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JEE Main · 2024 · 30 Jan 2024 · Shift 1 · Differentiation

If

f (x) = 2 cos^{4} x 3 + 2 cos^{4} x 2 cos^{4} x 2 sin^{4} x 2 sin^{4} x 3 + 2 sin^{4} x 3 + sin^{2} 2 x sin^{2} 2 x sin^{2} 2 x

then

\frac{1}{5} f^{'} (0)

is equal to

(A)0✓ Correct
(B)1
(C)2
(D)6

Answer: (A)

Solution

R_{2} \to R_{2} - R_{1}, R_{3} \to R_{3} - R_{1}

reduces the determinant to

2 cos^{4} x 30 2 sin^{4} x 03 3 + sin^{2} 2 x - 3 - 3

.

Expanding gives

f (x) = 45

, a constant, so

f^{'} (x) = 0

for all

x

; in particular

\frac{1}{5} f^{'} (0) = 0

.

JEE Main · 2024 · 30 Jan 2024 · Shift 1 · Single correct option

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