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

JEE Main · 2026 · 05 Apr 2026 · Shift 2 · Area Under Curves

Let

f : R \to R

be a function such that

f (x) + 3 f (\frac{π}{2} - x) = sin x, x \in R

. Let maximum value of

f

R

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

Answer: 16

Replacing

x

\frac{π}{2} - x

f (x) + 3 f (\frac{π}{2} - x) = sin x

gives

f (\frac{π}{2} - x) + 3 f (x) = cos x

; eliminating

f (\frac{π}{2} - x)

yields

8 f (x) = 3 cos x - sin x

, so

f (x) = \frac{1}{8} (3 cos x - sin x)

with maximum

α = \frac{10}{8}

. The curves

y = x^{2}

and

y = β x^{3}

meet at

x = 0

and

x = \frac{1}{β}

, enclosing area

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

. Setting

\frac{1}{12 β ^{3}} = α^{2} = \frac{10}{64}

gives

β^{3} = \frac{8}{15}

, so

30 β^{3} = 16

JEE Main · 2026 · 05 Apr 2026 · Shift 2 · Integer answer