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

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
$\frac{dy}{dx}$ = $\frac{d}{dx}$ sin (x2 + 5)
= cos (x2 + 5). $\frac{d}{dx}$(x2 + 5)
= cos(x2 + 5).(2x)
∴ $\frac{d}{dx}$[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
$\frac{dy}{dx}$ = $\frac{d}{dx}$ [cos(sin x)]
$\frac{dy}{dx}$ = - sin(sin x). $\frac{d}{dx}$(sin x)
∴ $\frac{d}{dx}$[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
$\frac{dy}{dx}$ = $\frac{d}{dx}$[sin (ax + b)]
$\frac{dy}{dx}$ = cos(ax + b) $\frac{d}{dx}$ (ax + b)
⇒ $\frac{dy}{dx}$ = cos (ax + b) × (a.1 × 0)
= cos (ax + b).(a)
∴ $\frac{dy}{dx}$ = 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.
$\frac{\sin (ax+b)}{\cos (cx+d)}$
Answer:
Let y = $\frac{\sin (ax+b)}{\cos (cx+d)}$
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
∴ $\frac{dy}{dx}$ = cos x3 $\frac{d}{dx}$ {sin2 (x5)} + sin2 (x5) $\frac{d}{dx}$ (cos x3)
= cos x3.2 sin x5.cos x5. $\frac{d}{dx}$ (x5) + sin2 x5.(- sin x3).$\frac{d}{dx}$(x3)
∴ $\frac{dy}{dx}$ = 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 ${lim}_{h\to 0}\frac{1}{h}$ 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 ${lim}_{h\to 0}\frac{1}{h}$ is not defined.
Thus, fraction is not differentiable at x = 2.
Hence Proved.

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