Definite IntegrationhardPYQ · JEE Main · 2 Apr 2026 · Shift 1 (Morning)Free

Function-and-Inverse Integrals: α² = 192 | JEE Main 2026

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

Question
If α=∫023log⁡2(x2+4) dx+∫242x−4  dx\alpha=\displaystyle\int_0^{2\sqrt3}\log_2\left(x^2+4\right)\,dx+\int_2^{4}\sqrt{2^x-4}\,\,dx, then α2\alpha^2 is equal to
Solution
Answer: 192 (± 0.01)
Step 1: Let f(x)=log⁡2(x2+4)f(x)=\log_2(x^2+4) for x≥0x\ge0. Its inverse is found from y=log⁡2(x2+4)⇒2y=x2+4⇒x=2y−4y=\log_2(x^2+4)\Rightarrow 2^y=x^2+4\Rightarrow x=\sqrt{2^y-4}, so f−1(y)=2y−4f^{-1}(y)=\sqrt{2^y-4}. The second integrand is exactly f−1f^{-1}. Step 2: Use the standard result for a function and its inverse:
∫abf(x) dx+∫f(a)f(b)f−1(x) dx=b f(b)−a f(a).\int_a^b f(x)\,dx+\int_{f(a)}^{f(b)} f^{-1}(x)\,dx=b\,f(b)-a\,f(a).
 Step 3: Here a=0, b=23a=0,\ b=2\sqrt3. Check the limits: f(0)=log⁡24=2f(0)=\log_2 4=2 and f(23)=log⁡2(12+4)=log⁡216=4f(2\sqrt3)=\log_2(12+4)=\log_2 16=4, matching the second integral's limits 22 to 44. So
α=b f(b)−a f(a)=(23)(4)−(0)(2)=83.\alpha=b\,f(b)-a\,f(a)=(2\sqrt3)(4)-(0)(2)=8\sqrt3.
 Step 4:
α2=(83)2=64⋅3=192.\alpha^2=(8\sqrt3)^2=64\cdot3=192.
 Correct answer: 192
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