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 ⁣(π2−x)=sin⁡xf(x)+3f\!\left(\dfrac{\pi}{2}-x\right)=\sin x, x∈Rx\in\mathbb{R}. Let α\alpha be the maximum value of ff. If the area bounded by g(x)=x2g(x)=x^2 and h(x)=βx3h(x)=\beta x^3, β>0\beta>0, equals α2\alpha^2, then 30β330\beta^3 equals
Solution
Answer: 16
Step 1: Solve the functional equation for f(x)f(x) The given equation
f(x)+3f ⁣(π2−x)=sin⁡x⋯(1)f(x)+3f\!\left(\tfrac{\pi}{2}-x\right)=\sin x \quad\cdots(1)
 Replacing xx with π2−x\frac{\pi}{2}-x:
f ⁣(π2−x)+3f(x)=cos⁡x⋯(2)f\!\left(\tfrac{\pi}{2}-x\right)+3f(x)=\cos x \quad\cdots(2)
 Eliminating f(π2−x)f(\frac{\pi}{2}-x): 9f(x)−f(x)=3cos⁡x−sin⁡x9f(x)-f(x) = 3\cos x-\sin x, giving
f(x)=3cos⁡x−sin⁡x8f(x) = \frac{3\cos x-\sin x}{8}
 Step 2: Find α=fmax⁡\alpha = f_{\max}
α=32+(−1)28=108  ⟹  α2=1064=532\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 gg and hh The curves y=x2y=x^2 and y=βx3y=\beta x^3 intersect at x=0x=0 and x=1βx=\frac{1}{\beta}. Since x2>βx3x^2>\beta x^3 on (0,1β)\left(0,\frac{1}{\beta}\right):
Area=∫01/β(x2−βx3) dx=[x33−βx44]01/β=13β3−14β3=112β3\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 30β330\beta^3
112β3=532  ⟹  β3=3260=815\frac{1}{12\beta^3} = \frac{5}{32} \implies \beta^3 = \frac{32}{60} = \frac{8}{15}
 $30β3=30⋅815=1630\beta^3 = 30\cdot\frac{8}{15} = 16. Answer: 16
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