Jee Mains 2018 practice problem -2

Calculus Level 3

If y = ( x + x 2 − 1 ) 15 + ( x − x 2 − 1 ) 15 y=\left(x+\sqrt{x^{2}-1}\right)^{15}+\left(x-\sqrt{x^{2}-1}\right)^{15} , what is ( x 2 − 1 ) d 2 y d x 2 + x d y d x \left(x^{2}-1\right)\dfrac{d^{2}y}{dx^{2}}+x\dfrac{dy}{dx} ?

225 y 2 225y^{2} 225 y 225y 224 y 2 224y^{2} 125 y 125y

Shivam Jadhav by Shivam Jadhav , 22, India

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1 solution

y = ( x + x 2 − 1 ) 15 + ( x − x 2 − 1 ) 15 Let x = sec ⁡ θ ⟹ x 2 − 1 = tan ⁡ θ = ( sec ⁡ θ + tan ⁡ θ ) 15 + ( sec ⁡ θ − tan ⁡ θ ) 15 d y d x = 15 ( sec ⁡ θ + tan ⁡ θ ) 14 ( sec ⁡ θ tan ⁡ θ + sec ⁡ 2 θ ) + 15 ( sec ⁡ θ − tan ⁡ θ ) 14 ( sec ⁡ θ tan ⁡ θ − sec ⁡ 2 θ ) sec ⁡ θ tan ⁡ θ Chain rule: d y d x = d y d θ ⋅ d θ d x = 15 ( ( sec ⁡ θ + tan ⁡ θ ) 15 − ( sec ⁡ θ − tan ⁡ θ ) 15 ) tan ⁡ θ \begin{aligned} y & = \left(x+\sqrt{x^2-1}\right)^{15} + \left(x-\sqrt{x^2-1}\right)^{15} & \small \color{#3D99F6} \text{Let }x = \sec \theta \implies \sqrt{x^2-1} = \tan \theta \\ & = \left(\sec \theta +\tan \theta \right)^{15} + \left(\sec \theta -\tan \theta \right)^{15} \\ \color{#3D99F6} \frac {dy}{dx} & = \frac {\color{#3D99F6}15\left(\sec \theta +\tan \theta \right)^{14}\left(\sec \theta\tan \theta + \sec^2 \theta \right) + 15\left(\sec \theta - \tan \theta \right)^{14}\left(\sec \theta\tan \theta - \sec^2 \theta \right)}{\color{#D61F06}\sec \theta \tan \theta} & \small \color{#3D99F6} \text{Chain rule: }\frac {dy}{dx} = \frac {dy}{d \theta} \cdot\color{#D61F06} \frac {d\theta}{dx} \\ & = \frac {15\left(\left(\sec \theta +\tan \theta \right)^{15} - \left(\sec \theta - \tan \theta \right)^{15}\right)}{\tan \theta} \end{aligned}

Therefore,

tan ⁡ θ d y d x = 15 ( ( sec ⁡ θ + tan ⁡ θ ) 15 − ( sec ⁡ θ − tan ⁡ θ ) 15 ) Differentiate both sides w.r.t x tan ⁡ θ ⋅ d 2 y d x 2 + sec ⁡ 2 θ sec ⁡ θ tan ⁡ θ ⋅ d y d x = 1 5 2 ( ( sec ⁡ θ + tan ⁡ θ ) 15 + ( sec ⁡ θ − tan ⁡ θ ) 15 ) tan ⁡ θ Multiply throughout by tan ⁡ θ tan ⁡ 2 θ ⋅ d 2 y d x 2 + sec ⁡ θ ⋅ d y d x = 225 y ( x 2 − 1 ) d 2 y d x 2 + x d y d x = 225 y \begin{aligned} \tan \theta \frac {dy}{dx} & = 15\left(\left(\sec \theta +\tan \theta \right)^{15} - \left(\sec \theta - \tan \theta \right)^{15}\right) & \small \color{#3D99F6} \text{Differentiate both sides w.r.t }x \\ \tan \theta \cdot \frac {d^2y}{dx^2} + \frac {\sec^2 \theta}{\sec \theta \tan \theta} \cdot \frac {dy}{dx} & = \frac {15^2\left(\left(\sec \theta +\tan \theta \right)^{15} + \left(\sec \theta - \tan \theta \right)^{15}\right)}{\tan \theta} & \small \color{#3D99F6} \text{Multiply throughout by }\tan \theta \\ \tan^2 \theta \cdot \frac {d^2y}{dx^2} + \sec \theta \cdot \frac {dy}{dx} & = 225 y \\ \left(x^2-1\right) \frac {d^2y}{dx^2} + x \frac {dy}{dx} & = \boxed{225 y} \end{aligned}

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