QJEEM24Maf38a: JEE Main 2024 Mathematics - Class 12 JEEM24Maf38a Solution: Let for all . Consider a function such that for all . Then the…

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:

$g'(f(x)) \cdot f'(x) = 1$

Now, we want to find the value of $8g'(2)$ . 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:

$g'(f(0)) \cdot f'(0) = 1$

$g'(2) \cdot f'(0) = 1$

Let's find $f'(0)$ :

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

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

Substituting this back into our equation:

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

$g'(2) = 2$

Finally, we can calculate $8g'(2)$ :

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