JEE Mathematics 2023 Question with Solution

Given:

Find: The value of $\left( \frac{k}{e} \right)^8 + \frac{k^8}{e^5} + k^8$ when the local maximum value is $\frac{k}{e}$.

Take natural logarithm:

Differentiate with respect to $x$:

So,

For a local extremum, set $\frac{dy}{dx} = 0$. Since $x \in \left(0, \frac{\pi}{2}\right)$, we have $\sin x \neq 0$ and $\cos x \neq 0$. Hence,

Therefore,

Using the working shown in the solution,

Since $x \in \left(0, \frac{\pi}{2}\right)$,

Now substitute in $y$:

So the local maximum value is

Hence,

and therefore

Now evaluate:

Thus the expression becomes

The solution concludes with The Correct Option is D and states the final expression as $e^3 + e^6 + e^{11}$, while the option labeled D is $e^3 + e^6 + e^{10}$. This is a discrepancy on the solution's. Following the solution authority, the marked answer is D.

The function is of the form $u(x)^{v(x)}$, so logarithmic differentiation is appropriate.

Write

Then apply the product rule to the right-hand side. The derivative of $\sin^2 x$ is

and the derivative of

is

So,

which matches the extracted working.