RBSE Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.2

Rajasthan Board RBSE Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.2 Textbook Exercise Questions and Answers.

RBSE Class 12 Maths Solutions Chapter 5 Continuity and Differentiability Ex 5.2

Question 1.
sin (x² + 5)
Answer:
Let y = sin (x² + 5)
∴ Differentiating both sides w.r.t. x, we gel
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\) sin (x² + 5)
= cos (x² + 5). \(\frac{d}{d x}\)(x² + 5)
= cos(x² + 5).(2x)
∴ \(\frac{d}{d x}\)[sin(x² + 5)] = 2x cos (x² + 5)

Question 2.
cos (sin x)
Answer:
Let y = cos (sin x)
Differentiating both sides w.r.t. ‘x’, we get
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\) [cos(sin x)]
\(\frac{d y}{d x}\) = - sin(sin x). \(\frac{d}{d x}\)(sin x)
∴ \(\frac{d}{d x}\)[cos (sin x) = - sin (sin x) cos x

Question 3.
sin (ax + b)
Answer:
Let y = sin (ax + b)
y = sin (ax + b)
Differentiating both sides w.r.t. ‘x’, we get
\(\frac{d y}{d x}\) = \(\frac{d}{d x}\)[sin (ax + b)]
\(\frac{d y}{d x}\) = cos(ax + b) \(\frac{d}{d x}\) (ax + b)
⇒ \(\frac{d y}{d x}\) = cos (ax + b) × (a.1 × 0)
= cos (ax + b).(a)
∴ \(\frac{d y}{d x}\) = a cos (ax + b)

Question 4.
sec (tan √x)
Answer:
Let y = sec [tan (√x)]
Differentiating both sides w.r.t. ‘x’, we get

Question 5.
\(\frac{\sin (a x+b)}{\cos (c x+d)}\)
Answer:
Let y = \(\frac{\sin (a x+b)}{\cos (c x+d)}\)
Differentiating both sides w.r.t. ‘x’, we get

= a cos (ax + b) sec (cx + d) + c sin (ax + d) sec (cx + d) tan (cx + d)

Question 6.
cos x³.sin² (x⁵)
Answer:
Let y = cos x³.sin² (x⁵)
Differentiating both sides w.r.t. ‘x’, we get
∴ \(\frac{d y}{d x}\) = cos x³ \(\frac{d}{d x}\) {sin² (x⁵)} + sin² (x⁵) \(\frac{d}{d x}\) (cos x³)
= cos x³.2 sin x⁵.cos x⁵. \(\frac{d}{d x}\) (x⁵) + sin² x⁵.(- sin x³).\(\frac{d}{d x}\)(x³)
∴ \(\frac{d y}{d x}\) = cos x³.2 sin (x⁵).cos (x⁵)
= 10x⁴ cos x³ cos x⁵ - 3x² sin x³.sin² x⁵

Question 7.
2\(\sqrt{\cot (x)^{2}}\)
Answer:
Let y = 2\(\sqrt{\cot (x)^{2}}\)
Differentiating both sides w.r.t. ‘x’, we get

Question 8.
cos √x
Answer:
Let y = cos √x
Differentiating both sides w.r.t. 'x', we get

Question 9.
Prove that the function f given by f(x) = |x - 1|; x ∈ R, is not differentiable at x = 1.
Answer:
f(x) = x + 1

∴ L.H.D. = - 1 ≠ 1 = R.H.D.
Thus, function is not Differentiable at x = 1.
Hence Proved.

Question 10.
Prove that the greatest integer function defined by f(x) = [x], 0 < x < 3, is not differentiable at x = 1 and x = 2.
Answer:
(i) For x = 1
Left hand derivative

[Since 1 is greater integer before (1 + h)]
∵ derivative of left side \(\lim _{h \rightarrow 0}\frac{1}{h}\) is not defined.
Thus, function is not differentiable at x = 1.
Hence Proved.

(ii) For x = 2
Left hand derivative

[Since maximum integer in 2 before (2 + h)]
∵ Derivative of LHS \(\lim _{h \rightarrow 0}\frac{1}{h}\) is not defined.
Thus, fraction is not differentiable at x = 2.
Hence Proved.