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← IIT-JEE 2008 · Paper 1
MathematicsSingle correct+4 / -1Question 18 of 69

IIT-JEE 2008 · Paper 1, Mathematics question 18

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Paragraph for Question Nos. 18 to 20

Consider the functions defined implicitly by the equation y3−3y+x=0y^3-3y+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.

If f(−102)=22f(-10\sqrt2)=2\sqrt2, then f′′(−102)=f''(-10\sqrt2)=

Choose one option
Answer and worked solution

Official answer: B

First derivative. Differentiating y3−3y+x=0y^3-3y+x=0:
f′=−13(f2−1).f'=-\frac1{3(f^2-1)}.

Second derivative.
f′′=2ff′3(f2−1)2.f''=\frac{2ff'}{3(f^2-1)^2}.

At x=−102x=-10\sqrt2, f=22f=2\sqrt2, f2−1=7f^2-1=7, f′=−121f'=-\frac1{21}:
f′′=2⋅22⋅(−1/21)3⋅49=−423087=−4273⋅32.f''=\frac{2\cdot2\sqrt2\cdot(-1/21)}{3\cdot49}=-\frac{4\sqrt2}{3087}=-\frac{4\sqrt2}{7^3\cdot3^2}.

(B)

Topics:Chain rule and implicit differentiationHigher-order derivatives
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