Question

If f(x+y) = 2f(x)f(y) where f’(0) = 3 and f(4) = 2, then f’(4) equals
[a] 6
[b] 12
[c] 4
[d] 3

Answer 1

Hint: Use the fact that $f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( x+h \right)-f\left( x \right)}{h}$. Using the functional equation write $f\left( x+h \right)$ as $2f\left( x \right)f\left( h \right)$. Put this value in the formula for $f'\left( x \right)$. Also, substitute x = 0 in the formula to get the relation for $f'\left( 0 \right)$. Using this find the equation for $f'\left( x \right)$and hence form a differential equation. Solve the differential equation for f(x) and hence find $f'\left( 4 \right)$
Alternatively put x = 0 and x = 4 alternately in the formula for $f'\left( x \right)$ and solve the system of equations.

Complete step-by-step solution -
We have $f\left( x+y \right)=2f\left( x \right)f\left( y \right)$
Put x = y = 0, we get
$f\left( 0 \right)=2f{{\left( 0 \right)}^{2}}$
Hence (0) = 0 or f(0) $=\dfrac{1}{2}$
If f(0) = 0
Put x = x and y = 0, we get
f(x+0) = 2f(x)f(0) = 0
Since f’(0) = 3, f(0) =0 is rejected.
Hence f(0) $=\dfrac{1}{2}$
Hence we have $f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( x+h \right)-f\left( x \right)}{h}$
Using $f\left( x+y \right)=2f\left( x \right)f\left( y \right)$, we get
$f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{2f\left( x \right)f\left( h \right)-f\left( x \right)}{h}=f\left( x \right)\underset{h\to 0}{\mathop{\lim }}\,\dfrac{2f\left( h \right)-1}{h}$
Hence we have
$f'\left( x \right)=2f\left( x \right)\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( h \right)-f\left( 0 \right)}{h}=2f\left( x \right)f'\left( 0 \right)$
Since $f'\left( 0 \right)=3$, we have
$f'\left( x \right)=6f\left( x \right)$
Dividing both sides by f(x) and integrating we get
$\begin{align}
  & \int{\dfrac{d\left( f\left( x \right) \right)}{f\left( x \right)}}=\int{6dx} \\
 & \Rightarrow \ln \left( f\left( x \right) \right)=6x+C \\
 & \Rightarrow f\left( x \right)=A{{e}^{6x}} \\
\end{align}$
f(0) $=\dfrac{1}{2}$, we have
$A=\dfrac{1}{2}$
But f(4) = 2,
Hence this function does not exist.
If such a function existed, we have $f'\left( 4 \right)=2f\left( 4 \right)f'\left( 0 \right)=2\times 2\times 3=12$
Hence option [b] is correct.

Note: We can derive the above equation directly.
Differentiating w.r.t x keeping y as constant, we get
$f'\left( x+y \right)=2f'\left( x \right)f\left( y \right)$
Put x = 0, we get
$f'\left( y \right)=2f'\left( 0 \right)f\left( y \right)$, which is the same equation as obtained above.
Although mentioned above, such a function does not exist.