Differentiation and its Application: If f(-10sqrt2)=2sqrt2 , then f''(-10sqrt2)= | JEE Advanced 2008 Paper 1

Comprehension · Questions 1–1

Read the passage given below and answer the questions that follow.

Consider the functions defined implicitly by the equation $y^3-3y+x=0$ 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 $g(0)=0$.

JEE Advanced 2008 Paper 1Question 1

If $f(-10\sqrt2)=2\sqrt2$, then $f''(-10\sqrt2)=$

Differentiation and its Application

$\dfrac{4\sqrt2}{7^33^2}$$-\dfrac{4\sqrt2}{7^33^2}$✓ Correct answer$\dfrac{4\sqrt2}{7^33}$$-\dfrac{4\sqrt2}{7^33}$

Solution

Concept. Here $y=f(x)$ is defined by $y^3-3y+x=0$ for $|x|>2$. Differentiate twice implicitly and use $y=2\sqrt2$. Formulas used. (1) $\dfrac{d}{dx}\,\dfrac{1}{u}=-\dfrac{u'}{u^2}$Step 1 - Check the point. $(2\sqrt2)^3-3(2\sqrt2)-10\sqrt2=16\sqrt2-6\sqrt2-10\sqrt2=0$. Step 2 - First derivative. $3y^2y'-3y'+1=0$, so $$y'=-\frac{1}{3(y^2-1)}.$$ Step 3 - Second derivative. By (1), $$y''=\frac{2yy'}{3(y^2-1)^2}=-\frac{2y}{9(y^2-1)^3}.$$ Step 4 - Value. $y=2\sqrt2$, $y^2-1=7$: $$y''=-\frac{4\sqrt2}{9\cdot7^3}=-\frac{4\sqrt2}{7^3\,3^2}.$$ Answer: (B)

Differentiation and its Application - JEE Mathematics question (JEE Advanced 2008 Paper 1)

More from Differentiation and its Application