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Q.

Let f:R→R be a one-one onto differentiable function such that f(2)=1 and f1(2)=3 then find the value of ddxf−1(x)x=1

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a

1

b

12

c

13

d

14

answer is C.

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Detailed Solution

let f−1(x)=g(x)f(g(x))=x⇒f′(g(x))g′(x)=1g′(x)=1f′(g(x))g′(1)=1f′(g(1))=1f′(2)=13

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