Let $f:[0, \infty) \rightarrow[0,3]$ be a function defined by $f(x)=\left\{\begin{array}{cc}\max \{\sin t: 0 \leq t \leq x\}, & 0 \leq x \leq \pi \\ 2+\cos x, & x>\pi\end{array}\right.$ Then which of the following is true?

(A) $f$ is differentiable everywhere in $(0, \infty)$

(B) f is not continuous exactly at two points in $(0, \infty)$

(C) f is continuous everywhere but not differentiable exactly at two points in $(0, \infty)$

(D) $f$ is continuous everywhere but not differentiable exactly at one point in $(0, \infty)$

JEE Main · 2021 · 27 Jul 2021 · Shift 2 · Continuity and Differentiability

Let

f : [0, \infty) \to [0, 3]

be a function defined by

f (x) = {max {sin t : 0 \leq t \leq x}, 2 + cos x, 0 \leq x \leq π x > π

Then which of the following is true?

(A) $f$ is differentiable everywhere in $(0, \infty)$ ✓ Correct
(B)f is not continuous exactly at two points in $(0, \infty)$
(C)f is continuous everywhere but not differentiable exactly at two points in $(0, \infty)$
(D) $f$ is continuous everywhere but not differentiable exactly at one point in $(0, \infty)$

Answer: (A)

[0, \frac{π}{2}]

the function

sin t

increases, so

max {sin t : 0 \leq t \leq x} = sin x

; on

[\frac{π}{2}, π]

the maximum is attained at

t = \frac{π}{2}

and equals

1

. Hence

f (x) = ⎩ ⎨ ⎧ sin x, 1, 2 + cos x, 0 \leq x < \frac{π}{2} \frac{π}{2} \leq x \leq π x > π

x = \frac{π}{2}

both one-sided limits equal

1

, and the derivatives from the two sides are

cos \frac{π}{2} = 0

and

0

. At

x = π

the left-hand value is

1

and

2 + cos π = 1

, while the left-hand derivative is

0

and the right-hand derivative is

- sin π = 0

. Each piece is differentiable inside its own interval, so

f

is differentiable at every point of

(0, \infty)

JEE Main · 2021 · 27 Jul 2021 · Shift 2 · Single correct option