Let M and m respectively be the maximum and the minimum values of $f(x)=\left|\begin{array}{ccc}1+\sin ^{2} x & \cos ^{2} x & 4 \sin 4 x \\ \sin ^{2} x & 1+\cos ^{2} x & 4 \sin 4 x \\ \sin ^{2} x & \cos ^{2} x & 1+4 \sin 4 x\end{array}\right|, x \in R$ Then $\mathrm{M}^{4}-\mathrm{m}^{4}$ is equal to :

JEE Main · 2025 · 29 Jan 2025 · Shift 1 · Determinants

Let M and m respectively be the maximum and the minimum values of

f (x) = 1 + sin^{2} x sin^{2} x sin^{2} x cos^{2} x 1 + cos^{2} x cos^{2} x 4 sin 4 x 4 sin 4 x 1 + 4 sin 4 x, x \in R

Then

M^{4} - m^{4}

is equal to :

Answer: (A)

Applying

R_{2} \to R_{2} - R_{1}

and

R_{3} \to R_{3} - R_{1}

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

gives

f (x) = 1 + sin^{2} x - 1 - 1 cos^{2} x 10 4 sin 4 x 01 .

Expanding along the second row:

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

Since

- 1 \leq sin 4 x \leq 1

, the maximum is

M = 6

and the minimum is

m = - 2

. Thus

M^{4} - m^{4} = 6^{4} - (- 2)^{4} = 1296 - 16 = 1280.

JEE Main · 2025 · 29 Jan 2025 · Shift 1 · Single correct option