Let $f:[a, b] \rightarrow[1, \infty)$ be a continuous function and let $g: \mathbb{R} \rightarrow \mathbb{R}$ be defined as \[ g(x)= \begin{cases}0 & \text { if } x b\end{cases} \] Then

(A) $g(x)$ is continuous but not differentiable at $a$; (C) $g(x)$ is continuous but not differentiable at $b$

(B) $g(x)$ is differentiable on $\mathbb{R}$

(D) $g(x)$ is continuous and differentiable at either $a$ or $b$ but not both

JEE Advanced · 2014 · 25 May 2014 · Paper 1 · Continuity and Differentiability

Let

f : [a, b] \to [1, \infty)

be a continuous function and let

g : R \to R

be defined as

g (x) = ⎩ ⎨ ⎧ 0 \int_{a}^{x} f (t) d t \int_{a}^{b} f (t) d t if x < a if a \leq x \leq b if x > b

Then

Answer: (A), (C)

x = a

lim_{x \to a^{-}} g (x) = 0

and

lim_{x \to a^{+}} g (x) = g (a) = \int_{a}^{a} f (t) d t = 0

, so

g

is continuous at

a

. However,

g^{'} (a^{-}) = 0

while

g^{'} (a^{+}) = f (a) \geq 1

, so

g

is not differentiable at

a

. Similarly, at

x = b

lim_{x \to b^{-}} g (x) = g (b) = lim_{x \to b^{+}} g (x) = \int_{a}^{b} f (t) d t

, so

g

is continuous at

b

. But

g^{'} (b^{-}) = f (b) \geq 1

while

g^{'} (b^{+}) = 0

, so

g

is not differentiable at

b

. Thus (A) and (C) are correct.

JEE Advanced · 2014 · 25 May 2014 · Paper 1 · One or more correct options