Algebra – Matrices and Determinants – JEE Main 29 Jan 2025 Shift 1

Solution:

Applying row operations $R_{2} \to R_{2}-R_{1}$ and $R_{3} \to R_{3}-R_{1}$:

$$ f(x)=\left|\begin{matrix}1+\sin^{2}x & \cos^{2}x & 4\sin 4x \\ -1 & 1 & 0 \\ -1 & 0 & 1\end{matrix}\right| $$

Expand along $R_{1}$:

$f(x) = (1+\sin^{2}x)(1-0) – \cos^{2}x(-1-0) + 4\sin 4x(0-(-1))$

$f(x) = 1+\sin^{2}x + \cos^{2}x + 4\sin 4x$

$f(x) = 1 + 1 + 4\sin 4x$

$f(x) = 2 + 4\sin 4x$

The maximum value of $\sin 4x$ is $1$ and the minimum value is $-1$.

$M = \text{max value of } f(x) = 2 + 4(1) = 6$

$m = \text{min value of } f(x) = 2 + 4(-1) = -2$

We need to find $M^{4} – m^{4}$:

$M^{4} – m^{4} = (6)^{4} – (-2)^{4}$

$= 1296 – 16$

$= 1280$

Ans. (1)