Q55 / 125 · BITSAT 2012 · Mathematics
Let $f(x+y)=f(x) \cdot f(y)$ for all $x, y$ where $f(0) \neq 0$ If $f(5)=2$ and $f^{\prime}(0)=3,$ then $f^{\prime}(5)$ is equal to-
- Correct answerA6
- B0
- C1
- DNone of these
Solution
$f^{\prime}(5)=\lim _{h \rightarrow 0} \frac{f(5+h)-f(5)}{h}$
$=\lim _{h \rightarrow 0} \frac{f(5+h)-f(5+0)}{h}$
$=\lim _{h \rightarrow 0} \frac{f(5) \cdot f(h)-f(5)+f(0)}{h}$
$(\because f(x+y)=f(x), f(y)$ for all $x, y)$
$=\left(\lim _{h \rightarrow 0} \frac{f(h)-f(0)}{h}\right) \cdot f(5)=f^{\prime}(0) \cdot f(5)$
$=3 \times 2=6$
Hence, the correct answer is option: 'A'.