JEE Main 2026 — Mathematics Calculus

Let $f(x) = \log_2(x^2 + 4)$f(x)=log2​(x2+4) for $x \ge 0$x≥0.

To find the inverse function $f^{-1}(x)$f−1(x), we set $y = \log_2(x^2 + 4)$y=log2​(x2+4) and solve for $x$x:

$$2^y = x^2 + 4$$2y=x2+4

$$x^2 = 2^y - 4$$x2=2y−4

$$x = \sqrt{2^y - 4}$$x=2y−4​

Thus, $f^{-1}(x) = \sqrt{2^x - 4}$f−1(x)=2x−4​.

The given expression is $\displaystyle \alpha = \displaystyle\int_0^{2\sqrt{3}} f(x)\,dx + \displaystyle\int_2^4 f^{-1}(x)\,dx$α=∫023​​f(x)dx+∫24​f−1(x)dx.

We observe the limits of the first integral and evaluate the function at these points:

$$f(0) = \log_2(0 + 4) = 2$$f(0)=log2​(0+4)=2

$$f(2\sqrt{3}) = \log_2((2\sqrt{3})^2 + 4) = \log_2(12 + 4) = \log_2(16) = 4$$f(23​)=log2​((23​)2+4)=log2​(12+4)=log2​(16)=4

These exactly match the limits of the second integral. Using the standard property of definite integrals for inverse functions:

$$\displaystyle\int_a^b f(x)\,dx + \displaystyle\int_{f(a)}^{f(b)} f^{-1}(x)\,dx = b f(b) - a f(a)$$∫ab​f(x)dx+∫f(a)f(b)​f−1(x)dx=bf(b)−af(a)

Substituting $a = 0$a=0 and $b = 2\sqrt{3}$b=23​:

$$\alpha = 2\sqrt{3} \cdot f(2\sqrt{3}) - 0 \cdot f(0)$$α=23​⋅f(23​)−0⋅f(0)

$$\alpha = 2\sqrt{3} \cdot 4 - 0 = 8\sqrt{3}$$α=23​⋅4−0=83​

Therefore, $\alpha^2 = (8\sqrt{3})^2 = 64 \times 3 = 192$α2=(83​)2=64×3=192.

Answer: $192$192