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Let f(x y)=f(x)f(y)AAx , y in R and f...

Let `f(x y)=f(x)f(y)AAx , y in R ` and ` f` is differentiable at `x=1` such that `f^(prime)(1)=1.` Also, `f(1)!=0,f(2)=3.` Then find `f^(prime)(2)`

Text Solution

Verified by Experts

The correct Answer is:
`3//2`
`f(xy)=f(x)f(y)" (1)"`
`f'(2)=underset(hrarr0)lim(f(2+h)-f(2))/(h)`
`=underset(hrarr0)lim(f(2(1+(h)/(2)))-f(2))/(h)`
`=underset(hrarr0)lim(f(2)f(1+(h)/(2))-f(2))/(h)`
`=(f(2))/(2)underset(hrarr0)lim(f(1+(h)/(2))-1)/((h)/(2))`
Replace x and y be 1 in equation (1). Then `f(1)=(f(1))^(2)`
`"or "f(1)=1`
`therefore" "f'(2)=(f(2))/(2)underset(hrarr0)lim(f(1+(h)/(2))-f(0))/((h)/(2))=(f(2)f'(1))/(2)=(3)/(2)`
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