JEE Main 2026 · 5 Apr · Shift 2, Mathematics question 17

The limit. As $y\to0$, $1-\cos xy\approx\tfrac{x^2y^2}2$ and $\tan xy\approx xy$, so $f(x)=\tfrac{x^3}2$.

Solve $\tfrac{x^3}2=\sin x$.

• $x=0$ is a root.
• For $x>0$: at $x=1$, $\tfrac12<\sin1$; at $x=1.5$, $1.69>1$. The cubic overtakes $\sin x$ exactly once, giving one positive root.
• By oddness there is one negative root.

Count: 3.

(C)