JEE Main 2024 — 30 January, Shift 1 — Question 17

Let the function $f(x)$ be defined by the determinant:

$$f(x) = \begin{vmatrix} 2\cos^4 x & 2\sin^4 x & 3+\sin^2 2x \\ 3+2\cos^4 x & 2\sin^4 x & 3+2\sin^2 2x \\ 2\cos^4 x & 2\sin^4 x & 3+\sin^2 2x \end{vmatrix}$$

We need to find $\frac{1}{5} f'(0)$.

First, observe the rows of the determinant. Let $R_1, R_2, R_3$ denote the first, second, and third rows respectively. We can see that $R_1 = R_3$. A fundamental property of determinants states that if two rows (or columns) of a matrix are identical, the determinant of that matrix is zero.

In this case, $R_1$ and $R_3$ are identical. $R_1 = [2\cos^4 x \quad 2\sin^4 x \quad 3+\sin^2 2x]$ $R_3 = [2\cos^4 x \quad 2\sin^4 x \quad 3+\sin^2 2x]$

Since $R_1 = R_3$, the determinant $f(x)$ is identically zero for all $x \in \mathbb{R}$. Therefore, $f(x) = 0$ for all $x$.

If $f(x) = 0$ for all $x$, then its derivative $f'(x)$ is also zero for all $x$. $f'(x) = \frac{d}{dx}(0) = 0$.

Now, we need to find $\frac{1}{5} f'(0)$: $\frac{1}{5} f'(0) = \frac{1}{5} (0)$ $\frac{1}{5} f'(0) = 0$.

The final answer is $\boxed{0}$.