If y = ( x + 1)( x ^ 2 - x ) x x + x + x + (1)/(15)(3 ^ 2 x - 5) ^ 3 x, then 96 y ^ ' ((π)/(6)) is equal to:

Correct answer: 0 Given: y = ( x + 1)( x ^ 2 - x ) x x + x + x + (1)/(15)(3 ^ 5 x - 5 ^ 3 x) ⇒ y = ( x + 1)( x ^ 2 - x )( x - 1) (x x + x + x )( x - 1) + (1)/(15)(3 ^ 5 x - 5 ^ 3 x) ⇒ y = ( x + 1)( x ^ 2 - x )( x - 1) x (x + x + 1)( x - 1) + (1)/(15)(3 ^ 5 x - 5 ^ 3 x) ⇒ y = ( x + 1)( x ^ 2 - x )( x - 1) x ( ( x ) ^ 3 - 1) + (1)/(15)(3 ^ 5 x - 5 ^ 3 x) ⇒ y = ( x + 1) x ( ( x ) ^ 3 - 1)( x - 1) x ( ( x ) ^ 3 - 1) + (1)/(15)(3 ^ 5 x - 5 ^ 3 x) ⇒ y = ( x + 1)( x - 1) + (1)/(15)(3 ^ 5 x - 5 ^ 3 x) ⇒ y = x - 1 + (1)/(15)(3 ^ 5 x - 5 ^ 3 x) ⇒ (dy)/(dx) = 1 + (1)/(15)(15 ^ 4 x(- x) + 15 ^ 2 xsinx) ⇒ (dy)/(dx) = 1 + ( ^ 4 x(- x) + ^ 2 xsinx) ⇒ y ^ ' ((π)/(6)) = 1 + ( ^ 4 ((π)/(6))(- ((π)/(6))) + ^ 2 ((π)/(6)) ((π)/(6))) ⇒ y ^ ' ((π)/(6)) = 1 + ( ( 3 ) ^ 4 2 ^ 4 (- (1)/(2)) + ( 3 ) ^ 2 2 ^ 2 × (1)/(2)) ⇒ y ^ ' ((π)/(6)) = 1 - (9)/(32) + (3)/(8) ⇒ y ^ ' ((π)/(6)) = (32 - 9 + 12)/(32) ⇒ y ^ ' ((π)/(6)) = (35)/(32) ⇒ 96 y ^ ' ((π)/(6)) = 35 × 3 ⇒ 96 y ^ ' ((π)/(6)) = 105

JEE Main 20241 FebruaryEvening Shift

Exam: JEE Main
Year: 2024
Date: 1 February
Shift: Evening Shift
Subject: Mathematics
Chapter: Differentiation

JEE Main 2024 · 1 February · Evening Shift · Mathematics · Differentiation

If y = ( x + 1)( x ^ 2 - x ) x x + x + x + (1)/(15)(3 ^ 2 x - 5) ^...

y = \dfrac{\left(\sqrt{x} + 1\right)\left({x}^{2} - \sqrt{x}\right)}{x\sqrt{x} + x + \sqrt{x}} + \dfrac{1}{15}\left(3\cos^{2}x - 5\right)\cos^{3}x,

then

96{y}^{'}\left(\dfrac{\pi}{6}\right)

is equal to:

Correct answer

Step-by-step solution

Given: $y = \dfrac{\left(\sqrt{x} + 1\right)\left({x}^{2} - \sqrt{x}\right)}{x\sqrt{x} + x + \sqrt{x}} + \dfrac{1}{15}\left(3\cos^{5}x - 5\cos^{3}x\right)$

$⇒ y = \dfrac{\left(\sqrt{x} + 1\right)\left({x}^{2} - \sqrt{x}\right)\left(\sqrt{x} - 1\right)}{\left(x\sqrt{x} + x + \sqrt{x}\right)\left(\sqrt{x} - 1\right)} + \dfrac{1}{15}\left(3\cos^{5}x - 5\cos^{3}x\right)$

$⇒ y = \dfrac{\left(\sqrt{x} + 1\right)\left({x}^{2} - \sqrt{x}\right)\left(\sqrt{x} - 1\right)}{\sqrt{x}\left(x + \sqrt{x} + 1\right)\left(\sqrt{x} - 1\right)} + \dfrac{1}{15}\left(3\cos^{5}x - 5\cos^{3}x\right)$

$⇒ y = \dfrac{\left(\sqrt{x} + 1\right)\left({x}^{2} - \sqrt{x}\right)\left(\sqrt{x} - 1\right)}{\sqrt{x}\left({\left(\sqrt{x}\right)}^{3} - 1\right)} + \dfrac{1}{15}\left(3\cos^{5}x - 5\cos^{3}x\right)$

$⇒ y = \dfrac{\left(\sqrt{x} + 1\right)\sqrt{x}\left({\left(\sqrt{x}\right)}^{3} - 1\right)\left(\sqrt{x} - 1\right)}{\sqrt{x}\left({\left(\sqrt{x}\right)}^{3} - 1\right)} + \dfrac{1}{15}\left(3\cos^{5}x - 5\cos^{3}x\right)$

$⇒ y = \left(\sqrt{x} + 1\right)\left(\sqrt{x} - 1\right) + \dfrac{1}{15}\left(3\cos^{5}x - 5\cos^{3}x\right)$

$⇒ y = x - 1 + \dfrac{1}{15}\left(3\cos^{5}x - 5\cos^{3}x\right)$

$⇒ \dfrac{dy}{dx} = 1 + \dfrac{1}{15}\left(15\cos^{4}x\left(- \sin x\right) + 15\cos^{2}xsinx\right)$

$⇒ \dfrac{dy}{dx} = 1 + \left(\cos^{4}x\left(- \sin x\right) + \cos^{2}xsinx\right)$

$⇒ {y}^{'}\left(\dfrac{\pi}{6}\right) = 1 + \left(\cos^{4}\left(\dfrac{\pi}{6}\right)\left(- \sin \left(\dfrac{\pi}{6}\right)\right) + \cos^{2}\left(\dfrac{\pi}{6}\right)\sin \left(\dfrac{\pi}{6}\right)\right)$

$⇒ {y}^{'}\left(\dfrac{\pi}{6}\right) = 1 + \left(\dfrac{{\left(\sqrt{3}\right)}^{4}}{{2}^{4}}\left(- \dfrac{1}{2}\right) + \dfrac{{\left(\sqrt{3}\right)}^{2}}{{2}^{2}} \times \dfrac{1}{2}\right)$

$⇒ {y}^{'}\left(\dfrac{\pi}{6}\right) = 1 - \dfrac{9}{32} + \dfrac{3}{8}$

$⇒ {y}^{'}\left(\dfrac{\pi}{6}\right) = \dfrac{32 - 9 + 12}{32}$

$⇒ {y}^{'}\left(\dfrac{\pi}{6}\right) = \dfrac{35}{32}$

$⇒ 96{y}^{'}\left(\dfrac{\pi}{6}\right) = 35 \times 3$

$⇒ 96{y}^{'}\left(\dfrac{\pi}{6}\right) = 105$

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