Let f : R rightarrow R be a function such that f ( x )+3 f (π/2- x )=sin x , x in R . Let maximum value of f on R be α . If the area of the region bounded by the curves g ( x )= x 2 and h ( x )=β x 3 , β 0 , is α 2 , then 30 β 3 is equal to \ \ \ \

Mathematics · Functions

JEE Main 2026 — 5 April, Evening Shift — Question 47

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

Answer: 16

Numerical answer — enter this value.

Step-by-step solution

$f(x)+3 f\left(\frac{\pi}{2}-x\right)=\sin x$

Put $\mathrm{x} \rightarrow \frac{\pi}{2}-\mathrm{x}$

\Rightarrow \mathrm{f}\left(\frac{\pi}{2}-\mathrm{x}\right)+3 \mathrm{f}(\mathrm{x})=\cos \mathrm{x} \end{gathered}$$ From (1) & (2) $\mathrm{f}(\mathrm{x})=\frac{1}{8}(3 \cos \mathrm{x}-\sin \mathrm{x})$ $\mathrm{f}_{\text {max }}=\frac{\sqrt{10}}{8}$ $\mathrm{y}=\mathrm{g}(\mathrm{x}) \& \mathrm{y}=\mathrm{h}(\mathrm{x})$ intersect as shown in the figure ∴ Area bounded $=\Delta=\left|\int_{0}^{\frac{1}{\beta}}\left(\beta \mathrm{x}^{3}-\mathrm{x}^{2}\right) \mathrm{dx}\right|$ $=\frac{1}{12 \beta^{3}}=\alpha^{2}$ (given) $\Rightarrow 30 \beta^{3}=16$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 5 April, Evening Shift
Subject: Mathematics
Chapter: Functions
Topic: Functional Equations