JEE Main 2026 · 5 Apr · Shift 1, Mathematics question 25

Official key: marks to all. The question cannot be answered as printed.

Solving the equation. Put $y=vx$:
$$x\sin v\,(v\,dx+x\,dv)=(vx\sin v-x)\,dx\ \Rightarrow\ \sin v\,dv=-\frac{dx}x\ \Rightarrow\ \cos v=\ln x+C.$$
$y(1)=\tfrac\pi2$ gives $C=0$, so $\cos\dfrac yx=\ln x$.

The flaw. Then $\alpha=\cos\dfrac{y(e^{12})}{e^{12}}=\ln e^{12}=12$, but a cosine cannot exceed 1. The solution exists only for $\tfrac1e\le x\le e$, so $y(e^{12})$ is undefined.

Taking $\alpha=12$ formally. The circle $x^2+y^2-2px+2py+14=0$ has $r^2=2p^2-14$. Requiring $0<r^2\le36$ gives $7<p^2\le25$, so $p=\pm3,\pm4,\pm5$: 6 values.