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Question

A function f:RR is defined as f(x)=x32x2+5x1 and g:RR be its inverse function. If k g(3)=1, then value(s) of k is/are

A
3
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B
17
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C
27
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D
7
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Solution

The correct option is D 7
Since, g and f are inverse function
gf=Ig[f(x)]=x
Differentiating both the sides,
g[f(x)]f(x)=1
If f(x)=3, then
g(3)f(x)=1 ...(1)
Given, k g(3)=1 ...(2)

From (1) and (2), we have
k=f(x) ...(3)
Now, f(x)=3
x32x2+5x1=3
x32x2+5x4=0
(x1)(x2x+4)=0
x=1

f(x)=6x24x+5
f(1)=64+5=7
k=f(1)=7

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