JEE Main 2024 — Differentiation Question with Solution

Given that $$(g\circ f)(x)=x$$ for all $$x\in \mathbf{R}$$. This means $$g(f(x))=x$$ for all $$x\in \mathbf{R}$$. Differentiating both sides with respect to $$x$$, we get:

$$g^{\prime }(f(x))\cdot f^{\prime }(x)=1$$

Now, we want to find the value of $$8g^{\prime }(2)$$. To do this, we need to find a value of $$x$$ such that $$f(x)=2$$. Let's solve for $$x$$:

$$x^5+2e^{x/4}=2$$

By inspection, we see that $$x=0$$ is a solution. Therefore, $$f(0)=2$$. Now, we can substitute this into our differentiated equation:

$$g^{\prime }(f(0))\cdot f^{\prime }(0)=1$$

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

Let's find $$f^{\prime }(0)$$:

$$f^{\prime }(x)=5x^4+\frac{1}{2}{e}^{x/4}$$

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

Substituting this back into our equation:

$$g^{\prime }(2)\cdot \frac{1}{2}=1$$

$$g^{\prime }(2)=2$$

Finally, we can calculate $$8g^{\prime }(2)$$:

$$8g^{\prime }(2)=8\cdot 2=\boxed{16}$$