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

Gujarat Board GSEB Textbook Solutions Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.2 Textbook Questions and Answers.

Gujarat Board Textbook Solutions Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.2

Question 1.
sin(x² + 5)
Solution:
Let y = sin(x² + 5)
Put x² + 5 = t
∴ y = sin t and t = x² + 5.
So, $\frac { dy }{ dx }$ = $\frac { dy }{ dt }$. $\frac { dt }{ dx }$ = cos t. $\frac { dt }{ dx }$
= cos(x² + 5)$\frac { d }{ dx }$(x² + 5)
= cos(x² + 5) x 2x = 2x cos((x² + 5).

Question 2.
cos (sin x)
Solution:
Let y = cos (sin x)
Put sin x = t
∴ y = sin t and t = sin x.
∴ $\frac { dy }{ dx }$ = – sin t, $\frac { dt }{ dx }$ = cos x.
∴ $\frac { dy }{ dx }$ = $\frac { dy }{ dt }$ . $\frac { dt }{ dx }$ = ( – sin t) x (cos x)
Putting value of t, we get
$\frac { dy }{ dx }$ = – sin (sin x) × cos x
= – [sin(sin x)] cos x.

Question 3.
sin (ax + b)
Solution:
Let y = sin (ax + b).
Put ax + b = t.
∴ y = sin t and t = ax + b.
∴ $\frac { dy }{ dx }$ = cos t, $\frac { dt }{ dx }$ = $\frac { d }{ dx }$ (ax + b) = a.
Now, $\frac { dy }{ dx }$ = $\frac { dy }{ dt }$ . $\frac { dt }{ dx }$ = ( cos t) x a
= a cos (ax +b)

Question 4.
sec(tan($\sqrt{x}$)
Solution:
Let y = sec(tan($\sqrt{x}$)
put $\sqrt{x}$ = t and s = tan t.
⇒ y = sec s, s = tan t and t = $\sqrt{x}$.
Now, $\frac { dy }{ dx }$ = $\frac { dy }{ ds }$ x $\frac { ds }{ dt }$ x $\frac { dt }{ dx }$ … (1)
So, y = sec s. ∴ $\frac { dy }{ ds }$ = sec s tan s
Also, s = tan t. ∴ $\frac { ds }{ dt }$ = sec² t.
Further, t = $\sqrt{x}$ ∴ $\frac { dt }{ dx }$ = $\frac{1}{2 \sqrt{x}}$
Putting these values in (1), we get
$\frac { dy }{ dx }$ = (sec (tan s) (sec² t)$\left(\frac{1}{2 \sqrt{x}}\right)$
= sec (tan t) tan (tan (t). sec²$\sqrt{x}$ . $\frac{1}{2 \sqrt{x}}$
= $\frac{1}{2 \sqrt{x}}$sec tan$\sqrt{x}$ tan (tan ($\sqrt{x}$)).sec²$\sqrt{x}$.

Question 5.
$\frac{\sin (a x+b)}{\cos (c x+d)}$
Solution:

Question 6.
cos x³. sin² (x⁵)
Solution:
Let y = cos x³ sin²(x⁵) = uv,
where u = cosx³ and v = sin2(x⁵).
To find $\frac { du }{ dx }$, put x³ = t.
∴ u = cos t, t = x³.
∴ $\frac { du }{ dx }$ = – sin t and $\frac { dt }{ dx }$ = 3x².
∴ $\frac { du }{ dx }$ = $\frac { du }{ dt }$ x $\frac { dt }{ dx }$ = (- sin t) (3x²)
= – sin x³ (3x²) = – 3x² sin x³.
To find $\frac { du }{ dx }$, put t = x⁵ and sin t = s.
∴ v = s², s = sin t and t = x⁵.
∴ $\frac { dv }{ ds }$ = 2s, $\frac { ds }{ dt }$ = cos t and $\frac { dt }{ dx }$ = 5x⁴
= 2s x cos t x 5a⁴ = 2 sin t cos t x 5x⁴
= 10x⁴ sin x⁵ cos x⁵
Now, y = uv = (cos x³) (sin2 x³)
∴ $\frac { dy }{ dx }$ = $\frac { du }{ dx }$ x v + u x $\frac { dv }{ dx }$
∴ $\frac { dy }{ dx }$ = (- 3x² sin x³) × sin²x⁵ + cos x³ x 10x⁴ sinx⁵ cos x⁵
= – 3x² sin x³ sin²x⁵ + 10x⁴ cos x³ sin x⁵ cos x⁵
= x² sin x⁵ (- 3 sin x³ sin x⁵ + 10x² cos x³ cos x³).

Question 7.
$\sqrt{\cot \left(x^{2}\right)}$
Solution:

Question 8.
cos$\sqrt{x}$
Solution:

Question 9.
Prove that the function f is given by f(x) = |x-1|, x ∈ R is not differentiable at x = 1.
Solution:
The given function may be written as

So, R.H.D ≠ L.H.D
⇒ f is not differentiable at x = 1.

Question 10.
Prove that the greatest integer function defined by f(x) = [x], 0 < x < 3 is not differentiable at x = 1 and x = 3.
Solution:
(i) At x = 1,

∴ f is not differentiable at x = 1.

(ii) At x = 3,

⇒ f is not differentiable at x = 3.