Question

A function $f:R\to [1,\infty )$ satisfies the equation $f\left(xy\right)=f\left(x\right)f\left(y\right)-f\left(x\right)-f\left(y\right)+2$. If f is differentiable on $R-0andf\left(2\right)=5,f^{\prime }\left(x\right)=\frac{f\left(x\right)-1}{x}.\lambda then\lambda$ =________

• A.
• B.
• C.
• D.

Answer 1

The correct option is D

$f^{\prime }\left(x\right)=\stackrel{It}{x\to 0}\frac{f(x+h)-f\left(x\right)}{h}=\stackrel{It}{x\to 0}\frac{f\left\{x(1+\frac{h}{x})\right\}-f\left(x\right)}{h}(x\ne 0given)=\stackrel{It}{x\to 0}\frac{f(1+\frac{h}{x})-2}{\frac{h}{x}}.\frac{f\left(x\right)-1}{x}$
Putting x=1, y=2 in the given functional equation, f(1)=2
$\therefore f^{\prime }\left(x\right)=\frac{f\left(x\right)-1}{x}.\stackrel{It}{x\to 0}\frac{f(1+\frac{h}{x})-f\left(1\right)}{\frac{h}{x}}=\frac{f\left(x\right)-1}{x}.f^{\prime }\left(1\right)$