Question

PLZZ. ANSWER THE QUESTION

JEE Main 2002

XX NO SPAMXX​

Answer 1

AnswEr :

Given Expression,

$\sf y = (x + \sqrt{1 + {x}^{2} } ) {}^{n}$

Differentiating w.r.t x , we get :

$\implies \: \sf \dfrac{dy}{dx} = {n(x + \sqrt{1 + {x}^{2} }) }^{n - 1} \times \dfrac{d(x + \sqrt{1 + x {}^{2} }) }{dx} \\ \\ \implies \: \sf \: \dfrac{dy}{dx} = {n(x + \sqrt{1 + {x}^{2} }) }^{n - 1} \times \bigg(1 + \dfrac{1}{2 \sqrt{ {x}^{2} + 1 } } \times \frac{d( {x}^{2} )}{dx} \bigg) \\ \\ \implies \: \sf \dfrac{dy}{dx} = {n(x + \sqrt{1 + {x}^{2} }) }^{n - 1} \times \bigg(1 + \dfrac{2x}{2 \sqrt{ {x}^{2} + 1 } } \bigg) \\ \\ \implies \sf \: \dfrac{dy}{dx} = {n(x + \sqrt{1 + {x}^{2} }) }^{n - 1} \times \bigg( \dfrac{x + \sqrt{1 + {x}^{2} } }{\sqrt{ {x}^{2} + 1 } } \bigg) \\ \\ \implies \sf \: \dfrac{dy}{dx} = \frac{n(x + \sqrt{ {x}^{2} + 1}) {}^{n} }{ \sqrt{1 + {x}^{2} } } - - - - - - (1) \\ \\ \implies \: \sf \: ( \sqrt{1 + {x}^{2} } ) \dfrac{dy}{dx} = ny$

Again differentiating w.r.t x on both sides,

$\implies \sf \: \cfrac{x}{ \sqrt{1 + {x}^{2} } } \dfrac{dy}{dx} + \sqrt{1 + {x}^{2} } \dfrac{ {d}^{2}y }{dx {}^{2} } = n \times \dfrac{dy}{dx}$

Using equation (1),

$\implies \: \sf \: \dfrac{x \dfrac{dy}{dx} + (1 + {x}^{2} ) \dfrac{ {d}^{2} y}{ {dx}^{2} } }{ \cancel{ \sqrt{1 + {x}^{2}} } } = \dfrac{ {n}^{2} (x + \sqrt{1 + {x}^{2} } ) {}^{n} }{ \cancel{\sqrt{1 + {x}^{2} } } } \\ \\ \implies \boxed{ \boxed{\sf \: x \dfrac{dy}{dx} + (1 + {x}^{2} ) \dfrac{ {d}^{2} y}{ {dx}^{2} } = {n}^{2} y}}$

Option (a) n²y is correct

Answer 2

Please refer to the attachment.

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