FunctionsmediumPYQ · JEE Main · 5 Apr 0006 · Shift 2 (Afternoon)Free

Functions: Let Let Maximum Value Area Bounded Equals Equals (JEE Main 2026)

JEE Maths question with a full step-by-step solution.

Question

Let

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

x\in\mathbb{R}

. Let

\alpha

be the maximum value of

f

. If the area bounded by

g(x)=x^2

and

h(x)=\beta x^3

\beta>0

, equals

\alpha^2

, then

30\beta^3

equals

Solution

Answer: 16

Step 1: Solve the functional equation for

f(x)

The given equation

f(x)+3f\!\left(\tfrac{\pi}{2}-x\right)=\sin x \quad\cdots(1)

Replacing

x

with

\frac{\pi}{2}-x

f\!\left(\tfrac{\pi}{2}-x\right)+3f(x)=\cos x \quad\cdots(2)

Eliminating

f(\frac{\pi}{2}-x)

9f(x)-f(x) = 3\cos x-\sin x

, giving

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

Step 2: Find

\alpha = f_{\max}

\alpha = \frac{\sqrt{3^2+(-1)^2}}{8} = \frac{\sqrt{10}}{8} \implies \alpha^2 = \frac{10}{64} = \frac{5}{32}

Step 3: Compute the enclosed area between

g

and

h

The curves

y=x^2

and

y=\beta x^3

intersect at

x=0

and

x=\frac{1}{\beta}

. Since

x^2>\beta x^3

\left(0,\frac{1}{\beta}\right)

\text{Area} = \int_0^{1/\beta}(x^2-\beta x^3)\,dx = \left[\frac{x^3}{3}-\frac{\beta x^4}{4}\right]_0^{1/\beta} = \frac{1}{3\beta^3}-\frac{1}{4\beta^3} = \frac{1}{12\beta^3}

Step 4: Solve for

30\beta^3

\frac{1}{12\beta^3} = \frac{5}{32} \implies \beta^3 = \frac{32}{60} = \frac{8}{15}

30\beta^3 = 30\cdot\frac{8}{15} = 16

. Answer: 16

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