Category: Class 11

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 11 Integral Calculus Ex 11.9 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 11 Integral Calculus Ex 11.9

Integrate the following with respect to x:

Question 1.
e^x (tan x + log sec x)
Answer:
Let I = ∫e^x (tan x + log sec x) dx
Take f(x) = log sec x
f'(x) = \(\frac{1}{\sec x}\) sec x tan x
f'(x) = tan x
[∫e^x [f (x) + f (x)] dx = e^x f(x) + c]
∴ I = e^x log |sec x| + c

Question 2.

Answer:

[∫ e^x [f (x) + f (x)] dx = e^x f(x) + c]

Question 3.
e^x sec x (1 + tan x)
Answer:
Let I = \(\mathrm{I}=\int e^{x}(\sec x+\sec x \tan x) d x\)
Take f(x) = sec x
f ‘ (x) = sec x tan x
This is of the form of \(\int e^{x}\left[f(x)+f^{\prime}(x)\right] d x\) = e^x f(x) + c
= e^x sec x + c

Question 4.

Answer:

Take f(x) = tan x
f'(x) = sec² x
[∫ e^x [f (x) + f (x)] dx = e^x f(x) + c]
I = e^x tan x + c

Question 5.

Answer:

Put tan^-1 x = t
\(\frac{1}{1+x^{2}}\)dx = dt
x = tan t
∴ I = ∫e^t [1 + tan t + tan²t] dt
= ∫e^t [1 + tan² t + tan t] dt
= ∫e^t [sec²t + tan t] dt
= ∫e^t [tan t + sec²t] dt
f(x) = tan t
f'(x) = sec²t
[∫ e^x [f (x) + f (x)] dx = e^x f(x) + c]
∴ I = e^t tan t + c
I = e^tan-1(x) . x + c
I = x e^tan-1(x) + c

Question 6.

Answer:

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 11 Integral Calculus Ex 11.8 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 11 Integral Calculus Ex 11.8

Integrate the following with respect to x:

Question 1.
(i) e^ax cos bx
Answer:

(ii) e^2x sin x
Answer:

(iii) e^-x cos 2x
Answer:

Question 2.
(i) e^-3x sin 2x
Answer:

(ii) e^-4x sin 2x
Answer:

(iii) e^-3x cos x
Answer:

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 11 Integral Calculus Ex 11.7 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 11 Integral Calculus Ex 11.7

Integrate the following with respect to x:

Question 1.
(i) 9x e^3x
Answer:

(ii) x sin 3x
Answer:

(iii) 25 x e^-5x
Answer:

(iv) x sec x tan x
Answer:
∫ x sec x tan x dx
u = x, u’ = 1 , u” = 0
dv = sec x tan x dx , v = ∫ sec x tan x dx = sec x
v₁ = ∫ v dx = ∫ sec x dx = log |sec x + tan x|
v₂ = ∫ v₁ dx = ∫ log |sec x + tan x| dx
∫ u dv = uv – u’ v₁ + u” v₂ – u”’ v₃ + ………….
∫x sec x tan x dx = x sec x – 1 × log |sec x + tan x| + 0 × ∫ log |x sec x tan x| + c
∫x sec x tan x dx = x sec x – log |sec x + tan x| + c

Question 2.
(i) x log x
Answer:

(ii) 27 x² e^3x
Answer:

(iii) x² cos x
Answer:

(iv) x³ sin x
Answer:
∫ x³ sin x
u = x³, u’ = 3x², u” = 6x , u”’ = 6 , u’^v = 0
dv = sin x dx ⇒ v = ∫ sin x dx = – cosx
v₁ = ∫v dx = ∫- cos x dx = – ∫ cos x dx = – sin x
v₂ = ∫ v₁ dx = ∫ – sin x dx = – ∫ sin x dx = – (- cos x) = cos x
v₃ = ∫ v₂ dx = ∫ cos x dx = sin x
v₄ = ∫ v₃ dx = ∫ sin x dx = – cos x
∫ u dv = uv – u’ v₁ + u” v₂ – u”’ v₃ + u’^vv₄ – ………..
∫x³ sin x dx = x³ (- cos x) – 3x² (- sin x) + 6x (cos x) – 6 sin x + 0 (- cos x) + c
= -x³ cos x + 3x² sin x + 6x cos x – 6 sin x + c

Question 3.
(i)
Answer:

(ii) x⁵ e^x2
Answer:

(iii)
Answer:

(iv)
Answer:

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 10 Differentiability and Methods of Differentiation Ex 10.5 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 10 Differentiability and Methods of Differentiation Ex 10.5

Choose the correct or the most suitable answer from the given four alternatives.

Question 1.

(1) \(\frac{\pi}{180}\) cos x°
(2) \(\frac{1}{90}\) cos x°
(3) \(\frac{\pi}{90}\) cos x°
(4) \(\frac{2}{\pi}\) cos x°
Answer:
(2) \(\frac{1}{90}\) cos x°

Explaination:

Question 2.
If y = f(x² + 2) and f'(3) = 5 , then \(\frac{\mathrm{dy}}{\mathrm{d} x}\) at x = 1 is
(1) 5
(2) 25
(3) 15
(4) 10
Answer:
(4) 10

Explaination:
y = f(x² + 2)
\(\frac{\mathrm{dy}}{\mathrm{d} x}\) = f’ (x² + 2) × 2x
\(\frac{\mathrm{dy}}{\mathrm{d} x} / x=1\) = f’ (1² + 2) × 2 × 1
= f’(3) × 2
= 5 × 2 = 10

Question 3.
If y = \(\frac{1}{4}\)u⁴, u = \(\frac{2}{3}\) x³ + 5, then \(\frac{\mathrm{dy}}{\mathrm{d} x}\) is
(1) \(\frac{1}{27}\) x² (2x³ + 15)³
(2) \(\frac{2}{27}=\) x (2x³ + 5)³
(3) \(\frac{2}{27}=\) x² (2x³ + 15)³
(4) – \(\frac{2}{27}=\) x (2x³ + 5)³
Answer:
(3) \(\frac{2}{27}=\) x² (2x³ + 15)³

Explaination:

Question 4.
If f(x) = x² – 3x, then the points at which f (x) = f’ (x) are
(1) both positive integers
(2) both negative integers
(3) both irrational
(4) one rational and another irrational
Answer:
(3) both irrational

Explaination:
f(x) = x² – 3x
f’(x) = 2x – 3
f(x) = f'(x)
⇒ x² – 3x = 2x – 3
x² – 3x – 2x + 3 = 0
x² – 5x + 3 = 0

Question 5.
If y = \(\frac{1}{a-z}\), then \(\frac{\mathrm{d} \mathrm{z}}{\mathrm{d} \mathrm{y}}\) is
(1) (a – z)²
(2) – (z – a)²
(3) (z + a)²
(4) – (z + a)²
Answer:
(1) (a – z)²

Explaination:

Question 6.
If y = cos (sin x²), then \(\frac{\mathrm{dy}}{\mathrm{d} x}\) at x = \(\sqrt{\frac{\pi}{2}}\) is
(1) – 2
(2) 2
(3) – 2 \(\sqrt{\frac{\pi}{2}}\)
(4) 0
Answer:
(4) 0

Explaination:

Question 7.
If y = mx + c and f(0) = f’(0) = 1, then f(2) is
(1) 1
(2) 2
(3) 3
(4) – 3
Answer:
(3) 3

Explaination:
y = mx+c
\(\frac{d y}{d x}\) = m
y = x + c (i.e.) f(x) = x + c
y(a tx = 0) = f(0) 0 + c = 1 ⇒ c = 1
y = x + 1 ⇒ f(x) = x + 1
f(2) = 2 + 1 = 3

Question 8.
If f(x) = x tan^-1 x then f'(x) is
(1) \(1+\frac{\pi}{4}\)
(2) \(\frac{1}{2}+\frac{\pi}{4}\)
(3) \(\frac{1}{2}-\frac{\pi}{4}\)
(4) 2
Answer:
(2) \(\frac{1}{2}+\frac{\pi}{4}\)

Explaination:

Question 9.
\(\frac{d}{d x}\) (e^{x + 5 log x}) is
(1) e^x . x⁴ (x + 5)
(2) e^x . x (x + 5)
(3) e^x + \(\frac{5}{x}\)
(4) e^x – \(\frac{5}{x}\)
Answer:
(1) e^x . x⁴ (x + 5)

Explaination:

Question 10.
If the derivative of (ax – 5)e at x = 0 is – 13 , then the value of a ¡s
(1) 8
(2) – 2
(3) 5
(4) 2
Answer:
(4) 2

Explaination:
y = (ax – 5)e^3x
\(\frac{d y}{d x}\) = y’ = (ax – 5) (3e^3x) + e^3x (a)
= e^3x[3ax – 15 + a]
Given \(\frac{d y}{d x}\) = -13 at x = 0
⇒ [-15 + a] = -13
⇒ a = -13 + 15
a = 2

Question 11.
x = \(\frac{1-t^{2}}{1+t^{2}}\), y = \(\frac{2 t}{1+t^{2}}\) then \(\frac{d y}{d x}\) is
(1) – \(\frac{\mathbf{y}}{x}\)
(2) \(\frac{\mathbf{y}}{x}\)
(3) – \(\frac{x}{y}\)
(4) \(\frac{x}{y}\)
Answer:
(3) – \(\frac{x}{y}\)

Explaination:

Question 12.
If x = a sin θ and y = b cos θ, then \(\) is
(1) \(\frac{\mathbf{a}}{\mathbf{b}^{2}}\) sec² θ
(2) \(-\frac{\mathbf{b}}{\mathbf{a}}\) sec² θ
(3) \(-\frac{b}{a^{2}}\) sec³ θ
(4) \(-\frac{b^{2}}{a^{2}}\) sec³ θ
Answer:
(3) \(-\frac{b}{a^{2}}\) sec³ θ

Explaination:

Question 13.
The differential coefficient of log₁₀ x with respect to log₁₀ x is
(1) 1
(2) – (log₁₀ x)²
(3) (log₁₀ x)²
(4) \(\frac{x^{2}}{100}\)
Answer:
(2) – (log₁₀ x)²

Explaination:

Question 14.
If f(x) = x + 2 , then f’ (f(x)) at x= 4 is
(1) 8
(2) 1
(3) 4
(4) 5
Answer:
(2) 1

Explaination:
f(x) x + 2
f’ (f(x)) = \(\frac{\mathrm{d}}{\mathrm{d} x}\) (f(x))
= \(\frac{\mathrm{d}}{\mathrm{d} x}\) (x + 2) = 1

Question 15.
If y = \(\frac{(1-x)^{2}}{x^{2}}\), then \(\frac{\mathrm{dy}}{\mathrm{d} x}\) is
(1) \(\frac{2}{x^{2}}+\frac{2}{x^{3}}\)
(2) \(-\frac{2}{x^{2}}+\frac{2}{x^{3}}\)
(3) \(-\frac{2}{x^{2}}-\frac{2}{x^{3}}\)
(4) \(-\frac{2}{x^{3}}+\frac{2}{x^{2}}\)
Answer:
(4) y = \(\frac{(1-x)^{2}}{x^{2}}\)

Explaination:

Question 16.
If pv = 81, then \(\frac{\mathbf{d} \mathbf{p}}{\mathbf{d v}}\) at v = 9 is
(1) 1
(2) – 1
(3) 2
(4) – 3
Answer:
(2) – 1

Explaination:

Question 17.

(1) 0
(2) 2
(3) 3
(4) 4
Answer:
(3) 3

Explaination:

Question 18.
It is given that f’ (a) exists, then

(1) f(a) – af'(a)
(2) f'(a)
(3) – f'(a)
(4) f(a) + af'(a)
Answer:
(1) f(a) – af'(a)

Explaination:

Question 19.
If f(x) = then f’ (2) is
(1) 0
(2) 1
(3) 2
(4) does not exist
Answer:
(3) 2

Explaination:

Question 20.
If g(x) = (x² + 2x + 3) f(x) and f(0) – 5 and then g'(0) is
(1) 20
(2) 14
(3) 18
(4) 12
Answer:

Question 21.
If f(x) = then at x = 3, f'(x) is
(1) 1
(2) – 1
(3) 0
(4) does not exist
Answer:
(2) – 1

Explaination:

From equations (1) and (2) we get
f’ (3^–) ≠ f’ (3⁺)
∴ limit of f(x) does not exist at x = 3
f’ (x) does not exist at x = 3

Question 22.
The derivative of f(x)= x|x| at x = – 3 is
(1) 6
(2) – 6
(3) does not exist
(4) 0
Answer:
(1) 6

Explaination:
f(x) = x|x|
f(x) = x(-x) ⇒ f(x) = – x²
f ‘(x) = -(2x)
f ‘(-3) = -(2) (-3) = 6

Question 23.
If f(x) = then which one of the following is true?
(1) f(x) is not differentiable at x = a
(2) f(x) is discontinuous at x = a
(3) f(x) is continuous for all x in R
(4) f(x) is differentiable for all x ≥ a
Answer:
(1) f(x) is not differentiable at x = a

Explaination:


f’ (a⁺) = 3 ………. (2)
From equations (1) and (2) we get
f'(a^–) ≠ f'(a⁺)
∴ f’ (x) does not exist at x = a
∴ f(x) is not differentiable at x = a

Question 24.
If f(x) = is differentiable at x = 1, then
(1) a = \(\frac{1}{2}\), b = \(\frac{-3}{2}\)
(2) a = \(\frac{-1}{2}\), b = \(\frac{3}{2}\)
(3) a = \(-\frac{1}{2}\), b = \(-\frac{3}{2}\)
(4) a = \(\frac{1}{2}\), b = \(\frac{3}{2}\)
Answer:
(3) a = \(-\frac{1}{2}\), b = \(-\frac{3}{2}\)

Explaination:

Question 25.
The number of points in R in which the function f(x) = |x – 1| + |x – 3| + sin x is not differentiable, is
(1) 3
(2) 2
(3) 1
(4) 4
Answer:
(2) 2

Explaination:
f(x) = |x – 1| + |x – 3| + sin x is not differentiable at x = 1, and x = 3

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 11 Integral Calculus Ex 11.1 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 11 Integral Calculus Ex 11.1

Integrate the following with respect to x:

Question 1.
(i) x¹¹
(ii) \(\frac{1}{x^{7}}\)
(iii) \(\sqrt[3]{x^{4}}\)
(iv) (x⁵)^1/8
Answer:
(i) x¹¹

(ii) \(\frac{1}{x^{7}}\)

(iii) \(\sqrt[3]{x^{4}}\)

(iv) (x⁵)^1/8

Question 2.
(i) \(\frac{1}{\sin ^{2} x}\)
(ii) \(\frac{\tan x}{\cos x}\)
(iii) \(\frac{\cos x}{\sin ^{2} x}\)
(iv) \(\frac{1}{\cos ^{2} x}\)
Answer:
(i) \(\frac{1}{\sin ^{2} x}\)
\(\frac{1}{\sin ^{2} x}\) = ∫cosec² x dx
= – cot x + c

(ii) \(\frac{\tan x}{\cos x}\)
\(\frac{\tan x}{\cos x}\) = ∫sec x tan x dx
= sec x + c

(iii) \(\frac{\cos x}{\sin ^{2} x}\)

(iv) \(\frac{1}{\cos ^{2} x}\)
\(\frac{1}{\cos ^{2} x}\) = ∫sec² x dx
= tan x + c

Question 3.
(i) 12³
(ii) \(\frac{x^{24}}{x^{25}}\)
(iii) e^x
Answer:
(i) 12³
∫12³ dx = 12³ ∫dx = 12³x + c

(ii) \(\frac{x^{24}}{x^{25}}\)

(iii) e^x
∫e^x dx = e^x + c

Question 4.
(i) (1 + x²)^-1
(ii) (1 – x²)^-1/2
Answer:
(i) (1 + x²)^-1

(ii) (1 – x²)^-1/2

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 11 Integral Calculus Ex 11.2 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 11 Integral Calculus Ex 11.2

Integrate the following functions with respect to x

Question 1.
(i) (x + 5)⁶
(ii) \(\frac{1}{(2-3 x)^{4}}\)
(iii) \(\sqrt{3 x+2}\)
Answer:
(i) (x + 5)⁶

(ii) \(\frac{1}{(2-3 x)^{4}}\)

(iii) \(\sqrt{3 x+2}\)

Question 2.
(i) sin 3x
(ii) cos (5 – 11x)
(iii) cosec² (5x – 7)
Answer:
(i) sin 3x
∫sin (ax + b) dx = – \(\frac{1}{a}\) cos (ax + b) + c
∫sin 3x dx = – \(\frac{1}{3}\) cos 3x + c

(ii) cos (5 – 11x)

(iii) cosec² (5x – 7)
[∫cosec² (ax + b) dx = – \(\frac{1}{a}\) cot (ax + b) + c]
∫cosec² (5x – 7) dx = – \(\frac{1}{5}\) cot (5x – 7) + c

Question 3.
(i) e^{3x – 6}
(ii) e^{8 – 7x}
(iii) \(\frac{1}{6-4 x}\)
Answer:
(i) e^{3x – 6}
[∫e^ax+b dx = \(\frac{1}{a}\) e^{ax + b} + c]
∫e^{3x – 6} dx = \(\frac{1}{3}\) e^{3x – 6} + c

(ii) e^{8 – 7x}
[∫e^ax+b dx = \(\frac{1}{a}\) e^{ax + b} + c]
∫e^{8 – 7x} = \(\frac{1}{-7}\) e^{8 – 7x} + c
∫e^{8 – 7x} = \(-\frac{1}{7}\) e^{8 – 7x} + c

(iii) \(\frac{1}{6-4 x}\)

Question 4.
(i) sec² \(\frac{x}{5}\)
(ii) cosec (5x + 3) cot (5x + 3)
(iii) 30 sec (2 – 15x) tan (2 – 15x)
Answer:
(i) sec² \(\frac{x}{5}\)
[∫sec² (ax + b)dx = \(\frac{1}{a}\) tan (ax + b) + c]

(ii) cosec (5x + 3) cot (5x + 3)
[∫cosec (ax + b) cot (ax + b) dx = – \(\frac{1}{a}\) cosec (ax + b) + c]
∫cosec (5x + 3) cot (5x + 3) dx = – \(\frac{1}{5}\) cosec (5x + 3) + c]

(iii) 30 sec (2 – 15x) tan (2 – 15x)
[∫sec (ax + b) tan (ax + b)dx = \(\frac{1}{a}\) sec (ax + b) + c]
∫30 sec (2 – 15x) tan (2 – 15x)dx = 30 × \(\frac{1}{-15}\) × sec (2 – 15x) + c
= – 2 sec (2 – 15x) + c

Question 5.
(i) \(\frac{1}{\sqrt{1-(4 x)^{2}}}\)
(ii) \(\frac{1}{\sqrt{1-81 x^{2}}}\)
(iii) \(\frac{1}{1+36 x^{2}}\)
Answer:
(i) \(\frac{1}{\sqrt{1-(4 x)^{2}}}\)

(ii) \(\frac{1}{\sqrt{1-81 x^{2}}}\)

(iii) \(\frac{1}{1+36 x^{2}}\)

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 11 Integral Calculus Ex 11.3 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 11 Integral Calculus Ex 11.3

Integrate the following with respect to x:

Question 1.
(x + 4)⁵ + \(\) – cosec² (3x – 1)
Answer:

Question 2.
4 cos (5 – 2x) + 9 e^{3x – 6} + \(\frac{24}{6-4 x}\)
Answer:

Question 3.
sec² \(\frac{x}{5}\) + 18 cos 2x + 10 sec (5x + 3) tan (5x + 3)
Answer:

Question 4.

Answer:

Question 5.

Answer:

Question 6.

Answer:

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 11 Integral Calculus Ex 11.4 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 11 Integral Calculus Ex 11.4

Question 1.
If f'(x) = 4x – 5 and f(2) = 1, find f(x)
Answer:
\(\int f^{\prime}(x) d x=\int(4 x-5) d x\)
f(x) = \(\frac{4 x^{2}}{2}\) – 5x + c
f(x) = 2x² – 5x + c
But f(2) = 1
2(2)² – 5(2) + c = 1
8 – 10 + c = 1
c = 3
Thus, f(x) = 2x² – 5x + 3

Question 2.
If f'(x) = 9x² – 6x and f(0) = – 3 find f(x).
Answer:
Given f’ (x) = 9x² – 6x and f(0) = – 3
\(\frac{\mathrm{d} f(x)}{\mathrm{d} x}\) = 9x² – 6x
df(x) = (9x² – 6x) dx
∫ df(x) = ∫ (9x² – 6x)
∫ df(x) = ∫ 9x² dx – ∫ 6x dx
f(x) = 9 ∫x² dx – 6 ∫ x dx
f(x) = 9 × \(\frac{x^{3}}{3}\) – 6 \(\frac{x^{2}}{2}\) + C
f(x) = 3x³ – 3x² + c —— (1)
f(o) = – 3
(1) ⇒ f(0) = 3 × – 3 . 0² + c
– 3 = 0 – 0 ⇒c = – 3
Substituting in equation (1) we get
f(x) = 3x³ – 3x² – 3
f(x) = 3(x³ – x² – 1)

Question 3.
If f” (x) = 12x – 6 and f(1) = 30, f’ (1) = 5, find f(x)
Answer:
\(\)
f'(x) = \(\) – 6x + c
f'(x) = 6x² – 6x + c
But f'(1) = 5
6(1)² – 6(1) + c = 5
c = 5
f” (x) = 6x² – 6x + 5
\(\int f^{\prime \prime}(x) d x=\int\left(6 x^{2}-6 x+5\right) d x\)
f(x) = \(\frac{6 x^{3}}{3}-\frac{6 x^{2}}{2}\) + 5x + c
f(x) = 2x³ – 3x² + 5x + c
But f(1) = 30
2(1)³ – 3(1)² + 5(1) + c = 30
2 – 3 + 5 + c = 30
c = 26
f(x) = 2x³ – 3x² + 5x + 26

Question 4.
A ball is thrown vertically upward from the ground with an initial velocity of 39.2 m / sec. If the only force considered is that attributed to the acceleration due to gravity, find
(i) how long will it take for the ball to strike the ground?
(ii) the speed with which will it strike the ground? and
(iii) how high the ball will rise?
Answer:
Initial velocity of the ball = 39.2 in / s
Let s be the distance of the ball from the ground at time t.
u = 39.2m/s.
s = ut – \(\frac{1}{2}\) . gt² where g = 9.8 m / sec.
s = 39.2 t – \(\frac{1}{2}\) × (9.8) t²
s = 39.2 t – 49 t²

(i) how long will it take for the ball to strike the ground?
For the ball to strike the ground, s = 0
∴ 39.2 t – 4.9 t² = 0
t(39.2 – 4.9 t) = 0
39.2 – 4.9 t = 0 since t ≠ 0
4.9 t = 39.2
t = \(\frac{39.2}{4.9}\) = 8
t = 8 sec.

(ii) The speed with which will it strike the ground?
At t = 8; (1) ⇒ v = -9.8(8) + 39.2
= -78.4 + 39.2
= -39.2
∴ The speed with which the ball will strike the ground is = 39.2 m/s
(iii) At maximum height v = 0
⇒ (2) ⇒ -9.8 t + 39.2 = 0
t = 4 sec
⇒ (3) ⇒ x = \(-\frac{9.8 \times 16}{2}\) + 39.2 × 4
= -78.4 + 156.8 = 78.4 m/s

Question 5.
A wound is healing in such a way that t days since Sunday the area of the of the wound has been decreasing at a rate of \(-\frac{3}{(t+2)^{2}}\) cm² per day. If on Monday the area of the wound was 2 cm²
(i) What was the area of the wound on Sunday?
Answer:

(ii) What is the anticipated area of the wound on Thursday if it continues to heal at the same rate?
Answer:
(ii) From Sunday to Thursday there are 4 days. Substituting t = 4 in equation (2) we get

∴ The anticipated area of the wound on Thursday = 1.5 sq. cm.

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 11 Integral Calculus Ex 11.5 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 11 Integral Calculus Ex 11.5

Integrate the following functions with respect to x

Question 1.

Answer:

Question 2.

Answer:

Question 3.
(2x – 5) (3x + 4x)
Answer:
∫(2x – 5) (3x + 4x) dx
= ∫(72 x + 8x² – 180 – 20x) dx
= ∫ 72 x dx + ∫82x² dx – ∫180 dx – ∫2o x dx
= 72∫x dx + 8∫x²dx – 180 ∫dx – 20 ∫x dx

Question 4.
cot² x + tan² x
Answer:
∫cot² x + tan² x
= ∫(cosec²x – 1 + sec²x – 1) dx
= ∫(cosec²x + sec²x – 2) dx
= ∫cosec²x dx + ∫sec²x dx – ∫2 dx
= – cot x + tan x – 2x + c
= tan x – cot x – 2x + c

Question 5.

Answer:

[cos 2x = cos²x – sin²x-1
cos 2x = 2 cos²x – 1]


= 2 ∫(cos x + cos α) dx
= 2 ∫ cos x dx + 2 ∫ cos α dx
= 2 ∫ cos x dx + 2 ∫ cos α ∫ dx
= 2 sin x + 2 cos α(x) + c
= 2 sin x + 2x cos α + c

Question 6.

Answer:

Question 7.
\(\frac{3+4 \cos x}{\sin ^{2} x}\)
Answer:

= 3 ∫ cosec² x dx + 4 ∫ cot x cosec x . dx
= 3 ∫ cosec² x dx + 4 ∫ cosec x cot x dx
= 3 × – cot x + 4 × – cosec x + c
= – 3 cot x – 4 cosec x + c

Question 8.
\(\frac{\sin ^{2} x}{1+\cos x}\)
Answer:

= ∫ (1 – cos x) dx
= ∫ dx – ∫ cos x dx
= x – sin x + cannot

Question 9.
\(\frac{\sin 4 x}{\sin x}\)
Answer:

[sin 2A = 2 sin A cos A]

= 4 ∫ cos 2x cos x . dx
= 2 ∫ 2 cos 2x cosx . dx
= 2 ∫ [cos(2x + x) + cos(2x – x)] dx
[2 cos A cos B = cos (A + B) + cos (A – B)]
= 2 ∫ (cos 3x + cos x) dx
= 2 ∫ cos 3x dx + 2 ∫ cos x dx
= 2 \(\frac{\sin 3 x}{3}\) + 2 sin x + c
= 2 [\(\frac{\sin 3 x}{3}\) + sin x] + c

Question 10.
cos 3x cos 2x
Answer:

Question 11.
sin² 5x
Answer:

Question 12.

Answer:

[sin 2A = 2 sin A cos A]

Question 13.
e^{x log a} e^x
Answer:
∫e^{x log a} e^x = ∫e ^{log ax} . e^x . dx
= ∫ a^x e^x . dx
= ∫ (ae)^x . dx
[∫a^x . dx = \(\frac{\mathrm{a}^{x}}{\log \mathrm{a}}\) + c]
= \(\frac{(\mathrm{ae})^{x}}{\log (\mathrm{ae})}\) + c

Question 14.

Answer:

Question 15.

Answer:

Question 16.

Answer:

Question 17.

Answer:

Put x = – 3
– 3 + 1 = A (- 3 + 3) + B (- 3 + 2)
– 2 = A × 0 + B (- 1)
B = 2

Put x = – 2
– 2 + 1 = A(- 2 + 3) + B(- 2 + 2)
– 1 = A × 1 + B × 0
A = – 1

= – log |x + 2| + 2 log |x + 3| + c
= 2 log |x + 3| – log |x + 2|+ c

Question 18.

Answer:

1 = A(x + 2)² + B (x – 1) (x + 2) + C (x – 1)

Put x= -2
1 = A(- 2 + 2)² + B(- 2 – 1) (- 2 + 2) + C(- 2 – 1)
1 = A × 0 + B × 0 + C × – 3
C = \(\frac{1}{3}\)

Put x = 1
1 = A(1 + 2)² + B (1 – 1) (1 + 2) + C(1 – 1)
1 = A × 3² + B × 0 + C × 0
A = \(\frac{1}{9}\)

Put x = 0
1 = A (0 + 2)² + B (0 – 1) (0 + 2) + C (0 – 1)
1 = A × 4 – 2B – C
1 = \(\frac{1}{9}\) × 4 – 2B + \(\frac{1}{3}\)
1 = \(\frac{4}{9}+\frac{1}{3}\) – 2B

Question 19.

Answer:

3x – 9 = A(x + 2) (x² + 1) + B(x – 1) (x² + 1) + (Cx + D) (x – 1) (x + 2)
3x – 9 = A (x + 2) (x² + 1) + B(x – 1) + (x² + 1) + Cx (x – 1) (x + 2) + D (x – 1) (x + 2)

Put x = – 2
3 × – 2 – 9 = A (- 2 + 2) ((2)² + 1) + B(- 2 – 1) ((- 2)² + 1) + C(- 2) (- 2 – 1) (- 2 + 2) + D(- 2 – 1) (- 2 + 2)
– 6 – 9 = A × 0 + B × (-3) (4 + 1) + C × 0 + D × 0
-15 = B ×- 3 × 5
-15 = – 15B ⇒ B = 1

Put x = 1
3 × 1 – 9 = A(1 + 2) (12 + 1) + B(1 – 1) (12 + 1) + C × 1 (1 – 1) (1 + 2) + D(1 – 1) (1 + 2)
3 – 9 = A × 3 × 2 + B × 0 + C × 0 + D × 0
– 6 = 6A ⇒ A = – 1

Put x = 0
3 × 0 – 9 = A(0 + 2) (0² + 1) + B(0 – 1) (0² + 1) + C × 0 (0 – 1) (0 + 2) + D(0 – 1) (0 + 2)
– 9 = 2A – B + 0 – 2D
– 9 = 2A – B – 2D
– 9 = – 2 × – 1 – 1 – 2D
– 9 = – 2 – 1 – 2D
9 = 3 + 2D
⇒ 2D = 9 – 3
⇒ 2D = 6 ⇒ D = 3

Put x = – 1
3 × – 1 – 9 = A(- 1 + 2) ((1)² + 1) + B(- 1 – 1) ((- 1)² + 1)) + C × – 1 × (- 1 – 1) (- 1 + 2) + D(- 1 – 1) (- 1 + 2)
– 3 – 9 = A × 1(1 + 1)+ B × (- 2) (1 + 1) – C × – 2 + D × – 2 × 1
– 12 = 2A – 4B + 2C – 2D
– 12 = 2 × -1 – 4 × 1 + 2C – 2 × 3
– 12 = – 2 – 4 + 2C – 6
– 12 = – 12 + 2C ⇒ C = 0

Question 20.

Answer:


1 = A(x – 2) + B(x – 1)

put x = 2
1 = A(2 – 2) + B(2 – 1)
1 = A × 0 + B × 1
B = 1

Put x = 1
1 = A(1 – 2) + B(1 – 1)
1 = A × – 1 + B × 0
A = – 1

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 10 Differentiability and Methods of Differentiation Ex 10.4 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 10 Differentiability and Methods of Differentiation Ex 10.4

Find the derivatives of the following:

Question 1.
y = x^{cos x}
Answer:
y = x^{cos x}
Taking log on both sides
log y = log x^{cos x}
log y = cos x log x
Differentiating with respect to x

Question 2.
y = x^{log x} + (log x)^x
Answer:
y = x^{log x} + (log x)^x
Let u = x^{log x}, v = (log x)^x
log u = log x^{log x}
log u = (log x) (log x)
log u = (log x)²

v = (log x)^x
log v = log (log x)^x
log v = x log (log x)

Question 3.
\(\sqrt{x y}\) = e^{(x – y)}
Answer:

Question 4.
x^y = y^x
Answer:
x^y = y^x
Taking log on both sides
log x^y = log y^x
y log x = x log y
Differentiate with respect to x

Question 5.
(cos x)^{log x}
Answer:
y = (cos x)^{log x}
Taking log on both sides
log y = log (cos x)^{log x}
log y = (log x) log (cos x)
Differentiating with respect to x

Question 6.
\(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\)
Answer:

Question 7.
\(\sqrt{x^{2}+y^{2}}=\tan ^{-1}\left(\frac{y}{x}\right)\)
Answer:

Question 8.
tan (x + y) + tan (x – y) = x
Answer:
tan (x + y) + tan (x – y) = x
Differentiating with respect to x

Question 9.
If cos(xy) = x, show that
\(\frac{d y}{d x}=\frac{-(1+y \sin (x y))}{x \sin x y}\)
Answer:
cos (xy) = x
Differentiating with respect to x

Question 10.
\(\tan ^{-1} \sqrt{\frac{1-\cos x}{1+\cos x}}\)
Answer:
Let y = \(\tan ^{-1} \sqrt{\frac{1-\cos x}{1+\cos x}}\)
[1 – cos 2θ = 2 sin²θ and 1 + cos 2θ = 2 sin² θ]

Question 11.
tan^-1 = \(\left(\frac{6 x}{1-9 x^{2}}\right)\)
Answer:

Question 12.
\(\cos \left(2 \tan ^{-1} \sqrt{\frac{1-x}{1+x}}\right)\)
Answer:

Question 13.
x = a cos^t ; y = a sin³t
Answer:
x = a cos^t , y = a sin³t

Question 14.
x = a (cos t + t sin t);
y = a (sin t – t cos t)
Answer:
x = a (cos t + t sin t) , y = a (sin t – t cos t)
\(\frac{d x}{d t}\) = a [- sin t + t cos t + sin t ]
\(\frac{d x}{d t}\) = at cos t —— (1)
y = a (sin t – t cos t)
\(\frac{d x}{d t}\) = a [cos t – (t × – sin t + cos t × 1)]
\(\frac{d x}{d t}\) = a[cos t + t sin t – cos t]
\(\frac{d x}{d t}\) = at sin t —— (2)
From equations (1) and (2) we get

Question 15.
x = \(\frac{1-t^{2}}{1+t^{2}}\) , y = \(\frac{2 t}{1+t^{2}}\)
Answer:

Question 16.
cos^-1\(\left(\frac{1-x^{2}}{1+x^{2}}\right)\)
Answer:
Let y = cos^-1\(\left(\frac{1-x^{2}}{1+x^{2}}\right)\)
Put x = tan θ

y = cos^-1 (cos 2θ)
y = 2θ
y = 2 tan^-1 x

Question 17.
sin^-1 (3x – 4x³)
Answer:
Let y = sin^-1 (3x – 4x³)
Put x = sin θ
y = sin^-1 (3 sin θ – 4 sin³ θ)
y = sin^-1 (sin 3θ)
y = 3θ
y = 3 sin^-1 x
\(\frac{d y}{d x}=\frac{3}{\sqrt{1-x^{2}}}\)

Question 18.
tan^-1 \(\left(\frac{\cos x+\sin x}{\cos x-\sin x}\right)\)
Answer:
Let y = tan^-1 \(\left(\frac{\cos x+\sin x}{\cos x-\sin x}\right)\)

Question 19.
Find the derivative of sin x² with respect to x².
Answer:
Let u = sin x²
\(\frac{\mathrm{d} \mathrm{u}}{\mathrm{d} x}\) = cos (x²) × 2x
\(\frac{\mathrm{d} \mathrm{u}}{\mathrm{d} x}\) = 2x cos (x²)
Let v = x²

Question 20.
Find the derivative of sin^-1\(\left(\frac{2 x}{1+x^{2}}\right)\) with respect to tan^-1 x.
Answer:

Question 21.
If u = tan^-1\(\frac{\sqrt{1+x^{2}}-1}{x}\) and v = tan^-1x, find \(\frac{\mathrm{du}}{\mathrm{dv}}\)
Answer:

Question 22.
Find the derivative with tan^-1\(\left(\frac{\sin x}{1+\cos x}\right)\) with respect to tan^-1\(\left(\frac{\cos x}{1+\sin x}\right)\)
Answer:

Question 23.
If y = sin^-1x then find y”.
Answer:
y = sin^-1 x

Question 24.
If y = e^tan-1x, show that (1 + x²) y” + (2x – 1) y’ = 0
Answer:
y = e^tan-1x
y = e^tan-1x \(\left(\frac{1}{1+x^{2}}\right)\)
⇒ y’ = \(\frac{y}{1+x^{2}}\) ⇒ y'(1 + x²) = y
differentiating w.r.to x
y’ (2x) + (1 + x²) (y”) = y’
(i.e.) (1 + x²) y” + y’ (2x) – y’ = 0
(i.e.) (1 + x²) y” + (2x – 1) y’ = 0

Question 25.
If y = \(\frac{\sin ^{-1} x}{\sqrt{1-x^{2}}}\), show that (1 – x²)y₂ – 3xy₁ – y = 0.
Answer:

Question 26.
If x = a (θ + sin θ), y = a (1 – cos θ) then prove that at θ = \(\frac{\pi}{2}\), y” = \(\frac{1}{a}\)
Answer:

Question 27.
If sin y = x sin (a + y), the prove that \(\frac{d y}{d x}=\frac{\sin ^{2}(a+y)}{\sin a}\), a ≠ nπ
Answer:
Given sin y = x sin (a + y) ——- (1)
Differentiating with respect to x , we get
cos y \(\frac{\mathrm{dy}}{\mathrm{d} x}\) = x cos (a + y) (0 + \(\frac{\mathrm{dy}}{\mathrm{d} x}\)) + sin (a + y) . 1

Question 28.
If y = (cos^-1 x)², prove that (1 – x²) \(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{d} x^{2}}\) – x \(\frac{\mathrm{dy}}{\mathrm{d} x}\) – 2 = 0. Hence find y₂ when x = 0.
Answer: