For a>0, t ∈(0, (π)/(2)) , let x= a^ ^ -1 t and y= a^ ^ -1 t , Then, 1+((d y)/(d x))^2 equals :

Correct answer: D. (x^2+y^2)/(x^2) aligned Let x= a^ ^ -1 t & x²=a^ ^ -1 t 2 x= ^ -1 t · a & (2)/(x)= a 1-t^2 · (d t)/(d x) & 2 1-t^2 x a =(d t)/(d x) aligned Now, let y= a^ ^ -1 t aligned & 2 y= ^ -1 t · a & (2)/(y) · (d y)/(d x)= - a 1-t^2 · (d t)/(d x) aligned (2)/(y) · (d y)/(d x)= - a 1-t^2 × 2 1-t^2 x a (from(1) (d y)/(d x)=-(y)/(x) Hence, 1+((d y)/(d x))^2=1+((-y)/(x))^2=(x^2+y^2)/(x^2)

JEE Main 201322 AprilOnline

Exam: JEE Main
Year: 2013
Date: 22 April
Mode: Online
Subject: Mathematics
Chapter: Differentiation

JEE Main 2013 · 22 April · Online · Mathematics · Differentiation

For a>0, t ∈(0, (π)/(2)) , let x= a^ ^ -1 t and y= a^ ^ -1 t ,...

For

a>0, t \in\left(0, \frac{\pi}{2}\right)

, let

x=\sqrt{a^{\sin ^{-1} t}}

and

y=\sqrt{a^{\cos ^{-1} t}}

, Then,

1+\left(\frac{d y}{d x}\right)^2

equals :

Options

A $\frac{x^2}{y^2}$
B $\frac{y^2}{x^2}$
C $\frac{x^2+y^2}{y^2}$
D $\frac{x^2+y^2}{x^2}$

Correct answer

D. $\frac{x^2+y^2}{x^2}$

Step-by-step solution

\begin{aligned} \text{ Let } x=\sqrt{a^{\sin ^{-1} t}}\\ & \Rightarrow \quad x²=a^{\sin ^{-1} t} \Rightarrow 2 \log x=\sin ^{-1} t \cdot \log a \\ & \Rightarrow \quad \frac{2}{x}=\frac{\log a}{\sqrt{1-t^2}} \cdot \frac{d t}{d x}\\ & \Rightarrow \quad \frac{2 \sqrt{1-t^2}}{x \log a}=\frac{d t}{d x} \end{aligned}

Now, let

y=\sqrt{a^{\cos ^{-1} t}}

\begin{aligned}& \Rightarrow 2 \log y=\cos ^{-1} t \cdot \log a \\ & \Rightarrow \quad \frac{2}{y} \cdot \frac{d y}{d x}=\frac{-\log a}{\sqrt{1-t^2}} \cdot \frac{d t}{d x} \end{aligned}

\Rightarrow \frac{2}{y} \cdot \frac{d y}{d x}=\frac{-\log a}{\sqrt{1-t^2}} \times \frac{2 \sqrt{1-t^2}}{x \log a} \quad

(from(1) \Rightarrow

\frac{d y}{d x}=-\frac{y}{x}

Hence,

1+\left(\frac{d y}{d x}\right)^2=1+\left(\frac{-y}{x}\right)^2=\frac{x^2+y^2}{x^2}

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