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Its an old IIT-JEE question [Paper 1,JEE 2008,Q.20] Consider the function defined implicitly by y^3 - 3y + x =0 on various intervals. If x belongs to (-2,2)the equation implicitly defines a unique real valued differentiable function y=g(x) satisfying g(0)=0 then (I'm writing in words because i can't get the symbols) Integral(-1 to +1) g'(x) = [g'(x) is derivative of g(x)] answer given is 2g(1). How to get it?

Its an old IIT-JEE question [Paper 1,JEE 2008,Q.20]

Consider the function defined implicitly by
y^3 - 3y + x =0 on various intervals.

If x belongs to (-2,2)the equation implicitly defines a unique real valued differentiable function y=g(x) satisfying g(0)=0 then

(I'm writing in words because i can't get the symbols)

Integral(-1 to +1) g'(x) =

[g'(x) is derivative of g(x)]


answer given is 2g(1).

How to get it?

Grade:12

1 Answers

SAGAR SINGH - IIT DELHI
878 Points
13 years ago

Dear student,

y3 - 3y + x = 0 Draw graph

{first draw y = 3x - x3 & then reflect in y = x line}

Please feel free to ask your queries here. We are all IITians and here to help you in your IIT JEE preparation.

All the best.

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Sagar Singh

B.Tech, IIT Delhi

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