JEE Advanced 2026 · Paper 1 · Mathematics Q10

← JEE Advanced 2026 · Paper 1

MathematicsNumerical+4Question 10 of 48

Consider the function $f:\left(-\dfrac\pi2,\dfrac\pi2\right)\to(-\infty,\infty)$ defined by
$f(x)=(|x|+|x-1|)\sin x+[x\sin x],$
where $[x\sin x]$ is the greatest integer less than or equal to $x\sin x$ .

Let $\alpha$ be the total number of points in the interval $\left(-\dfrac\pi2,\dfrac\pi2\right)$ at which $f$ is NOT continuous, and let $\beta$ be the total number of points in the interval $\left(-\dfrac\pi2,\dfrac\pi2\right)$ at which $f$ is NOT differentiable.

Then the value of $\alpha+\beta$ is ________.

Your answer

Answer and worked solution

Official answer: 5

Continuity. The first term is continuous. The function $x\sin x$ is even, equals 0 at $x=0$ and increases to $\frac\pi2\approx1.57$ at the ends. So $[x\sin x]$ is 0 near 0 and jumps to 1 where $x\sin x=1$ , at $x\approx\pm1.114$ . Hence $\alpha=2$ .

Differentiability. Besides those two points:

Near $x=0$ : $h(x)=(|x|+|x-1|)\sin x$ is $(1-2x)\sin x$ for $x<0$ and $\sin x$ for $0<x<1$ . Both have derivative $1$ at $0$ , so $f$ is differentiable at 0.
At $x=1$ (inside the interval, as $1<\frac\pi2$ ): for $x>1$ , $h=(2x-1)\sin x$ with slope $2\sin1+\cos1$ ; for $x<1$ , slope $\cos1$ . Not equal, so not differentiable.

So $\beta=3$ and $\alpha+\beta=5$ .

Topics:Differentiability Continuity

← Previous question Next question →

JEE Advanced 2026 · Paper 1, Mathematics question 10