JEE Mathematics 2023 Question with Solution

Given: $f(x) = \lfloor a + 13\sin x \rfloor$ for $x \in (0,\pi)$, where $a$ is an integer.

Find: The number of points in $(0,\pi)$ where $f(x)$ is not differentiable.

Since $0 < x < \pi$, we have $0 < \sin x \leq 1$, and hence

$$0 < 13\sin x \leq 13$$

The greatest integer function is discontinuous at integer values of its argument. Therefore, $\lfloor a + 13\sin x \rfloor$ is discontinuous, and hence not differentiable, whenever $a + 13\sin x$ is an integer.

Because $a$ is an integer, this happens exactly when $13\sin x$ is an integer.

So we solve

$$13\sin x = k$$

for integer values of $k$ from $1$ to $13$.

For each integer $k = 1,2,\dots,12$, we have $0 < \frac{k}{13} < 1$, so the equation $\sin x = \frac{k}{13}$ has two solutions in $(0,\pi)$.

For $k = 13$, the equation becomes

$$\sin x = 1$$

which has only one solution in $(0,\pi)$, namely $x = \frac{\pi}{2}$.

Therefore, the total number of such points is

$$2 \times 12 + 1 = 25$$

Therefore, the function is not differentiable at $25$ points.