Question

Let $ f(x) = x|x| $ , \[g(x) = \sin x\] and $ h(x) = (gof)(x) $ . Then
A. h(x) is not differentiable at x = 0.
B. h(x) is differentiable at x = 0, but h’(x) is not continuous at x = 0
C. h’(x) is differentiable at x = 0.
D. h’(x) is continuous at x = 0 but it is not differentiable at x = 0

Answer 1

Hint: To solve the above question, first we will derive h(x) from given data. Then we will check continuity of it by checking the limit of the differentiation of it and differentiability by checking differentiation of it. Then again we will check the same for the $h'(x)$.

Complete step-by-step answer:
According to the question,
 $ f(x) = x|x| $
That means,
 $ f(x) $ is $ - {x^2} $ for x value less than 0.
 $ f(x) = - {x^2},x < 0 $
 $ f(x) $ is $ {x^2} $ for x value greater and equal to 0.
 $ f(x) = {x^2},x \geqslant 0 $
Again it is given in the question that,
\[g(x) = \sin x\]
And, $ h(x) = (gof)(x) $
That means,
 $ h(x) = g(f(x)) = \sin ( - {x^2}),x < 0 $ …………..(1)
And, $ h(x) = g(f(x)) = \sin {x^2},x \geqslant 0 $ ……………(2)
For above two functions, if $ x \to 0,h(x) = 0 $
Hence, $ h(0) = 0 $
i.e. \[h\left( x \right)\]is continuous at x = 0……………………….Conclusion (1)
For \[h\left( x \right)\], at x = 0,
Differentiation of equation 1,
 $ h'(x) = ( - 2x) \times \cos ( - {x^2}) = 0 $
And differentiation of equation 2,
 $ h'(x) = (2x) \times \cos ({x^2}) = 0 $
Hence, \[h\left( x \right)\]is differentiable at x = 0…………………………. Conclusion (2)
And for $ h'(x) $ , limit of the both of the side is equal i.e. 0
Hence, $ h'(x) $ is continuous at x = 0…………………………. Conclusion (3)
We have, $ h'(x) = ( - 2x) \times \cos ( - {x^2}),x < 0 $
And, $ h'(x) = (2x) \times \cos ({x^2}),x \geqslant 0 $
Again taking derivative of above equations and putting x = 0 we get,
 $ h''(x) = - 2\cos ( - {x^2}) + 4{x^2}\sin ( - {x^2}) = - 2 $
And $ h''(x) = 2\cos {x^2} - 4{x^2}\sin {x^2} = 2 $
Hence, $ h'(x) $ is not differentiable at x = 0. ……………………………….. Conclusion (4)
From conclusion 1, 2, 3 and 4 we get that, \[h\left( x \right)\] is continuous and differentiable at x = 0, but $ h'(x) $ is continuous but not differentiable at x = 0.

So, the correct answer is “Option D”.

Note: A continuous function is a function that does not have any abrupt changes in value.
A function f is continuous when, for every value of c in its domain, f(c) is defined and the limit of the f is f(c), for $ x \to c $ . I.e. $ \mathop {\lim }\limits_{x \to c} f(x) = f(c) $
In the above question, for the function to be continuous, the limit value of the function must be 0 in both the cases.
A differentiable function of one real variable is a function whose derivative exists at each point in its domain.
A function f is differentiable when the function is continuous and the derivative of the function must be equal in both the cases and c for $ x \to c $ . i.e. $ f'(x) = f(c) $