A function`f: Rvec[1,oo)` satisfies the equation `f(x y)=f(x)f(y)-f(x)-f(y)+2.` If differentiable on `R-{0}a n df(2)=5,f^(prime)(x)=(f(x)-1)/xdotlambdat h e nlambda=` `2^(prime)f(1)` b. `3f^(prime)(1)` c. `1/2f^(prime)(1)` d. `f^(prime)(1)`

Options

A. `2f'(1)`
B. `3f'(1)`
C. `(1)/(2)f'(1)`
D. `f'(1)`

Answer

D

Explanation

`f(xy)=f(x)f(y)-f(x)-f(y)+2" (1)"`
`f'(x)=underset(hrarr0)(lim)(f(x+h)-f(x))/(h)`
`=underset(hrarr0)(lim)(f{x(1+(h)/(x))}-f(x))/(h)(x ne 0" given"")`
`=underset(hrarr0)(lim)(f(1+(h)/(x))-2)/(h).(f(x)-1)/(x)`
Putting x = 1, y = 2 in (1), we get f(1) = 2
`therefore" "f'(x)=f'(1).(f(x)-1)/(x)`