If {x²+y²=a ^{-1} ≤ft( y/x ), a > 0, then d² y/dx² at x=0 is

Correct Answer is : None of these

Correct Answer is : None of these

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CGPET 2011 Solved Paper

Section: Mathematics

Question : 124 of 150

Marks: +1, -0

\sqrt{x^2+y^2}=a \tan^{-1} \left( \frac{y}{x} \right), a > 0

\frac{d^2 y}{dx^2}

x=0

Solution:

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

. . . (i)

Putting

x=0

, we get

y=a \tan^{-1}(\infty)

y=\frac{a\pi}{2}

Now, differentiating Eq. (i) w.r.t. x, we get

\frac{2x+2y\frac{dy}{dx}}{2\sqrt{x^2+y^2}}=a \cdot \frac{1}{1+\frac{y^2}{x^2}} \cdot \left\{ \frac{x\frac{dy}{dx}-y}{x^2} \right\}

\Rightarrow \frac{x+y\frac{dy}{dx}}{\sqrt{x^2+y^2}} = \frac{a x^2}{x^2+y^2} \left\{ \frac{x\frac{dy}{dx}-y}{x^2} \right\}

\Rightarrow \sqrt{x^2+y^2} \left(x+y\frac{dy}{dx}\right) = a \left(x\frac{dy}{dx}-y\right) \dots (ii)

\text{ At } x=0, y=\frac{a\pi}{2},

\frac{a\pi}{2} \left[ \frac{a\pi}{2} \cdot \frac{dy}{dx} \right] = a \left[ -\frac{a\pi}{2} \right]

\frac{\pi}{2} \cdot \frac{dy}{dx} = -1

\frac{dy}{dx} = -\frac{2}{\pi}

Again, differentiating Eq. (ii), w.r.t.

x

, we get

\frac{2x+2y\frac{dy}{dx}}{2\sqrt{x^2+y^2}} \cdot \left(x+y\frac{dy}{dx}\right)

+\sqrt{x^2+y^2} \left[ 1 + y \frac{d^2 y}{dx^2} + \left( \frac{dy}{dx} \right)^2 \right]

= a \left[ x \frac{d^2 y}{dx^2} + \frac{dy}{dx} - \frac{dy}{dx} \right]

\Rightarrow \frac{\left(x+y\frac{dy}{dx}\right)^2}{\sqrt{x^2+y^2}}+\sqrt{x^2+y^2}

\left[ 1+y\frac{d^2 y}{dx^2} + \left( \frac{dy}{dx} \right)^2 \right] = a x \frac{d^2 y}{dx^2}

Putting

x=0, y=\frac{a\pi}{2}, \frac{dy}{dx}=-\frac{2}{\pi}

\frac{a\pi}{2} \left(-\frac{2}{\pi}\right)^2 + \frac{a\pi}{2} \left[ 1 + \frac{a\pi}{2} \left( \frac{d^2 y}{dx^2} \right) + \left(-\frac{2}{\pi}\right)^2 \right] = 0

\Rightarrow \frac{2a}{\pi} + \frac{a\pi}{2} + \frac{a^2 \pi^2}{4} \cdot \frac{d^2 y}{dx^2} + \frac{4}{\pi^2} \cdot \frac{a\pi}{2} = 0

\Rightarrow \frac{4a}{\pi} + \frac{a\pi}{2} + \frac{a^2 \pi^2}{4} \cdot \frac{d^2 y}{dx^2} = 0

\Rightarrow \frac{4}{\pi} + \frac{\pi}{2} + \frac{a \pi^2}{4} \cdot \frac{d^2 y}{dx^2} = 0

\Rightarrow \frac{d^2 y}{dx^2} = \frac{-\frac{4}{\pi} - \frac{\pi}{2}}{\frac{a \pi^2}{4}} = \frac{-2(8+\pi^2)}{a \pi^3}

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