Area Ques 5
Passage Based Questions
Consider the functions defined implicity 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 $g(0)=0$.
$(2008, \mathrm{M})$
- If $f(-10 \sqrt{2})=2 \sqrt{2}$, then $f^{\prime \prime}(-10 \sqrt{2})$ is equal to
(a) $\frac{4 \sqrt{2}}{7^3 3^2}$
(b) $-\frac{4 \sqrt{2}}{7^3 3^2}$
(c) $\frac{4 \sqrt{2}}{7^3 3}$
(d) $-\frac{4 \sqrt{2}}{7^3 3}$
Show Answer
Answer:
Correct Answer: 5.(b)
Solution: (b) Given, $ y^3-3 y+x=0 $
$\Rightarrow \quad 3 y^2 \frac{d y}{d x}-3 \frac{d y}{d x}+1=0 $ $\quad$ ……..(i)
$\Rightarrow \quad 3 y^2\left(\frac{d^2 y}{d x^2}\right)+6 y\left(\frac{d y}{d x}\right)^2-3 \frac{d^2 y}{d x^2}=0$ $\quad$ ……..(ii)
At $x=-10 \sqrt{2}, y=2 \sqrt{2}$
On substituting in Eq. (i) we get
$3(2 \sqrt{2})^2 \cdot \frac{d y}{d x}-3 \cdot \frac{d y}{d x}+1 =0$
$\Rightarrow \quad \frac{d y}{d x} =-\frac{1}{21}$
Again, substituting 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 \quad & \frac{d^2 y}{d x^2}=\frac{-12 \sqrt{2}}{(21)^3}=\frac{-4 \sqrt{2}}{7^3 \cdot 3^2} \end{aligned} $
BETA
