Question

If $f\left( x+y \right)=2f\left( x \right)f\left( y \right)$ , $f'\left( 5 \right)=1024\left( \log 2 \right)$ , $f\left( 2 \right)=8$ , then the value of $f'\left( 3 \right)$ is:

1. $64\left( \log 2 \right)$
2. $128\left( \log 2 \right)$
3. $256$
4. $256\left( \log 2 \right)$

Answer 1

Here in this question we have been asked to find the value of $f'\left( 3 \right)$ given that$f\left( x+y \right)=2f\left( x \right)f\left( y \right)$ , $f'\left( 5 \right)=1024\left( \log 2 \right)$ and $f\left( 2 \right)=8$. For answering this question we will use the concept given as $f'\left( x \right)=\displaystyle \lim_{h \to 0}\dfrac{f\left( x+h \right)-f\left( x \right)}{h}$ and simplify using the given information.

Complete step-by-step solution:
Now considering from the question we have been asked to find the value of $f'\left( 3 \right)$ given that$f\left( x+y \right)=2f\left( x \right)f\left( y \right)$ , $f'\left( 5 \right)=1024\left( \log 2 \right)$ and $f\left( 2 \right)=8$.
From the basic concepts of derivations, we know that we can say that $f'\left( x \right)=\displaystyle \lim_{h \to 0}\dfrac{f\left( x+h \right)-f\left( x \right)}{h}$ we will use this to answer the question.
Here in this question as we know that $f\left( x+y \right)=2f\left( x \right)f\left( y \right)$ by using this in the above expression we can conclude that
$\begin{aligned} & f'\left( x \right)=\displaystyle \lim_{h \to 0}\dfrac{f\left( x+h \right)-f\left( x \right)}{h} \\\ & \Rightarrow f'\left( x \right)=\displaystyle \lim_{h \to 0}\dfrac{2f\left( x \right)f\left( h \right)-f\left( x \right)}{h} \\\ & \Rightarrow f'\left( x \right)=f\left( x \right)\displaystyle \lim_{h \to 0}\dfrac{2f\left( h \right)-1}{h} \\\ \end{aligned}$ .
As we need the value of $f'\left( 3 \right)$ we can say that $f'\left( 3 \right)=\displaystyle \lim_{h \to 0}\dfrac{f\left( 3+h \right)-f\left( 3 \right)}{h}$ .
Now as we know that $f\left( x+y \right)=2f\left( x \right)f\left( y \right)$ by using this in the above expression we will have $\Rightarrow f'\left( 3 \right)=\displaystyle \lim_{h \to 0}\dfrac{2f\left( 3 \right)f\left( h \right)-f\left( 3 \right)}{h}$ .
Now we will take out $f\left( 3 \right)$ as it is a constant. By doing that we will have $\Rightarrow f'\left( 3 \right)=f\left( 3 \right)\displaystyle \lim_{h \to 0}\dfrac{2f\left( h \right)-1}{h}$ .
Now we will try to evaluate $f'\left( 5 \right)$ by doing that we will have $\begin{aligned} & \Rightarrow f'\left( 5 \right)=f\left( 5 \right)\displaystyle \lim_{h \to 0}\dfrac{2f\left( h \right)-1}{h} \\\ & \Rightarrow 1024\left( \log 2 \right)=2f\left( 3 \right)f\left( 2 \right)\left[ \displaystyle \lim_{h \to 0}\dfrac{2f\left( h \right)-1}{h} \right] \\\ \end{aligned}$.
Now by using $f\left( 2 \right)=8$ in the above expression we will have $\begin{aligned} & \Rightarrow 1024\left( \log 2 \right)=16f\left( 3 \right)\left[ \displaystyle \lim_{h \to 0}\dfrac{2f\left( h \right)-1}{h} \right] \\\ & \Rightarrow 64\left( \log 2 \right)=f\left( 3 \right)\left[ \displaystyle \lim_{h \to 0}\dfrac{2f\left( h \right)-1}{h} \right] \\\ \end{aligned}$ .
Therefore we can conclude that the value of $f'\left( 3 \right)$ is given as $64\left( \log 2 \right)$ given that$f\left( x+y \right)=2f\left( x \right)f\left( y \right)$ , $f'\left( 5 \right)=1024\left( \log 2 \right)$ and $f\left( 2 \right)=8$.
Hence we will mark the option “1” as correct.

Note: While answering questions of this type we just need to clearly interpret the given information in the correct place. Here by simply analyzing the given information in a correct method will strive for the answer. Very few mistakes are possible.