Q105 / 200 · Manipal (MET) 2013 · Mathematics
Let $f(x y)=f(x) \cdot f(y)$ for all $x, y \in R$. If $f^{\prime}(1)=2$ and $f(4)=4$, then $f^{\prime}(4)$ equal to
- A$4$
- B$1$
- C$\frac{1}{2}$
- Correct answerD$8$
Solution
We have $f(x y)=f(x) f(y)$ for all $x, y \in R$.
Putting $x=y=1$, we get
$f(1)=f(1) f(1)$
$\Rightarrow f(1)[1-f(1)]=0$
$\Rightarrow \quad f(1)=1 \quad[\because f(1) \neq 0] \quad \ldots(i)$
Now, $f^{\prime}(1)=2$
$\Rightarrow \quad \lim _{h \rightarrow 0} \frac{f(1+h)-f(1)}{h}=2$
$\Rightarrow \quad f(1) \lim _{h \rightarrow 0} \frac{f(h)-1}{h}=2$
$\Rightarrow \quad \lim _{h \rightarrow 0} \frac{f(1) f(h)-f(1)}{h}=2 \quad[$ using $f(1)=1]$ ...(ii)
$\Rightarrow \quad \lim _{h \rightarrow 0} \frac{f(h)-1}{h}=2$
Now $f^{\prime}(4)=\lim _{h \rightarrow 0} \frac{f(4+h)-f(4)}{h}$
$=\lim _{h \rightarrow 0} \frac{f(4) \cdot f(h)-f(4)}{h}$
$=\left\{\lim _{h \rightarrow 0} \frac{f(h)-1}{h}\right\} \cdot f(4)=2 f(4) \quad\{$ from $(i)\}$
$=2 \times 4=8$
Hence, the correct answer is option: 'D'.