If \(y=\frac{(\sqrt{x}+1)(x^{2}-\sqrt{x})}{x \sqrt{x}+x+\sqrt{x}}+\frac{1}{15}(3 \cos ^{2} x-5) \cos ^{3} x\), then \(96 y^{\prime}\left(\frac{\pi}{6}\right)\) is equal to :

JEE Main · 2024 · 01 Feb 2024 · Shift 2 · Differentiation

y = \frac{( x + 1 ) ( x ^{2} - x )}{x x + x + x} + \frac{1}{15} (3 cos^{2} x - 5) cos^{3} x

, then

96 y^{'} (\frac{π}{6})

is equal to :

Answer: 105

Write

t = x

in the first term:

\frac{( x + 1 ) ( x ^{2} - x )}{x x + x + x} = \frac{( t + 1 ) ( t ^{4} - t )}{t ^{3} + t ^{2} + t} = \frac{t ( t + 1 ) ( t ^{3} - 1 )}{t ( t ^{2} + t + 1 )} = \frac{t ( t + 1 ) ( t - 1 ) ( t ^{2} + t + 1 )}{t ( t ^{2} + t + 1 )} = (t + 1) (t - 1) = t^{2} - 1 = x - 1.

The second term:

\frac{1}{15} (3 cos^{2} x - 5) cos^{3} x = \frac{1}{5} cos^{5} x - \frac{1}{3} cos^{3} x .

y = (x - 1) + \frac{1}{5} cos^{5} x - \frac{1}{3} cos^{3} x .

Differentiate:

y^{'} = 1 + \frac{1}{5} \cdot 5 cos^{4} x (- sin x) - \frac{1}{3} \cdot 3 cos^{2} x (- sin x) = 1 - cos^{4} x sin x + cos^{2} x sin x .

x = \frac{π}{6}

sin x = \frac{1}{2}, cos x = \frac{3}{2}, cos^{2} x = \frac{3}{4}, cos^{4} x = \frac{9}{16}

y^{'} (\frac{π}{6}) = 1 - \frac{9}{16} \cdot \frac{1}{2} + \frac{3}{4} \cdot \frac{1}{2} = 1 - \frac{9}{32} + \frac{12}{32} = 1 + \frac{3}{32} = \frac{35}{32} .

96 y^{'} (\frac{π}{6}) = 96 \cdot \frac{35}{32} = 3 \cdot 35 = 105.

JEE Main · 2024 · 01 Feb 2024 · Shift 2 · Integer answer