JEE Main · 2026 · 02 Apr 2026 · Shift 1 · Application of Derivatives

The function $f(x)=\left|\frac{\sin x}{x}\right|$ is even and continuous on $(-2\pi ,2\pi )$ with $f(0)=1$. Away from the zeros of $\sin x$, $f^{\prime }(x)=\mathrm{sgn}\left(\frac{\sin x}{x}\right)\frac{x\cos x-\sin x}{x^2}$, which vanishes where $x\cos x-\sin x=0$, i.e. $\tan x=x$. In $(-2\pi ,2\pi )$ the equation $\tan x=x$ has the three solutions $x=0$ and $x=\pm x_0$ with $x_0\approx 4.49$. At $x=\pm \pi$ the ratio $\frac{\sin x}{x}$ changes sign, so the absolute value forms corners where $f^{\prime }$ does not exist, giving two more. In total there are $3+2=5$ critical points.