Question

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

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

Hint:

To check for differentiability of a piecewise function at a point, you must check for continuity first. Then, calculate the left-hand derivative and the right-hand derivative at that point. The function is differentiable if and only if these two derivatives are equal and finite.

Answer 1

The Correct Option is B

For \( 0 \le x \le \pi \): \[ f(x)= \begin{cases} \sin x, & 0 \le x \le \tfrac{\pi}{2} \\ 1, & \tfrac{\pi}{2} < x \le \pi \end{cases} \] At \( x=\tfrac{\pi}{2} \): \[ \text{LHD}=\cos\!\left(\tfrac{\pi}{2}\right)=0, \quad \text{RHD}=0 \] Hence differentiable. At \( x=\pi \): \[ \text{LHD}=0, \quad \text{RHD}=-\sin(\pi)=0 \] Hence differentiable. For \( x > \pi \), \( f(x)=2+\cos x \) is differentiable. Therefore, \( f(x) \) is differentiable for all \( x > 0 \).