1
JEE Main 2026 (Online) 5th April Evening Shift
Numerical
+4
-1
Change Language

From the point $(-1,-1)$, two rays are sent making angles of $45^{\circ}$ with the line $x+y=0$. These rays get reflected from the mirror $x+2 y=1$. If the equations of the reflected rays are $\mathrm{a} x+\mathrm{b} y=9$ and $c x+d y=7, a, b, c, d \in \mathbf{Z}$, then the value of $a d+b c$ is $\_\_\_\_$ .

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2
JEE Main 2026 (Online) 5th April Evening Shift
Numerical
+4
-1
Change Language

If $S=\left\{\theta \in[-\pi, \pi]: \cos \theta \cos \frac{5 \theta}{2}=\cos 7 \theta \cos \frac{7 \theta}{2}\right\}$, then $n(S)$ is equal to $\_\_\_\_$ .

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3
JEE Main 2026 (Online) 5th April Evening Shift
Numerical
+4
-1
Change Language

Let $f: \mathbf{R} \rightarrow \mathbf{R}$ be a function such that $f(x)+3 f\left(\frac{\pi}{2}-x\right)=\sin x, x \in \mathbf{R}$. Let the maximum value of $f$ on $\mathbf{R}$ be $\alpha$. If the area of the region bounded by the curves $g(x)=x^2$ and $h(x)=\beta x^3, \beta>0$, is $\alpha^2$, then $30 \beta^3$ is equal to $\_\_\_\_$ .

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4
JEE Main 2026 (Online) 5th April Evening Shift
Numerical
+4
-1
Change Language

Let $y=y(x)$ be the solution of the differential equation $(\tan x)^{1 / 2} \mathrm{~d} y=\left(\sec ^3 x-(\tan x)^{3 / 2} y\right) \mathrm{d} x, 0 < x <\frac{\pi}{2}, y\left(\frac{\pi}{4}\right)=\frac{6 \sqrt{2}}{5}$. If $y\left(\frac{\pi}{3}\right)=\frac{4}{5} \alpha$, then $\alpha^4$ equals

$\_\_\_\_$ .

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