Question

If $f(x+y)=f\left(x\right)f\left(y\right)\forall x,y$ and $f\left(5\right)=2,f^{\prime }\left(0\right)=3$; then $f^{\prime }\left(5\right)$ is equal to-

• A. $2$
• B. $4$
• C. $6$
• D. $8$

Answer 1

The correct option is C $6$

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

Also, $f\left(5\right)=2,f^{\prime }\left(0\right)=3$

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

Putting $x,y=0,$ we get

$f\left(0\right)=0$ or $1.$

$f\left(0\right)=1$ not possible.

So, $f\left(0\right)=0$

Now, differentiating both sides w.r.t. x, we have $f^{\prime }(x+y)=f^{\prime }\left(x\right)f\left(y\right)+f\left(x\right)f^{\prime }\left(y\right)$

Thus, substituting $x=0$ and $y=5,f^{\prime }\left(5\right)=f^{\prime }\left(0\right)f\left(5\right)+f\left(0\right)f^{\prime }\left(5\right)$

$\Rightarrow f^{\prime }\left(5\right)=3\times 2=6$