JEE Main 2026 Apr 5 Shift 2, Mathematics Q24: Area as Definite Integral

JEE Main2026Apr 5, Shift 2Mathematics

Let $f : R \to R$ be a function such that $f (x) + 3 f (\frac{π}{2} - x) = sin x$ , $x \in R$ . Let the 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 ______.

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Solution

Replace $x$ by $π /2 - x$ in the functional equation: $f (π /2 - x) + 3 f (x) = cos x$ .

Solving the two linear equations: $f (x) = \frac{1}{8} (3 cos x - sin x)$ .

Maximum value: $\frac{9 + 1}{8} = \frac{10}{8}$ . So $α = \frac{10}{8}$ and $α^{2} = \frac{10}{64} = \frac{5}{32}$ .

Curves $y = x^{2}$ and $y = β x^{3}$ meet where $x^{2} = β x^{3}$ , i.e. $x = 0$ or $x = 1/ β$ . On $(0, 1/ β)$ , $x^{2} > β x^{3}$ .

Area $= \int_{0}^{1/ β} (x^{2} - β x^{3}) d x = \frac{1}{3 β ^{3}} - \frac{1}{4 β ^{3}} = \frac{1}{12 β ^{3}}$ .

Setting equal to $α^{2}$ : $\frac{1}{12 β ^{3}} = \frac{5}{32}$ would give $β^{3} = \frac{32}{60} = \frac{8}{15}$ , so $30 β^{3} = 16$ .

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JEE Main 2026 Apr 5 Shift 2, Mathematics Q24: Area as Definite Integral

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