AP Inter 2nd Year Maths Exercise 5b Solutions

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

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

I.

Question 1.
Differentiate the function sin(x² + 5) with respect to x.
Solution:
Let y = sin(x² + 5)
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{\mathrm{d}}{\mathrm{dx}}\) sin(x² + 5) = cos(x² + 5) \(\frac{\mathrm{d}}{\mathrm{dx}}\)(x² + 5) [∵ \(\frac{\mathrm{d}}{\mathrm{dx}}\) sin f(x) = cos f(x) \(\frac{\mathrm{d}}{\mathrm{dx}}\)f(x)]
= cos(x² + 5) (2x + 0) [∵ \(\frac{\mathrm{d}}{\mathrm{dx}}\) xⁿ = nx^n-1 and \(\frac{\mathrm{d}}{\mathrm{dx}}\) (c) = 0]
= 2x cos(x² + 5)

Question 2.
Differentiate the function cos(sin x) with respect to x.
Solution:
Let y = cos(sin x)
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{\mathrm{d}}{\mathrm{dx}}\) cos(sin x) = -sin(sin x) \(\frac{\mathrm{d}}{\mathrm{dx}}\)(sin x) [∵ \(\frac{\mathrm{d}}{\mathrm{dx}}\) cos f(x) = -sin f(x) \(\frac{\mathrm{d}}{\mathrm{dx}}\)f(x)]
= -sin(sin x) cos x = -cos x sin(sin x)

Question 3.
Differentiate the function sin(ax + b) with respect to x.
Solution:
Let y = sin(ax + b)
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{\mathrm{d}}{\mathrm{dx}}\)sin(ax + b) = cos(ax + b) \(\frac{\mathrm{d}}{\mathrm{dx}}\)(ax + b)
= cos(ax + b)[a\(\frac{\mathrm{d}}{\mathrm{dx}}\)(x) + \(\frac{\mathrm{d}}{\mathrm{dx}}\)(b)]
= cos(ax + b)[a(1) + 0] = a cos(ax + b)

Question 4.
Differentiate the function sec(tan(\(\sqrt{x}\))) with respect to x.
Solution:
Let y = sec(tan(\(\sqrt{x}\)))

Question 5.
Differentiate the function \(\frac{\sin (a x+b)}{\cos (c x+d)}\) with respect to x.
Solution:

Question 6.
Differentiate the function cosx³ . sin²(x⁵) with respect to x.
Solution:
Let y = cosx³ . sin²(x⁵) = cos x³ (sin x⁵)² [∵ sin²f(x) = [sin f(x)]²]
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = cos x³ \(\frac{\mathrm{d}}{\mathrm{dx}}\) (sin x⁵)² + (sin x⁵)² \(\frac{\mathrm{d}}{\mathrm{dx}}\) cos x³ [∵ \(\frac{\mathrm{d}}{\mathrm{dx}}\) (uv) = u\(\frac{\mathrm{dv}}{\mathrm{dx}}\) + v\(\frac{\mathrm{du}}{\mathrm{dx}}\)]
= cos x³ 2(sin x⁵)\(\frac{\mathrm{d}}{\mathrm{dx}}\)sin x⁵ +(sin x⁵)² (-sin x³ )\(\frac{\mathrm{d}}{\mathrm{dx}}\)x³
= cos x³2(sin x⁵)cos x⁵ (5x⁴ + (sin x⁵)²(-sin x³)(3x²) [∵ \(\frac{\mathrm{d}}{\mathrm{dx}}\) sin x⁵ = cos x⁵ \(\frac{\mathrm{d}}{\mathrm{dx}}\) x⁵ = cos x⁵(5x⁴)]
= 10x⁴ cos x³ sin x⁵ cos x⁵ – 3x² sin² x⁵ sin x³
= x² sin x⁵[10x² cos x³ cos x⁵ – 3 sin x⁵ sin x³].

Question 7.
Differentiate the function \(2 \sqrt{\cot \left(x^2\right)}\) with respect to x.
Solution:

Question 8.
Differentiate the function cos(\(\sqrt{x}\)) with respect to x.
Solution:
Let y = cos(\(\sqrt{x}\))
∴ \(\frac{\mathrm{dy}}{\mathrm{dx}}\) = \(\frac{\mathrm{d}}{\mathrm{dx}}\) cos(\(\sqrt{x}\))
= -sin \(\sqrt{x}\) \(\frac{\mathrm{d}}{\mathrm{dx}}\) \(\sqrt{x}\) [∵ \(\frac{\mathrm{d}}{\mathrm{dx}}\) cos f(x) = -sin f(x) \(\frac{\mathrm{d}}{\mathrm{dx}}\) f(x)]
= -sin \(\sqrt{x} \frac{1}{2 \sqrt{x}}\) [∵ \(\frac{\mathrm{d}}{\mathrm{dx}} \sqrt{x}=\frac{1}{2 \sqrt{x}}\)

II.

Question 1.
Prove that the function f given by f(x) = |x – 1|, x ∈ R k is not differentiable at x = 1.
Solution:
Given f(x) = |x – 1| , x ∈ R ……….. (i)
To prove: f(x) is not differentiable at x = 1
Putting x = 1 in (i), f'(1) = |1 – 1| = |0| = 0

Here, L.H.L of f'(1) ≠ R.H.L of f'(1)
∴ f(x) is not differentiable at x = 1

Question 2.
Prove (hat the greatest nleger function defined by f(x) = |x|, 0 < x < 3 is not differentiable at x = 1 and x = 2.
Solution:
Given f(x) = [x], 0 < x < 3 ……….(i)
(a) Differentiability at x = 1
Putting x = 1 in (i), f(1) = [1] = 1 .
Left Hand derivative of f'(1) = \(\underset{\mathrm{x} \rightarrow 1-}{\mathrm{Lt}}\) \(\frac{f(x)-f(1)}{x-1}\) = \(\underset{\mathrm{x} \rightarrow 1-}{\mathrm{Lt}}\)\(\frac{[x]-1}{x-1}\)
Put x = 1 – h, h → 0+
\(\underset{\mathrm{h} \rightarrow 0+}{\mathrm{Lt}}\)\(\frac{[1-\mathrm{h}]-1}{1-\mathrm{h}-1}\) = \(\underset{\mathrm{h} \rightarrow 0+}{\mathrm{Lt}}\)\(\frac{0-1}{-h}\)
= \(\underset{\mathrm{h} \rightarrow 0+}{\mathrm{Lt}}\)\(\frac{1}{h}\)
We know that as h → 0+, [c – h] = c – 1 if c is an integer
∴ [1 – h] = 1 – 1 = 0
Put h = 0, \(\frac{1}{h}\) = \(\frac{1}{0}\) = ∞ does not exist
∴ f(x) is not differentiable at x = 1
(We need not find R f'(1) as L f'(1) does not exist).

(b) Differentiability at x = 2
Putting x = 2 in (i), f(2) = [2] = 2 .
Left Hand derivative of f'(2) = \(\underset{\mathrm{x} \rightarrow 1-}{\mathrm{Lt}}\) \(\frac{f(x)-f(2)}{x-2}\) = \(\underset{\mathrm{x} \rightarrow 1-}{\mathrm{Lt}}\)\(\frac{[x]-2}{x-2}\)
Put x = 2 – h, h → 0+
\(\underset{\mathrm{h} \rightarrow 0+}{\mathrm{Lt}}\)\(\frac{[2-\mathrm{h}]-2}{2-\mathrm{h}-2}\) = \(\underset{\mathrm{h} \rightarrow 0+}{\mathrm{Lt}}\)\(\frac{1-2}{-h}\)
= \(\underset{\mathrm{h} \rightarrow 0+}{\mathrm{Lt}}\)\(\frac{-1}{-h}\)
(For h → 0+, [2 – h] = 2 – 1 = 1
\(\underset{\mathrm{h} \rightarrow 0+}{\mathrm{Lt}}\)\(\frac{1}{h}\) = \(\frac{1}{0}\) = ∞ does not exist
∴ f(x) is not differentiable at x = 2
Note. For h → 0+, [c + h] = c if c is an integer