JEE Main 2026 · 5 Apr · Shift 2 · Mathematics Q24

← JEE Main 2026 · 5 Apr · Shift 2

MathematicsInteger+4 / -1Question 24 of 75

Let $f:\mathbf R\to\mathbf R$ be a function such that $f(x)+3f\left(\dfrac\pi2-x\right)=\sin x$ , $x\in\mathbf R$ . 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 ________.

Your answer

Answer and worked solution

Official answer: 16

Find f. Replacing x by $\tfrac\pi2-x$ gives $f\left(\tfrac\pi2-x\right)+3f(x)=\cos x$ . Solving with the original,
$f(x)=\frac{3\cos x-\sin x}8,\qquad\alpha=\frac{\sqrt{10}}8,\qquad\alpha^2=\frac5{32}.$

Area. The curves meet at $x=\tfrac1\beta$ :
$\int_0^{1/\beta}(x^2-\beta x^3)\,dx=\frac1{12\beta^3}=\frac5{32}.$

So $\beta^3=\tfrac8{15}$ and $30\beta^3=16$ .

Answer: 16

Topics:Area between curves Types of functions

← Previous question Next question →

JEE Main 2026 · 5 Apr · Shift 2, Mathematics question 24