Question

Assertion :Let $f(x+y)=f\left(x\right)f\left(y\right)$ for all $x,y$ where $f\left(0\right)\ne 0$. If $f^{\prime }\left(0\right)=2$ then $f\left(x\right)=A{e}^{2x}$, where $A$ is a constant Reason: $f^{\prime }\left(x\right)=f\left(x\right)$

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Both Assertion and Reason are incorrect

Answer 1

The correct option is C Assertion is correct but Reason is incorrect

$f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{f(x+h)-f\left(x\right)}{h}$

$=\underset{h\to 0}{lim}\frac{f(x+h)-f(x+0)}{h}$

$=\underset{h\to 0}{lim}\frac{f\left(x\right).f\left(h\right)-f\left(x\right).f\left(0\right)}{h}$

$=\underset{h\to 0}{lim}\frac{f\left(h\right)-f\left(0\right)}{h}.f\left(x\right)$

$=f^{\prime }\left(0\right).f\left(x\right)=2f\left(x\right).(\because f^{\prime }\left(0\right)=2)$

Now, $\frac{df}{dx}=2f$ or $\frac{df}{f}=2dx\Rightarrow d\left(\log f-2x\right)=0$

$\therefore \log f-2x=c,\therefore f=e^{2x+c}=e^c.e^{2x}=A{e}^{2x},$

where $A=e^c=$ constant.