Question

If $f\left(x+y\right)=f\left(x\right)\times f\left(y\right)\forall x,y\in R$ and $f\left(5\right)=2,f^{\prime }\left(0\right)=3,$ then $f^{\prime }\left(5\right)$ is

• A. $0$
• B. $1$
• C. $6$
• D. $2$

Answer 1

The correct option is A $0$

$f(x+y)=f\left(x\right)f\left(y\right)$

$f(5+0)=f\left(5\right)f\left(0\right)$

$f\left(0\right)=\frac{2}{2}=1$

$f^{\prime }(x+y)=f^{\prime }\left(x\right)f\left(y\right)+f\left(x\right)f^{\prime }\left(y\right)$

$f^{\prime }(0+5)=f^{\prime }\left(0\right)f\left(5\right)+f\left(0\right)f^{\prime }\left(5\right)$

$f^{\prime }\left(5\right)=3\times 2+f^{\prime }\left(5\right)$

$f^{\prime }\left(5\right)=0$