Question

Let $f\left(xy\right)=f\left(x\right)\cdot f\left(y\right)$ for all $x,y\in R$. If $f^{{}^{\prime }}\left(1\right)=2$ and $f\left(4\right)=4$, then $f^{{}^{\prime }}\left(4\right)$ equal to

• A. $4$
• B. $1$
• C. $\frac{1}{2}$
• D. $8$

Answer 1

The correct option is D $8$

We have $f(x,y)=f\left(x\right)f\left(y\right)$ for all $x,y\in R$.
Putting $x=y=1$, we get
$f\left(1\right)=f\left(1\right)f\left(1\right)$
$\Rightarrow f\left(1\right)[1-f\left(1\right)]=0$
$\Rightarrow f\left(1\right)=1$ $[\because f\left(1\right)\ne 0]$
Now, $f^{{}^{\prime }}\left(1\right)=2$
$\Rightarrow \underset{h\to 0}{lim}\frac{f(1+h)-f\left(1\right)}{h}=2$
$\Rightarrow f\left(1\right)\underset{h\to 0}{lim}\frac{f\left(h\right)-1}{h}=2$
$\Rightarrow \underset{h\to 0}{lim}\frac{f\left(1\right)f\left(h\right)-f\left(1\right)}{h}=2$ [using $f\left(1\right)=1$]
$\Rightarrow \underset{h\to 0}{lim}\frac{f\left(h\right)-1}{h}=2$ ....(i)
Now, $f^{{}^{\prime }}\left(4\right)=\underset{h\to 0}{lim}\frac{f(4+h)-f\left(4\right)}{h}$
$=\underset{h\to 0}{lim}\frac{f\left(4\right)\cdot f\left(h\right)-f\left(4\right)}{h}$
$=\left\{\underset{h\to 0}{lim}\frac{f\left(h\right)-1}{h}\right\}\cdot f\left(4\right)=2f\left(4\right)$ ....{from (i)}
$=2\times 4=8$