Question

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

• A. $64\left(\log 2\right)$
• B. $128\left(\log 2\right)$
• C. $256$
• D. $256\left(\log 2\right)$

Answer 1

The correct option is A

$64\left(\log 2\right)$

Explanation for the correct option:

Step 1: Find the value of $f\text{'}\left(5\right)$:

Given that,

$f\left(x+y\right)=2f\left(x\right)f\left(y\right)$,

Put $x=0=y$

$$\begin{gathered} f\left(0\right)=2f^2\left(0\right) \\ \Rightarrow f\left(0\right)=\frac{1}{2} \end{gathered}$$

Substitute $x=2,y=3$ in the above function, we get

$f\left(5\right)=2f\left(2\right)f\left(3\right)...\left(1\right)$

By using the derivative of $f\left(5\right)$,

$\begin{array}{rcl}f\text{'}\left(5\right)& =& \underset{h\to 0}{\lim }\frac{f\left(5+h\right)-f\left(5\right)}{h}\\ & =& \underset{h\to 0}{\lim }\frac{2f\left(5\right)f\left(h\right)-f\left(5\right)}{h}\\ & =& 2f\left(5\right)\underset{h\to 0}{\lim }\frac{\left(f\left(h\right)-{\displaystyle \frac{1}{2}}\right)}{h}\end{array}$

Step 2: Find the value of $f\left(3\right)f\text{'}\left(0\right)$:

It is given that $f\text{'}\left(5\right)=1024\left(\log 2\right)$, so

$$\begin{gathered} 1024\left(\log 2\right)=2f\left(5\right)f\text{'}\left(0\right)\mathbf{[}\mathbf{\because }\mathbf{}\underset{\mathbf{h}\mathbf{\to }\mathbf{0}}{\mathbf{f}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{l}\mathbf{i}\mathbf{m}}\frac{\mathbf{f}\mathbf{\left(}\mathbf{h}\mathbf{\right)}\mathbf{-}{\displaystyle \frac{1}{2}}}{\mathbf{h}}\mathbf{]} \\ \Rightarrow 1024\left(\log 2\right)=2\times 2f\left(2\right)f\left(3\right)f\text{'}\left(0\right)\mathbf{[}\mathbf{from}\mathbf{}\mathbf{1}\mathbf{]} \\ \Rightarrow f\left(3\right)f\text{'}\left(0\right)=\frac{1024\left(\log 2\right)}{4\times 8} \\ =32\log 2...\left(2\right) \end{gathered}$$

Step 3: Find the value of $f\text{'}\left(3\right)$:

$\begin{array}{rcl}f\text{'}\left(3\right)& =& \underset{h\to 0}{\lim }\frac{f\left(3+h\right)-f\left(3\right)}{h}\\ & =& \underset{h\to 0}{\lim }\frac{2f\left(3\right)f\left(h\right)-f\left(3\right)}{h}\\ & =& 2f\left(3\right)\underset{h\to 0}{\lim }\frac{\left[f\left(h\right)-{\displaystyle \frac{1}{2}}\right]}{h}\\ & =& 2f\left(3\right)f\text{'}\left(0\right)\mathbf{[}\mathbf{\because }\mathbf{}\underset{\mathbf{h}\mathbf{\to }\mathbf{0}}{\mathbf{lim}}\frac{\left[f\left(h\right)-{\displaystyle \frac{1}{2}}\right]}{\mathbf{h}}\mathbf{]}\\ & =& 2\left(32\log 2\right)\left[\mathrm{from}2\right]\\ & =& 64\log 2\end{array}$

Hence, the correct option is A.