JEE Main 2025 Jan 29 Shift 1, Mathematics Q2: Maxima & Minima

JEE Main2025Jan 29, Shift 1Mathematics

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

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

Then $M^{4} - m^{4}$ is equal to :

A
$1280$
B
$1295$
C
$1040$
D
$1215$

Solution

Subtract the first row from each of the other two: $R_{2} \to R_{2} - R_{1}$ and $R_{3} \to R_{3} - R_{1}$ .

$f (x) = 1 + sin^{2} x cos^{2} x 4 sin 4 x - 110 - 101 .$

Expanding along the first row, $f (x) = (1 + sin^{2} x) (1) - cos^{2} x (- 1) + 4 sin 4 x (1) = 1 + sin^{2} x + cos^{2} x + 4 sin 4 x = 2 + 4 sin 4 x .$

Since $sin 4 x \in [- 1, 1]$ : $M = 2 + 4 = 6$ and $m = 2 - 4 = - 2$ .

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

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JEE Main 2025 Jan 29 Shift 1, Mathematics Q2: Maxima & Minima

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