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.