JEE Advanced 2025 — Paper 2 — Question 30

Let $h(x)=f\left(g^{-1}(x)\right)$ and $g(0)=2$

$h^{\prime}(x)=f^{\prime}\left(g^{-1}(x)\right) \cdot\left(g^{-1}(x)\right)^{\prime}$

$h^{\prime}(2)=f^{\prime}\left(g^{-1}(2)) \cdot\left(g^{-1}\right)^{\prime}(2)\right.$

$=\mathrm{f}^{\prime}(0) \cdot\left(\mathrm{g}^{-1}\right)^{\prime}(2)$

Now $f(x)=\log _{e}\left(x^{2}+2 x+4\right)$

$f^{\prime}(x)=\frac{2 x+2}{x^{2}+2 x+4}$

$f^{\prime}(0)=\frac{1}{2}$

$g(x)=\frac{4}{1+e^{-2 x}}, g(0)=2$

$\mathrm{g}^{-1}(\mathrm{~g}(\mathrm{x}))=\mathrm{x}$

$\left(\left(g^{-1}\right)^{\prime}(g(x))\right) g^{\prime}(x)=1$

$g(x)=\frac{4}{1+e^{-2 x}}$

$\left(g^{-1}\right)^{\prime}(2)=\frac{1}{g^{\prime}(0)}$

$g^{\prime}(x)=\frac{8 e^{-2 x}}{\left(1+e^{-2 x}\right)^{2}}$

$g^{\prime}(0)=\frac{8}{4}=2$

$\left(g^{-1}\right)^{\prime}(2)=\frac{1}{2}$

So $h^{\prime}(2)=\frac{1}{4}=0.25$