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JEE Advanced 2008Paper 1
Exam
JEE Advanced
Year
2008
Paper
Paper 1
Subject
Mathematics
Chapter
Differentiation

JEE Advanced 2008 · Paper 1 · Mathematics · Differentiation

Paragraph: Consider the functions defined implicitly by the...

Paragraph: Consider the functions defined implicitly by the equation y3−3y+x=0y^3-3 y+x=0 on various intervals in the real line. If x∈(−∞,−2)∪(2,∞)x \in(-\infty,-2) \cup(2, \infty), the equation implicitly defines a unique real valued differentiable function y=f(x)y=f(x). If x∈(−2,2)x \in(-2,2), the equation implicitly defines a unique real valued differentiable function y=g(x)y=g(x), satisfying g(0)=0g(0)=0.Question: If f(−102)=22f(-10 \sqrt{2})=2 \sqrt{2}, then f′′(−102)f^{\prime \prime}(-10 \sqrt{2}) is equal to

Options

  1. A4273⋅32\frac{4 \sqrt{2}}{7^3 \cdot 3^2}
  2. B−4273⋅32-\frac{4 \sqrt{2}}{7^3 \cdot 3^2}
  3. C4273⋅3\frac{4 \sqrt{2}}{7^3 \cdot 3}
  4. D−4273⋅3-\frac{4 \sqrt{2}}{7^3 \cdot 3}

Correct answer

B. −4273⋅32-\frac{4 \sqrt{2}}{7^3 \cdot 3^2}

Step-by-step solution

Given, y3−3y+x=0⇒3y2dydx−3dydx+1=0y^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 y²(d2ydx2)+\left(\frac{d^2 y}{d x^2}\right)+6 y\left(\frac{d y}{d x}\right)^2−32-3\frac{d^2 y}{d x^2}=0Onsubstituting0 On substitutingx=-10 \sqrt{2}, y=2 \sqrt{2}inEq.(i),weget3(2in Eq. (i), we get 3(2\sqrt{2})^22\cdot \frac{d y}{d x}-33\cdot \frac{d y}{d x}+1=0⇒1=0 \Rightarrow\frac{d y}{d x}=\frac{-1}{21}AgainonsubstitutingAgain on substitutingx=-10 \sqrt{2}, y=2 \sqrt{2}inEq.(ii),wegetin Eq. (ii), we get3(22)2d2ydx2+6(22)⋅(−121)2−3⋅d2ydx2=0⇒21⋅d2ydx2=−122(21)2⇒d2ydx2=−122(21)3=−4273⋅32\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}$
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