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JEE Main 20178 AprilOnline
Exam
JEE Main
Year
2017
Date
8 April
Mode
Online
Subject
Mathematics
Chapter
Differentiation

JEE Main 2017 · 8 April · Online · Mathematics · Differentiation

If y = [x + x ^ 2 - 1 ] ^ 15 + [x - x ^ 2 - 1 ] ^ 15 , then ( x ^ 2...

If y=[x+x2−1]15+[x−x2−1]15y = {\left[x + \sqrt{{x}^{2} - 1}\right]}^{15} + {\left[x - \sqrt{{x}^{2} - 1}\right]}^{15}, then \left({x}^{2} - 1\right)\dfrac{{d}^{2}y}{d{x}^{2}} + x\dfrac{dy}{dx} is equal to

Options

  1. A224y2224{y}^{2}
  2. B125y125y
  3. C225y225y
  4. D225y2225{y}^{2}

Correct answer

C. 225y225y

Step-by-step solution

y={x+x2−1}15+{x−x2−1}15y = {\left\{x + \sqrt{{x}^{2} - 1}\right\}}^{15} + {\left\{x - \sqrt{{x}^{2} - 1}\right\}}^{15}

dydx=15{x+x2−1}14(1+xx2−1)+15{x−x2−1}14(1−xx2−1)\dfrac{dy}{dx} = 15{\left\{x + \sqrt{{x}^{2} - 1}\right\}}^{14}\left(1 + \dfrac{x}{\sqrt{{x}^{2} - 1}}\right) + 15{\left\{x - \sqrt{{x}^{2} - 1}\right\}}^{14}\left(1 - \dfrac{x}{\sqrt{{x}^{2} - 1}}\right)

dydx=15{x+x2−1}14(x+x2−1x2−1)+15{x−x2−1}14(x−x2−1x2−1)\dfrac{dy}{dx} = 15{\left\{x + \sqrt{{x}^{2} - 1}\right\}}^{14}\left(\dfrac{x + \sqrt{{x}^{2} - 1}}{\sqrt{{x}^{2} - 1}}\right) + 15{\left\{x - \sqrt{{x}^{2} - 1}\right\}}^{14}\left(\dfrac{x - \sqrt{{x}^{2} - 1}}{\sqrt{{x}^{2} - 1}}\right)

dydx=15[{x+x2−1}15−{x−x2−1}15]x2−1\dfrac{dy}{dx} = \dfrac{15\left[{\left\{x + \sqrt{{x}^{2} - 1}\right\}}^{15} - {\left\{x - \sqrt{{x}^{2} - 1}\right\}}^{15}\right]}{\sqrt{{x}^{2} - 1}}

⇒⇒ x2−1.dydx=15[{x+x2−1}15−{x−x2−1}15]\sqrt{{x}^{2} - 1} . \dfrac{dy}{dx} = 15\left[{\left\{x + \sqrt{{x}^{2} - 1}\right\}}^{15} - {\left\{x - \sqrt{{x}^{2} - 1}\right\}}^{15}\right]

⇒⇒ xx2−1.dydx+x2−1d2ydx2=15×15yx2−1\dfrac{x}{\sqrt{{x}^{2} - 1}} . \dfrac{dy}{dx} + \sqrt{{x}^{2} - 1}\dfrac{{d}^{2}y}{d{x}^{2}} = 15 \times \dfrac{15y}{\sqrt{{x}^{2} - 1}}

⇒⇒ xdydx+(x2−1)d2ydx2=15x2−1×15yx2−1=225yx\dfrac{dy}{dx} + \left({x}^{2} - 1\right)\dfrac{{d}^{2}y}{d{x}^{2}} = 15\sqrt{{x}^{2} - 1} \times \dfrac{15y}{\sqrt{{x}^{2} - 1}} = 225y

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