\) on R, where a, b, c are real numbers c ≠ 0 and a > 0.
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) III Q18

Question 19.
\(\int \frac{d x}{1+e^x}\), x ∈ R.
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) III Q19

Question 20.
\(\int \frac{x^2}{(a+b x)^2} d x\), x ∈ I ⊂ R – {\(-\frac{a}{b}\)}, where a, b are real numbers, b ≠ 0.
Solution:
Let a + bx = t then b dx = dt
⇒ dx = \(\frac{1}{b}\) dt
Also x = \(\frac{t-a}{b}\)
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) III Q20

Question 21.
\(\int \frac{x^2}{\sqrt{1-x}} d x\), x ∈ (-∞, 1).
Solution:
TS Inter 2nd Year Maths 2B Solutions Chapter 6 Integration Ex 6(b) III Q21

TS 10th Class Maths Solutions Chapter 3 Polynomials Ex 3.4

February 21, 2024February 20, 2024 by Mahesh

Students can practice TS 10th Class Maths Solutions Chapter 3 Polynomials Ex 3.4 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 3 Polynomials Exercise 3.4

Question 1.
Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following.
i) p(x) = x³ – 3x² + 5x – 3, g(x) = x² – 2
ii) p(x) = x⁴ – 3x² + 4x + 5, g(x) = x² + 1 – x
iii) p(x) = x⁴ – 5x + 6 and g(x) = 2 – x²
Solution:
i) p(x) = x³ – 3x² + 5x – 3 and
g(x) = x² – 2
The given polynomials are in standard form.
TS 10th Class Maths Solutions Chapter 3 Polynomials Ex 3.4 1
The degree of x² – 2 is 2.
The degree of 7x – 9 is 1.
∴ We stop here since the degree of (7x – 9) < degree of (x² – 2)
So, the quotient is x – 3 and the remainder is 7x – 9.

ii) p(x) = x⁴ – 3x² + 4x + 5,
g(x) = x² + 1 – x
p(x) = x⁴ – 3x² + 4x + 5, it is in standard form.
g(x) = x² + 1 – x, it is not in standard form. Writing it in standard form, we have x² – x + 1. Now we apply the division algorithm to the given polynomials.
TS 10th Class Maths Solutions Chapter 3 Polynomials Ex 3.4 2
The degree of x² – x + 1 is 2
The degree of 8 is 0.
∴ We stop here since the degree of (8) < degree of (x² – x + 1)
So, the quotient is x² + x – 3 and the remainder is 8.

iii) p(x) = x⁴ – 5x + 6 and g(x) = 2 – x²
p(x) = x⁴ – 5x + 6 → it is in standard form
g(x) = 2 – x² → it is not in standard form writing it in standard form, we have -x² + 2.
Now we apply the division algorithm to the given polynomials
TS 10th Class Maths Solutions Chapter 3 Polynomials Ex 3.4 3
The degree of -x² + 2 is 2 and that of (-5x + 10) is 1
The degree of (-x² + 2) > degree of (-5x + 10)
So, we stop here
So, the quotient is -x² – 2 and the remainder is -5x + 1o.

Question 2.
Check in which case the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial:
i) t² – 3, 2t⁴ + 3t³ – 2t² – 9t – 12
ii) x² + 3x + 1, 3x⁴ + 5x³ – 7x² + 2x + 2
iii) x³ – 3x + 1, x⁵ – 4x³ + x² + 3x + 1
Solution:
Divide the second polynomial by the first polynomial. If the remainder is zero, then the first polynomial is a factor of the second one. The given polynomials are in standard form.
i)
TS 10th Class Maths Solutions Chapter 3 Polynomials Ex 3.4 4
Since the remainder is ‘0’, t² – 3 is a factor of 2t⁴ + 3t³ – 2t² – 9t – 12.

ii)
TS 10th Class Maths Solutions Chapter 3 Polynomials Ex 3.4 5
Since the remainder is ‘0’, x² + 3x + 1 is a factor of 3x⁴ + 5x³ – 7x² + 2x + 2.

iii)
TS 10th Class Maths Solutions Chapter 3 Polynomials Ex 3.4 6
Here, the remainder is 2.
∴ x³ – 3x + 1 is not a factor of x⁵ – 4x³ + x² + 3x + 1.
x⁵ – 4x³ + x² + 3x + 1

Question 3.
Obtain all other zeroes of 3x⁴ + 6x³ – 2x² – 10x – 5, if two of its zeroes are
\(\sqrt{\frac{5}{3}}\) and –\(\sqrt{\frac{5}{3}}\).
Solution:
The given polynomial is
3x⁴ + 6x³ – 2x² – 10x – 5.
Two of its zeroes are \(\sqrt{\frac{5}{3}}\) and –\(\sqrt{\frac{5}{3}}\).
∴ [x – \(\sqrt{\frac{5}{3}}\)][x + \(\sqrt{\frac{5}{3}}\)] = [x² – \(\sqrt{\frac{5}{3}}\)] is a factor of the given polynomial.
Now, we apply the division algorithm to the given polynomial & x² – \(\frac{5}{3}\).
TS 10th Class Maths Solutions Chapter 3 Polynomials Ex 3.4 7
3x⁴ + 6x³ – 2x² – 10x – 5 = (x² – \(\frac{5}{3}\)) (3x² + 6x + 3)
3x² + 6x + 3 = 3(x² + 2x + 1) = 3(x + 1)²
So, the zeroes of 3(x + 1)² are -1 and -1.
Therefore, the zeroes of the given polynomial are and \(\sqrt{\frac{5}{3}}\) and –\(\sqrt{\frac{5}{3}}\), -1 and -1

Question 4.
On dividing x³ – 3x² + x + 2 by a polynomial g(x), the quotient and remainder were x – 2 and -2x + 4 respectively. Find g(x).
Solution:
By division algorithm, we have
p(x) = g(x) × q(x) + r(x)
g(x) × q(x) = p(x) – r(x)
g(x) × (x – 2) = x³ – 3x² + x + 2 – (-2x + 4)
= x³ – 3x² + x + 2 + 2x – 4
= x³ – 3x² + 3x – 2
∴ g(x) = (x³ – 3x² + 3x – 2) ÷ (x – 2)
TS 10th Class Maths Solutions Chapter 3 Polynomials Ex 3.4 8
∴ g(x) = x² – x + 1

Question 5.
Give examples of polynomials p(x), g(x), q(x) and r(x), which satisfy the division algorithm and
i) deg p(x) = deg q(x)
ii) deg q(x) = deg r(x)
iii) deg r(x) = 0
Solution:
i) p(x) = 6x² – 12x + 15, g(x) = 3,
q(x) = 2x² – 4x + 5, r(x) = 0
ii) p(x) = x³ + 2x² + x – 6, g(x) = x² + 2
q(x) = x + 2, r(x) = -x – 10
iii) p(x) = x³ + 5x² – 3x – 10, g(x) = x² – 3
q(x) = x + 5, r(x) = 5

TS 6th Class Maths Solutions Chapter 13 Practical Geometry Ex 13.5

February 21, 2024February 20, 2024 by Mahesh

Students can practice Telangana SCERT Class 6 Maths Solutions Chapter 13 Practical Geometry Ex 13.5 to get the best methods of solving problems.

TS 6th Class Maths Solutions Chapter 13 Practical Geometry Exercise 13.5

Question 1.
Construct ∠ABC = 60° without using protractor.
Answer:
TS 6th Class Maths Solutions Chapter 13 Practical Geometry Ex 13.5 1

Steps of construction:

Draw a line l and choose a point B on it.
Place the pointer of the compasses at B and draw an arc of convenient radius which cuts the line / at a point say C.
Take C as centre and with the same radius (as in step 2) in draw an arc.
Now take B as centre and with the same radius (as in step 2), draw another arc cutting the previous arc (drawn in step 2) at A.
Join AB, we get ∠ABC whose measure is 60°.

Question 2.
Construct an angle of 120° with using protractor and compasses.
Answer:
With using compasses:
TS 6th Class Maths Solutions Chapter 13 Practical Geometry Ex 13.5 2

Steps of construction:

Draw any ray OA.
Place the pointer of the compasses at O with O as centre and any convenient radius draw an arc cutting OA at M.
With M as centre and without altering radius (as in step 2) draw an arc which cuts the first arc at P.
With P as centre and without altering the radius (as in step 2) draw an arc which cuts the first arc at Q.
Join OQ. Then ∠AOQ is the required angle.

With Using protractor:
TS 6th Class Maths Solutions Chapter 13 Practical Geometry Ex 13.5 4

Steps of construction :

Draw a ray QR of any length.
Place the centre point of the protractor at
Q and the line aligned with the \(\overline{\mathrm{QR}}\).
Mark at a point P at 120°.
Join QP. ∠PQR is the required angle.

Question 3.
Construct the following angles using ruler and compasses. Write the steps of construction in each case.
(i) 75°
Answer:
TS 6th Class Maths Solutions Chapter 13 Practical Geometry Ex 13.5 3

Steps of construction:

Draw any ray OA.
Place the pointer of the compasses at O. With ‘O’ as centre and any convenient radius draw an arc cutting OA at M.
With M as centre and without altering radius (as in step 2) draw an arc which, cuts the first arc at P.
With P as centre and without altering the radius (as in step 2) draw an arc which cuts the first arc at Q.
Join OQ. ∠AOQ = 120° (∵∠AOP + ∠POQ = 60°)
∠AOQ = ∠AOP + ∠POQ = 60° + 60° = 120°
Draw OB the perpendicular bisector of ∠POQ.
Now ∠POB = ∠QOB = 30°
∴∠AOB = 90°
Draw OC the perpendicular bisector of ∠POB.
Now ∠POC = ∠BOC = 15°
∠AOC = ∠AOP + ∠OC
= 60° + 15°
= 75°
(or)
∠AOC = ∠AOB – ∠BOC
= 90° – 15°
= 75°
∠AOC is the required angle whose measure is 75°.

(ii) 15°
Answer:
TS 6th Class Maths Solutions Chapter 13 Practical Geometry Ex 13.5 5

Steps of construction:

Draw a line l and mark a point O on it.
Place the pointer of the compasses at O and draw an arc of convenient radius which cuts the line / at a point say A.
With the pointer at A (as centre) and the same radius as in the step – 2, now draw an arc that passes through O.
Let the two arcs intersect at B. Join OB. We get ∠BOA whose measure is 60°.
Draw the bisector OC of ∠AOB. Now ∠AOC = 30°.
[∠AOC = ∠BOC = 30° ]
Draw the bisector OD of ∠AOC. Now ∠AOD = 15°.
[∠AOD = ∠COD = 15°]
∠AOD = 15° is the required angle.

(iii) 105°
Answer:
TS 6th Class Maths Solutions Chapter 13 Practical Geometry Ex 13.5 6

Steps of construction:

Draw any ray OM.
Place the pointer of the compasses at O. With O as centre and any convenient radius draw an arc cutting OM at A.
With A as centre and without altering radius (as in step 2) draw an arc which cuts the first arc at B.
With B as centre and without altering radius (as in step 2) draw an arc which cuts the first arc at B.
Join QC. ∠AOC = 120°.
We know that ∠AOB = ∠BOC = \(\frac{1}{2}\) ∠AOC = \(\frac{1}{2}\) × 120° = 60°
Draw OD the bisector of ∠BOC.
∴ ∠BOD = ∠DOC = \(\frac{1}{2}\) ∠BOC = \(\frac{1}{2}\) × 60° = 30°
Draw OE the bisector of ∠DOC.
∴ ∠DOE = ∠EOC = \(\frac{1}{2}\) ∠DOC = \(\frac{1}{2}\) × 30° = 15°
Now ∠AOE = ∠AOD + ∠DOE = 90° + 15° = 105°
∠AOE = 105° is the required angle.

Question 4.
Draw the angles given in Q. 3 using a protractor.
Answer:
The angles given in Q. 3 are 75°, 105° and 15°.

(i) 75°
TS 6th Class Maths Solutions Chapter 13 Practical Geometry Ex 13.5 7

Steps of construction:

Draw a ray QR of any length.
Place the centre point of the protractor at Q and the line aligned with the \(\overline{\mathrm{QR}}\).
Mark a point P at 75°.
Join QP. ∠RQP is the required angle.

(ii) 105°
TS 6th Class Maths Solutions Chapter 13 Practical Geometry Ex 13.5 8

Steps of construction:

Draw a ray QR of any length.
Place the centre point of the protractor at Q and the line aligned with the \(\overline{\mathrm{QR}}\).
Mark a point P at 105°
Join QP. ∠RQP is the required angle.

(iii) 15°
TS 6th Class Maths Solutions Chapter 13 Practical Geometry Ex 13.5 9

Steps of construction:

Draw a ray QR of any length.
Place the centre point of the protractor at Q and the line aligned with the \(\overline{\mathrm{QR}}\).
Mark a point P at 15°
Join QP. ∠RQP is the required angle.

Question 5.
Construct ∠ABC = 50° and then draw another angle ∠XYZ equal to ∠ABC without using a protactor.
Answer:
TS 6th Class Maths Solutions Chapter 13 Practical Geometry Ex 13.5 10

Steps of construction:

Make an angle ∠ABC = 50° v
Draw a line l and choose a point Y on it.
Use the same compasses setting to draw an arc with Y as centre, cutting l in Z.
Set your compasses to the length MN.
With the same radius place the compasses pointer at Z and draw an arc to cut the arc drawn earlier at X.
Join XY. This gives us ∠XYZ. It has the same measure as ∠ABC i.e,, 50°.

Question 6.
Construct ∠DEF = 60°. Bisect it, measure each half by using a protractor.
Answer:
TS 6th Class Maths Solutions Chapter 13 Practical Geometry Ex 13.5 11

Steps of construction :

Draw a line P and mark a point ‘E’ on it.
Place the pointer of the compasses at E and draw an arc of convenient radius which cuts the line \(\overleftrightarrow{\mathrm{PQ}}\) at a point say F.
With the pointer at F (as centre), now draw an arc that passes through E.
Let the two arcs intersect at D. Join ED. We get ∠DEF whose measure is 60°.
Draw OC the bisector of ∠DEF.
Now ∠FEC = ∠CED = 30°.

TS 6th Class Maths Solutions Chapter 13 Practical Geometry Ex 13.4

February 21, 2024February 20, 2024 by Mahesh

Students can practice Telangana SCERT Class 6 Maths Solutions Chapter 13 Practical Geometry Ex 13.4 to get the best methods of solving problems.

TS 6th Class Maths Solutions Chapter 13 Practical Geometry Exercise 13.4

Question 1.
Draw the following angles with the help of a protractor.
(i) ∠ABC = 65°
Answer:
∠ABC = 65°
TS 6th Class Maths Solutions Chapter 13 Practical Geometry Ex 13.4 1

Steps of construction:

Draw a ray BC of any length.
Place the centre point of the protractor at B and the line aligned with the \(\overrightarrow{\mathrm{BC}}\).
Mark a point A at 65°.
Join BA. ∠ABC is the required angle.

(ii) ∠PQR = 136°
Answer:
∠ PQR = 136°
TS 6th Class Maths Solutions Chapter 13 Practical Geometry Ex 13.4 2

Steps of construction:

Draw a ray QR of any length.
Place the centre point of the protractor at Q and the line aligned with the \(\overrightarrow{\mathrm{QR}}\).
Mark a point P at 136°.
Join QP. ∠PQR is the required angle.

(iii) ∠Y = 45°
Answer:
∠Y = 45°
TS 6th Class Maths Solutions Chapter 13 Practical Geometry Ex 13.4 3

Steps of construction:

Draw a ray YZ of any length.
Place the centre point of the protractor
at Y and the line aligned with the \(\overrightarrow{\mathrm{YZ}}\).
Mark a point X at 45°.
Join YX. ∠XYZ is the required angle.

(iv) ∠O =172°
Answer:
∠O = 172°
TS 6th Class Maths Solutions Chapter 13 Practical Geometry Ex 13.4

Steps of construction:

Draw a ray OP of any length.
Place the centre point of the protractor at 0 and the line aligned with the OP
Mark a point N at 172°.
Join ON. ∠NOP is the required angle.

Question 2.
Copy the following angles in your notebook and find their bisector.
TS 6th Class Maths Solutions Chapter 13 Practical Geometry Ex 13.4 5
Answer:

Steps of construction :

Draw a line l and choose a point P on it.
Now place the compasses at A and draw an arc to cut the rays AC and AB.
Use the same compasses setting to draw an arc with P as centre, cutting / at Q.
Set your compasses with \(\overline{\mathrm{BC}}\) as the radius.
Place the compasses pointer at Q and draw an arc to cut the existing arc at R.
Join PR. This gives us ∠RPQ. It has the same measure as ∠CAB.
This means ∠QPR has same measure as ∠BAC.
Take Q as centre and with radius more than half of the length PQ, draw an arc. Now take R as centre and with the same radius, draw another arc intersecting the previous arc at S. Join PS. PS is the bisector of the given angle.

TS 10th Class Maths Solutions Chapter 3 Polynomials Ex 3.3

February 21, 2024February 19, 2024 by Mahesh

Students can practice TS 10th Class Maths Solutions Chapter 3 Polynomials Ex 3.3 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 3 Polynomials Exercise 3.3

Question 1.
Find the zeroes of the following quadratic polynomial and verify the relationship between the zeroes and coefficients.

i) x² – 2x – 8
ii) 4s² – 4s + 1
iii) 6x² – 3 – 7x
iv) 4u² + 8u
v) t² – 15
vi) 3x² – x – 4
Solution:
i) Let p(x) = x² – 2x – 8
= x² – 4x + 2x – 8
= x(x – 4) + 2(x – 4)
= (x + 2) (x – 4)
The zeroes of p(x) are given by P(x) = 0
⇒ (x + 2) (x – 4) = 0
⇒ x + 2 = 0 (or) x – 4 = 0
⇒ x = -2 (or) x = 4
Hence the zeroes of x² – 2x – 8 are -2 and 4.
Sum of the zeroes = -2 + 4 = 2
TS 10th Class Maths Solutions Chapter 3 Polynomials Ex 3.3 2
Product of the zeroes = -2 × 4 = -8

ii) Let p(s) = 4s² – 4s + 1
= (2s)² – 2(2s) (1) + (1)²
= (2s – 1)²
= (2s – 1) (2s – 1)
The zeroes of p(s) are given by p(s) = 0
⇒ (2s – 1) (2s – 1) = 0
⇒ 2s – 1 =0 (or) 2s – 1 = 0
⇒ s’ = \(\frac{1}{2}\) (or) s = \(\frac{1}{2}\)
Hence the zeroes of 4s² – 4s + 1 are \(\frac{1}{2}\), \(\frac{1}{2}\)
Sum of the zeroes = \(\frac{1}{2}\) + \(\frac{1}{2}\) = 1;
TS 10th Class Maths Solutions Chapter 3 Polynomials Ex 3.3 4

(iii) Let p(x) = 6x² – 7x – 3
= 6x² – 9x + 2x – 3
= 3x(2x – 3) + 1 (2x – 3)
= (3x + 1) (2x – 3)
The zeroes of p(x) are given by p(x) = 0
⇒ (3x + 1) (2x – 3) = 0
⇒ 3x + 1 = 0 (or) 2x – 3 = 0
⇒ 3x = -1 (or) 2x = 3
⇒ x = \(\frac{-1}{3}\) (or) x = \(\frac{3}{2}\)
Hence the zeroes of 6x – 7x – 3 are \(\frac{-1}{3}\) and \(\frac{3}{2}\)
Sum of the zeroes = \(\frac{-1}{3}\) + \(\frac{3}{2}\)
TS 10th Class Maths Solutions Chapter 3 Polynomials Ex 3.3 1

iv) Let p(u) = 4u² + 8u
= 4u(u + 2)
The zeroes of p(u) are given by
⇒ p(u) = 0
⇒ 4u(u + 2) = 0
⇒ 4u = 0 (or) u + 2 = 0
⇒ u = 0 (or) u = -2
Hence the zeroes of 4u + 8u are 0 and -2.
Sum of the zeroes = 0 + (-2) = -2
TS 10th Class Maths Solutions Chapter 3 Polynomials Ex 3.3 5

v) Let p(t) = t² – 15
The zero of p(t) is given by p(t) = 0
⇒ t² – 15 = 0 ⇒ t² = 15
TS 10th Class Maths Solutions Chapter 3 Polynomials Ex 3.3 6

vi) Let P(x) = 3x²
The zero of p(x) is given by P(x) = 0
⇒ 3x² – x – 4 = 0
⇒ 3x² + 3x – 4x – 4 = 0
⇒ 3x(x + 1) – 4(x + 1) = 0
⇒ (x + 1) (3x – 4) = 0
⇒ x + 1 = 0 (or) 3x – 4 = 0
⇒ x = -1 (or) x = 4/3
Hence the zeroes of 3x² – x – 4 are -1 (or) 4/3
Sum of the zeroes of 3x² – x – 4 are -1 (or) 4/3.
Sum of the zeroes =
TS 10th Class Maths Solutions Chapter 3 Polynomials Ex 3.3 7

Question 2.
Find the quadratic polynomial in each case, with the given numbers as the sum and product of its zeroes respectively.

i) \(\frac{1}{4}\), -1
ii) \(\sqrt{2}\), \(\frac{1}{3}\)
iii) 0, \(\sqrt{5}\)
iv) 1, 1
v) –\(\frac{1}{4}\), \(\frac{1}{4}\)
vi) 4, 1 (A.P. Mar. ’15)
Solution:
i) \(\frac{1}{4}\), -1
Let α, β be the zeroes of the quadratic polynomial.
Sum of the zeroes = α + β = \(\frac{1}{4}\)
Product of the zeroes = αβ = -1
The required quadratic polynomial will be
k[x² – x (α + β) + αβ] where k is a constant
⇒ k[x² – x(\(\frac{1}{4}\)) + (-1)]
⇒ k(x² – \(\frac{x}{4}\) – 1)
If k = 4, then the polynomial will be 4(x² – \(\frac{x}{4}\) – 1) = 4x² – x – 4

ii) \(\sqrt{2}\), \(\frac{1}{3}\)
Let α, β be the zeroes of the quadratic polynomial.
Sum of the zeroes = α + β = \(\sqrt{2}\)
Product of the zeroes = αβ = \(\frac{1}{3}\)
∴ The required quadratic polynomial will be
k[x² – x(α + β) + αβ] where k is a constant
⇒ k[x² – x(\(\sqrt{2}\)) + \(\frac{1}{3}\)]
⇒ k[x² – \(\sqrt{2}\)x + \(\frac{1}{3}\)]
when k = 3, then the polynomial will be
3[x² – \(\sqrt{2}\)x + \(\frac{1}{3}\)] (or) 3x² – 3\(\sqrt{2}\)x + 1

iii) 0, \(\sqrt{5}\)
Let α, β be the zeroes of the quadratic polynomial.
Sum of the zeroes = α + β = 0
Product of the zeroes = αβ = \(\sqrt{5}\)
The required quadratic polynomial will be k[x² – x(α + β) + αβ where k is a constant
⇒ k[x² -x(0) + \(\sqrt{5}\)]
⇒ k[x² + \(\sqrt{5}\)].
Where k = 1, the polynomial will be x² + \(\sqrt{5}\).

iv) 1, 1
Let α, β be the zeroes of the quadratic polynomial.
Sum of the zeroes = α + β = 1
Product of the zeroes = αβ = 1
∴ The required quadratic polynomial will be
k[x² – x(α + β) + αβ] where k is a constant
⇒ k[x² – x(α – β) + 1]
⇒ k[x² – x + 1]
Where k = 1, the polynomial will be x² – x + 1.

v) –\(\frac{1}{4}\), \(\frac{1}{4}\)
Let α, β be the zeroes of the quadratic polynomial.
Sum of the zeroes = α + β = –\(\frac{1}{4}\)
Product of the zeroes = αβ = \(\frac{1}{4}\)
The required quadratic polynomial will be k[x² – x(α + β) + αβ] where k is a constant
⇒ k[x² – x(-\(\frac{1}{4}\)) + \(\frac{1}{4}\)]
⇒ k[x² + \(\frac{x}{4}\) + \(\frac{1}{4}\)]
Where k = 4, the polynomial will be
4[x² + \(\frac{x}{4}\) + \(\frac{1}{4}\)] = 4x² + x + 1

Question 3.
Find the quadratic polynomial for the zeroes α, β given in each case. (A.P.Mar.’16)

i) 2, -1
ii) \(\sqrt{3}\), –\(\sqrt{3}\)
iii) \(\frac{1}{4}\), -1
iv) \(\frac{1}{2}\), \(\frac{3}{2}\)
Solution:
i) 2,-1
Let the quadratic polynomial be ax² + bx + c, a ≠ 0 and its zeroes be α & β
Here α = 2, β = -1
Sum of the zeroes = α + β = 2 + (-1) = 1
Product of the zeros = αβ = 2 × (-1)
= -2
∴ The required quadratic polynomial
ax² + bx + c is k[x² – x(α + β) + αβ]
where k is a constant
⇒ k [x² – x(1) + (-2)]
⇒ k[x² – x – 2]
Where k = 1, the quadratic polynomial will be x² – x – 2.

ii) \(\sqrt{3}\), –\(\sqrt{3}\)
Let the quadratic polynomial be
ax² + bx + c, a ≠ 0 and its zeroes be α & β
Here α = \(\sqrt{3}\) , β = –\(\sqrt{3}\)
Sum of the zeroes = α + β
= \(\sqrt{3}\) + (-\(\sqrt{3}\)) = o
Product of the zeros
= αβ = \(\sqrt{3}\) × –\(\sqrt{3}\) = -3
Therefore the quadratic polynomial
ax² + bx + c is k[x² – x(α + β) + αβ]
where k is a constant
⇒ k [x² – x(0) + (-3)]
⇒ k[x² – 3]
Where k = 1, the quadratic polynomial will be [x² – 3].

iii) \(\frac{1}{4}\), -1
Let the quadratic polynomial be
ax² + bx + c, a ≠ 0 and its zeroes be α & β
Here α = \(\frac{1}{4}\), β = -1
Sum of the zeroes = α + β
= \(\frac{1}{4}\) + (-1) = \(\frac{-3}{4}\)
Product of the zeroes = αβ = \(\frac{1}{4}\) × (-1)
= –\(\frac{1}{4}\)
Therefore, the quadratic polynomial
ax² + bx + c is k[x² – x(α + β) + αβ]
where k is a constant
Where k = 4, the quadratic polynomial will be [4x² + 3x – 1]

iv) \(\frac{1}{2}\), \(\frac{3}{2}\)
Let the quadratic polynomial be
ax² + bx + c, a ≠ 0 and its zeroes be α & β
Here α = \(\frac{1}{2}\), β = \(\frac{3}{2}\)
Sum of the zeroes = α + β
= \(\frac{1}{2}\) + \(\frac{3}{2}\) = \(\frac{4}{2}\) = 2
Product of the zeroes = αβ
= \(\frac{1}{2}\) × \(\frac{3}{2}\) = \(\frac{3}{4}\)
Therefore the quadratic polynomial
ax² + bx + c is k[x² – x(α + β) + αβ]
where k is a constant
TS 10th Class Maths Solutions Chapter 3 Polynomials Ex 3.3 8
Where k = 4, the quadratic polynomial will be [4x² – 8x + 3].

Question 4.
Verify that 1, -1 and -3 are the zeroes of the cubic polynomial x³ + 3x² – x – 3 and check the relationship between zeroes and the coefficients. (A.P. Mar. ’15)
Solution:
The given polynomial is x³ + 3x² – x – 3
Comparing the given polynomial with ax³ + bx² + cx + d
We get a = 1, b = 3, c = -1, d = -3
Let p(x) = x³ + 3x² – x – 3
Then p(1) = (1)³ + 3(1)² – 1 – 3
= 1 + 3 – 1 – 3
= 4 – 4 = 0
p(-1) = (-1)³ + 3(-1)² – (-1) – 3
= -1 + 3 + 1 – 3 = 0
p(-3) = (-3)³ + 3(-3)² – (-3) – 3
= -27 + 27 + 3 – 3 = 0
Therefore 1, -1 and -3 are the zeroes of
x³ + 3x² – x – 3.
So α = 1, β = -1, γ = -3
α + β + γ = 1 – 1 – 3 = -3 = -3/1 = \(\frac{-b}{a}\)
αβ + βγ + αγ = 1(-1) + (-1)(-3) + (-3)(1)
= -1 + 3 – 3 = -1 = -1/1 = c/a
αβγ = (1)(-1)(-3) = – \(\left(\frac{-3}{1}\right)\) = \(\frac{-\mathrm{d}}{\mathrm{a}}\)

TS 6th Class Maths Solutions Chapter 9 Introduction to Algebra InText Questions

February 21, 2024February 19, 2024 by Mahesh

Students can practice TS SCERT Class 6 Maths Solutions Chapter 9 Introduction to Algebra InText Questions to get the best methods of solving problems.

TS 6th Class Maths Solutions Chapter 9 Introduction to Algebra InText Questions

Try These

Question 1.
Can you now write the rule to form the following pattern with match-sticks ?
TS 6th Class Maths Solutions Chapter 9 Introduction to Algebra InText Questions 1
Answer:
The rule for the above figures is 3n.
[∵ 3 × 1, 3 × 2, 3 × 3, ………………]

Question 2.
Find the rule for required number of matchsticks to form a pattern repeating ‘H’. How would the rule be for repeating the shape ‘L’ ?
Answer:
The required pattern of repeating ‘H’ is
‘3n’. (3 × 1, 3 × 2, ………………..)
The rule for ‘L’ is ‘2n’.

Question 3.
A line of shapes is constructed using matchsticks.
TS 6th Class Maths Solutions Chapter 9 Introduction to Algebra InText Questions 2
(i) Find the rule that shows how many sticks are needed to make a group of such shapes ?
(ii) How many matchsticks are needed to form a group of 12 shapes ?
Answer:
(i) Number of matchsticks that are used in each figure are
3 ; 5 ; 7 ; 9 ………………
Shape-1; Shape-2; Shape-3; Shape-4
2 × 1 + 1; 2 × 2 + 1 ; 2 × 3 + 1 ; 2 × 4 + 1 ………
In general (2n + 1) is the rule for the above shapes.

(ii) 2n + 1 = 2 × 12 + 1 = 25 [ ∵ n = 12]

(Try These)

Question 1.
Find the general rule for the perimeter of a rectangle. Use variables ‘l’ and ‘b’ for length and breadth of the rectangle respectively.
Answer:
TS 6th Class Maths Solutions Chapter 9 Introduction to Algebra InText Questions 3
Perimeter of a rectangle
= l + b + l + b
= 2l + 2b = 2 (l + b)

Question 2.
Find the general rule for the area of a square by using the variable ’s’ for the side of a square ?
Answer:
TS 6th Class Maths Solutions Chapter 9 Introduction to Algebra InText Questions 4
The rule for the area of a square is A = side × side
= s × s
A = s²

Question 3.
What would be the rule for perimeter of an Isosceles triangle ?
Answer:
TS 6th Class Maths Solutions Chapter 9 Introduction to Algebra InText Questions 5
The rule for the perimeter of an Isosceles triangle
= 3 × side
= 3 × a
P = 3a

Do This

Question 1.
Find the n^th term in the following sequences.
(i) 3, 6, 9, 12, …………………
(ii) 2, 5, 8, 11, …………………
(iii) 1, 8, 27, 64, 125, …………………
Answer:
(i) The n^th> term of the above series is
3 × 1, 3 × 2, 3 × 3, ……………….. = 3n.

(ii) The n^th term is 3 × 1 – 1, 3 × 2 – 1, 3 × 3 – 1, 3 × 4 – 1,
i.e., 3n – 1

(iii) The nth term is 1 × 1 × 1, 2 × 2 × 2, 3 × 3 × 3, ……. is
1³, 2³, 3³ ………………….
i.e., n³

Question 2.
Write LHS and RHS of the following simple equations:
(i) 2x + 1 = 10
Answer:
2x + 1 = 10
LHS = 2x + 1, RHS = 10

(ii) 9 = y – 2
Answer:
9 = y – 2 ;
LHS = 9, RHS = y – 2

(iii) 3p + 5 = 2p + 10
Answer:
3p + 5 = 2p + 10
LHS = 3p + 5, RHS = 2p + 10

Question 3.
Write any two simple equations and give their LHS and RHS.
Answer:

Simple equation	LHS	RHS
1. 4x – 10 = 7x + 6	4x – 10	7x + 6
2. t – 2= 5	t – 2	5

Question 4.
Find the solution of the equation ‘x – 4 = 2’ by Trail and Error method.
Answer:
Given equation is x – 4 = 2
If x = 1 ⇒ x – 4 = 2
⇒ 1 – 4 = 2
– 3 ≠ 2
If x = 3 ⇒ 3 – 4 = 2
– 1 ≠ 2
If x = 5 ⇒ 5 – 4 = 2 .
1 ≠ 2
If x = 6 ⇒ 6 – 4 = 2
2 = 2
x = 6 satisfies the given equation.
∴ x = 6 is the solution of the equation.

TS 10th Class Maths Solutions Chapter 3 Polynomials Ex 3.2

February 21, 2024February 19, 2024 by Mahesh

Students can practice TS 10th Class Maths Solutions Chapter 3 Polynomials Ex 3.2 to get the best methods of solving problems.

TS 10th Class Maths Solutions Chapter 3 Polynomials Exercise 3.2

Question 1.
The graphs of y = p(x) are given in the figure below, for some polynomials p(x). In each case, find the number of zeroes
of p(x).
Solution:
TS 10th Class Maths Solutions Chapter 3 Polynomials Ex 3.2 1
Solution:
Six figures are given and they are numbered.
In the first figure, there are no zeroes.
In the second figure, there is one zero.
In the third figure, there are three zeroes.
In the fourth figure, there are two zeroes.
In the fifth figure, there are four zeroes.
In the sixth figure, there are three zeroes.

Question 2.
Find the zeroes of the given polynomials.
(i) p(x) = 3x
(ii) p(x) = x² + 5x + 6
(iii) p(x) = (x + 2) (x + 3)
(iv) p(x) = x⁴ – 16
Solution:
i) p(x) = 3x
p(0) = 3 × 0 = 0
∴ The zero of p(x) = 3x is 0

ii) p(x) = x² + 5x + 6
= x² + 3x + 2x + 6
= x(x + 3) + 2(x + 3) = (x + 2) (x + 3)
To find zeroes, let p(x) = 0
⇒ (x + 2) (x + 3) = 0
So, x + 2 = 0 (or) x + 3 = 0
⇒ x = -2 (or) x = -3
Therefore, the zeroes of x + 5x + 6 are – 2 and -3.

iii) p(x) = (x + 2) (x + 3)
⇒ x² + 2x + 3x + 6
⇒ x² + 5x + 6
To find zeroes, let p(x) = 0
⇒ (x + 2) (x + 3) = 0
So, x + 2 = 0 (or) x + 3 = 0
⇒ x = -2 (or) x = -3
∴ The zeroes of (x + 2) (x + 3) are -2 and -3.

iv) p(x) = x⁴ – 16
= (x²)² – (4)² = (x² + 4) (x² – 4)
= (x² + 4)(x + 2)(x – 2)
To find zeroes, let p(x) = O
⇒ (x² + 4) (x + 2)(x – 2) = 0
(i.e.,) x² + 4 = 0 (or) x + 2 = 0 (or) x – 2 = 0
If x² + 4 = 0, then x² = -4 ⇒
x = ± \(\sqrt{-4}\)
If x + 2 = 0, then x = -2
If x – 2 = 0, then x = 2
∴ The zeroes of the polynomial x⁴ – 16 are -2, 2 and ± \(\sqrt{-4}\).

Question 3.
Draw the graphs of the given polynomials and find the zeroes. Justify the answers. (A.P.June’ 15)
(i) p(x) = x² – x – 12
(ii) p(x) = x² – 6x + 9
(iii) p(x) = x² – 4x + 5
(iv) p(x) = x² + 3x – 4
(v) p(x) = x² – 1
Solution:
i) p(x) = x² – x – 12
TS 10th Class Maths Solutions Chapter 3 Polynomials Ex 3.2 2
∴ The zeros of x² – x – 12 are 4 and -3.

Scale:
On X-axis: 1 cm = 1 unit
OnY-axis: 1 cm = 2 units
TS 10th Class Maths Solutions Chapter 3 Polynomials Ex 3.2 17
The graph intersects the X-axis at (4, 0) and (-3, 0)

ii) p(x) = x² – 6x + 9
TS 10th Class Maths Solutions Chapter 3 Polynomials Ex 3.2 11

The graph intersects one the X-axis at only point (3, 0)

iii) p(x) = x² – 4x + 5
TS 10th Class Maths Solutions Chapter 3 Polynomials Ex 3.2 12

Scale:
On X-axis: 1 cm = 1 unit
OnY-axis: 1 cm = 2 units

The graph does not intersect at the X-axis.
There are no zeroes of the polynomial x² – 4x + 5

iv) p(x) = x² + 3x – 4
TS 10th Class Maths Solutions Chapter 3 Polynomials Ex 3.2 14

∴ The graph intersects the X-axis at (1, 0) and (-4, 0)
∴ The zeroes of the polynomial x² + 3x – 4 are 1 and -4.

v) p(x) = x² – 1
TS 10th Class Maths Solutions Chapter 3 Polynomials Ex 3.2 8

The graph intersects the X-axis at (-1, 0) and (1, 0)
∴ The zeroes of the polynomial x² – 1 are -1 and 1.

Question 4.
Why are \(\frac{1}{4}\) and -1 zeroes of the polynomial p(x) = 4x² + 3x – 1?
Solution:
p(x) = 4x² + 3x – 1
∴ p(\(\frac{1}{4}\)) = 4(\(\frac{1}{4}\))² + 3(\(\frac{1}{4}\)) – 1
= \(\frac{1}{4}\) + \(\frac{3}{4}\) + \(\frac{1}{1}\)
= \(\frac{1+3-4}{4}\) = \(\frac{0}{4}\) = 0
∴ p(-1) = 4(-1)² + 3(-1) – 1
= 4(1) – 3 – 1
= 4 – 3 – 1
= 4 – 4 = 0
Since p(\(\frac{1}{4}\)) and p(-1) are equal to zero.
\(\frac{1}{4}\) and -1 are the zeroes of the polynomial.

TS 6th Class Maths Solutions Chapter 13 Practical Geometry Ex 13.3

February 21, 2024February 19, 2024 by Mahesh

Students can practice Telangana SCERT Class 6 Maths Solutions Chapter 13 Practical Geometry Ex 13.3 to get the best methods of solving problems.

TS 6th Class Maths Solutions Chapter 13 Practical Geometry Exercise 13.3

Question 1.
Draw a line segment PQ = 5.8 cm and construct its perpendicular bisector using ruler and compasses.
Answer:
TS 6th Class Maths Solutions Chapter 13 Practical Geometry Ex 13.3 1

Steps of construction:

Draw a line segment \(\overline{\mathrm{PQ}}\) = 5.8 cm.
Take P as centre and radius more than half of PQ draw two arcs above and below the line segment \(\overline{\mathrm{PQ}}\).
Take Q as centre and with the same radius, draw two more arcs intersecting the previous arcs at A and B.
Join A and B. This line intersects PQ at O.
AB is the required perpendicular bisector of the line PQ.

Question 2.
Ravi made a line segment of length 8.6 Find the length of AC and BC.
Answer:
TS 6th Class Maths Solutions Chapter 13 Practical Geometry Ex 13.3 2

Steps of construction :

Draw a line segment AB = 8.6 cm.
Take A as centre and radius more
than half of the length AB, draw two arcs above and below the line segment AB. ,
Take B as cfentre and with the same radius, draw two more arcs intersecting the previous- arcs at P and Q.
Join PQ. This line intersects AB at C.
Measure AC and BC. On measuring, it is noticed that AC = BC = 4.3 cm.

Question 3.
Using ruler and compasses, draw AB = 6.4 cm. Find its mid point.
Answer:
We can find the mid point of the line segment AB = 6.4 cm by drawing its perpendicular bisector.
TS 6th Class Maths Solutions Chapter 13 Practical Geometry Ex 13.3 3

Steps of construction:

Draw a line segment AB = 6.4 cm.
Take A as centre and radius more than half of the length AB draw two arcs above and below the line segment AB.
Take B as centre and with the same radius draw two more arcs intersecting the previous arcs at M and N.
Join MN. This line intersects AB at O. ‘O’ is the mid point of AB.

TS 6th Class Maths Solutions Chapter 13 Practical Geometry Ex 13.2

February 21, 2024February 19, 2024 by Mahesh

Students can practice Telangana SCERT Class 6 Maths Solutions Chapter 13 Practical Geometry Ex 13.2 to get the best methods of solving problems.

TS 6th Class Maths Solutions Chapter 13 Practical Geometry Exercise 13.2

Question 1.
Construct a circle with centre M and radius 4 cm.
Answer:
TS 6th Class Maths Solutions Chapter 13 Practical Geometry Ex 13.2 1

Steps of construction:

Open the compasses for 4 cm radius.
Mark a point with a sharp pencil. This is the centre. Mark it as ‘O’.
Place the pointer of the compasses firmly at ‘O’.
Without moving its metal point, now slowly rotate the pencil till it comes back to the starting point.

Question 2.
Construct a circle with centre X and diameter 10 cm.
Answer:
Diameter of the required circle = 10 cm
Radius of the circle = \(\frac{10}{2}\) = 5 cm
TS 6th Class Maths Solutions Chapter 13 Practical Geometry Ex 13.2 2

Steps of construction:

Open the compasses for 5 cm radius.
Mark ’a point with a sharp pencil. This is the centre. Mark it as ‘X’. ‘
Place the pointer of the compasses firmly at X.
Without moving its metal point, now slowly rotate the pencil till it comes back to the starting point.

Question 3.
Draw four circles of radii 2 cm, 3 cm, 4 cm and 5 cm with the same centre P.
Answer:
TS 6th Class Maths Solutions Chapter 13 Practical Geometry Ex 13.2 3

Steps of construction:

Mark a point with sharp pencil. This is the centre mark it as ‘P’.
Place the pointer of the compasses firmly at P.
Without moving its metal point, slowly rotate the pencil till it comes back to the, starting point.
With ‘P’ as centre and radii 3, 4 and 5 cm, repeat the procedure given in the steps 2 and 3.
Now we have four circles with radii 2 cm, 3 cm, 4 cm and 5 cm having centre ‘P’. These circles are called concentric circles.

Question 4.
Draw any circle and mark three points A, B and C such that
(i) A Is on the circle.
(ii) B is in the interior of the circle.
(iii) C is in the exterior of the circle.
Answer:
‘O’ is the centre of the circle.
(i) A is on the circle.
(ii) B is in the interior of the circle.
(iii) C is in the exterior of the circle.
TS 6th Class Maths Solutions Chapter 13 Practical Geometry Ex 13.2 4

TS 6th Class Maths Solutions Chapter 13 Practical Geometry Ex 13.1

February 21, 2024February 19, 2024 by Mahesh

Students can practice Telangana SCERT Class 6 Maths Solutions Chapter 13 Practical Geometry Ex 13.1 to get the best methods of solving problems.

TS 6th Class Maths Solutions Chapter 13 Practical Geometry Exercise 13.1

Question 1.
Construct a line segment of length 6.9 cm using a ruler and compasses.
Answer:
TS 6th Class Maths Solutions Chapter 13 Practical Geometry Ex 13.1 1

Steps of construction:

Draw a line l. Mark a point A on the line l.
Place the metal pointer of the compasses on the zero mark of the ruler. Open the compasses so that pencil point touches 6.9 cm mark on the ruler.
Place the pointer on A on the line l and draw an arc to cut the line. Mark the point where the arc cuts the line as B.
On the line l, we got the line segment AB of required length.

Question 2.
Construct a line segment of length 4.3 cm using the ruler.
Answer:

Steps of construction:

Place the ruler on paper and hold it firmly. Mark a point with a sharp edged pencil against 0 cm mark of the ruler. Name the point as A.
Mark another point against 3 small divisions just after the 4 cm mark. Name this point as B.
Join points A and B along the edge of the ruler.
AB is the required line segment of length 4.3 cm.

Question 3.
Construct a line segment MN of length 6 cm. Mark any point O on it. Measure MO, ON and MN. What do you observe ?
Answer:
TS 6th Class Maths Solutions Chapter 13 Practical Geometry Ex 13.1 3

Steps of construction :

Draw a line l. Mark a point M on the line l.
Place the metal pointer of the compasses on the zero mark of the ruler. Open it to
place the pencil point upto the 6 cm mark.
Taking caution that the opening of the compasses has not changed, place the pointer on M and swing an arc to cut / at N.
MN is a line segment of required length.
Mark any point ‘O’ on MN.
Measure the length of MO and ON. It is found that MO = 3.8 cm and ON = 2.2 cm MO + ON = 3.8 +2.2 = 6 cm
It is noticed that \(\overline{\mathrm{MO}}+\overline{\mathrm{ON}}\) = \(\overline{\mathrm{MN}}\)

Question 4.
Draw a line segment \(\overline{\mathbf{A B}}\) of length 12 cm. Mark a point C on the line segment \(\overline{\mathbf{A B}}\), such that \(\overline{\mathbf{A C}}\) = 5.6 cm. What should be the length of \(\overline{\mathbf{C B}}\)? Measure the length of \(\overline{\mathbf{C B}}\).
Answer:
TS 6th Class Maths Solutions Chapter 13 Practical Geometry Ex 13.1 4

Steps of construction:

Draw a line l. Mark a point A on the line l.
Place the metal pointer of the compasses on the zero mark of the ruler. Open it to place the pencil point upto the 12 cm mark.
Taking caution that the opening of the compasses has not changed, place the pointer on A and swing an arc to cut / at B.
\(\overline{\mathbf{A B}}\) is a line segment of required length.
Similarly mark the point C on l such that \(\overline{\mathbf{A C}}\) = 5.6 cm.
The length of \(\overline{\mathbf{C B}}\) should be 6.4 cm = (12 – 5.6) on measuring, the length of \(\overline{\mathbf{C B}}\) = 6.4 cm.

Question 5.
Given that AB = 12 cm

(i) From the figure measure the lengths of the following line segments.
(a) \(\overline{\mathrm{CD}}\)
(b) \(\overline{\mathrm{DB}}\)
(c) \(\overline{\mathrm{EA}}\)
(d) \(\overline{\mathrm{AD}}\)
Answer:
(a) \(\overline{\mathrm{CD}}\) = 2.8 cm
(b) \(\overline{\mathrm{DB}}\) = 4.3 cm
(c) \(\overline{\mathrm{EA}}\) = 17.3 cm
(d) \(\overline{\mathrm{AD}}\) = 16.3 cm

(ii) Verify \(\overline{\mathbf{A E}}-\overline{\mathbf{C E}}\) = \(\overline{\mathbf{A C}}\)
Answer:
\(\overline{\mathbf{A E}}-\overline{\mathbf{C E}}\) = 17.3 cm – 3.8 cm = 13.5 cm
\(\overline{\mathrm{BC}}\) = 1.5 cm
\(\overline{\mathrm{AC}}\) = \(\overline{\mathbf{A B}}+\overline{\mathbf{B C}}\)
= 12 cm + 1.5 cm
= 13.5 cm
∴ \(\overline{\mathrm{AE}}-\overline{\mathrm{CE}}\) = \(\overline{\mathrm{AC}}\)

Question 6.
\(\overline{\mathrm{AB}}\) = 3.8 cm. Construct \(\overline{\mathrm{MN}}\) by compasses such that the length of \(\overline{\mathrm{MN}}\) = 3AB’is thrice that of \(\overline{\mathrm{AB}}\). Verify this with the help of a ruler.
Answer:
TS 6th Class Maths Solutions Chapter 13 Practical Geometry Ex 13.1 6

Steps of construction:

Draw a line l. Mark a point M on it.
Place the compasses pointer on the zero mark of the ruler.-Open it to place the pencil upto the 3.8 cm mark.
Taking caution that the opening of the compasses has not changed, place the pointer on M and swing an arc to cut l at A.
\(\overline{\mathrm{MA}}\) is a line segment of 3.8 cm.
Similarly place the pointer on A and swing an arc to cut l at B such that MA = AB.
Again place the pointer on B and swing an arc to cut l at N such that AB = BN.
Now \(\overline{\mathrm{MN}}\) is a line segment of 11.4 cm.