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.