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
Still stuck on this question?Ask your doubt on WhatsApp
Similar questions
Functions · medium
A={1,2,3}A = \{1,2,3\}, B={1,3,5,7,9}B = \{1,3,5,7,9\}. Then No. of one-one functionsNo. of strictly monotonic functions\dfrac{\text{No. of one-one functions}}{\text{No. of strictly monotonic functions}}.
Functions · easy
The range of the function y=[x2]−[x]2y=[x^{2}]-[x]^{2}, x∈[0,2]x\in[0,2] (where [⋅][\cdot] denotes the integral part) is:
Functions · expert
Let [k][k] and {k}\{k\} denote the greatest integer and fractional part of kk. A real-valued function ff is defined for all real x≠−1x \neq -1 by f(x)=([log⁡e(1+{x})]∫0x2+1sin⁡(et) dt+2026−x5x5+1)1/5.f(x) = \left(\left[\log_e(1 + \{x\})\right]\int_{0}^{x^2+1}\sin(e^t)\,dt + \frac{2026 - x^5}{x^5 + 1}\right)^{1/5}. Find the value of f−1(200)−f(200)+f(f(2026))f(f(2)).\frac{f^{-1}(200) - f(200) + f(f(2026))}{f(f(2))}.
Functions · hard
If the domain of the function f(x)f(x) which is defined as f(x)=1sin⁡(cos⁡x)+sin⁡−1(2xπ)+1{−x}+1ln⁡(1−[tan⁡x2]−[−tan⁡x2])f(x) = \frac1{\sqrt{\sin\left(\cos x\right)}}+\sin^{-1}\left(\frac{2x}\pi\right)+\frac1{\left\{-x\right\}} +\frac1{\ln\left(1-\left[\tan\dfrac x2\right]-\left[-\tan\dfrac x2\right]\right)} is x∈(a,b)−{c,d,e}x \in \left(a,b\right)-\left\{c,d,e\right\}, then the value of (b+c+d+e)−a\left(b+c+d+e\right)-a is equal to (Where [ . ][\,.\,] represents the greatest integer function and {x}=x−[x]\left\{x\right\} = x-[x])

Solve more, learn faster

Sign up free to solve more JEE Maths questions and explore doMath — timed drills, mastery sprints, bookmarks, and chapter-wise progress tracking.

Sign up freeExplore doMath