Paragraph: Consider the functions defined implicitly by the equation $y^3-3 y+x=0$ on various intervals in…

Paragraph: Consider the functions defined implicitly by the equation $y^3-3 y+x=0$ on various intervals in the real line. If $x \in(-\infty,-2) \cup(2, \infty)$, the equation implicitly defines a unique real valued differentiable function $y=f(x)$. If $x \in(-2,2)$, the equation implicitly defines a unique real valued differentiable function $y=g(x)$, satisfying $g(0)=0$.Question: If $f(-10 \sqrt{2})=2 \sqrt{2}$, then $f^{\prime \prime}(-10 \sqrt{2})$ is equal to
  1. $\frac{4 \sqrt{2}}{7^3 \cdot 3^2}$
  2. $-\frac{4 \sqrt{2}}{7^3 \cdot 3^2}$
  3. $\frac{4 \sqrt{2}}{7^3 \cdot 3}$
  4. $-\frac{4 \sqrt{2}}{7^3 \cdot 3}$

Solution

Given, $y^3-3 y+x=0 \Rightarrow 3 y^2 \frac{d y}{d x}-3 \frac{d y}{d x}+1=0$ $ \Rightarrow \quad 3 y^2\left(\frac{d^2 y}{d x^2}\right)+6 y\left(\frac{d y}{d x}\right)^2-3 \frac{d^2 y}{d x^2}=0 $ On substituting $x=-10 \sqrt{2}, y=2 \sqrt{2}$ in Eq. (i), we get $ 3(2 \sqrt{2})^2 \cdot \frac{d y}{d x}-3 \cdot \frac{d y}{d x}+1=0 \Rightarrow \frac{d y}{d x}=\frac{-1}{21} $ Again on substituting $x=-10 \sqrt{2}, y=2 \sqrt{2}$ in Eq. (ii), we get $ \begin{aligned} & 3(2 \sqrt{2})^2 \frac{d^2 y}{d x^2}+6(2 \sqrt{2}) \cdot\left(\frac{-1}{21}\right)^2-3 \cdot \frac{d^2 y}{d x^2}=0 \\ & \Rightarrow \quad 21 \cdot \frac{d^2 y}{d x^2}=-\frac{12 \sqrt{2}}{(21)^2} \Rightarrow \frac{d^2 y}{d x^2}=-\frac{12 \sqrt{2}}{(21)^3}=-\frac{4 \sqrt{2}}{7^3 \cdot 3^2} \end{aligned} $

Asked in: JEE Advanced 2008 (Paper 1)

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