Question

Let $f\left(x+y\right)=f\left(x\right)f\left(y\right)f\left(x\right)=1+\left(\sin 2x\right)g\left(x\right)$ where $g\left(x\right)$ is continuous. Then $f^{\prime }\left(x\right)$ is equal to

• A. f(x) g(0)
• B. 2 f(x) g(0)
• C. 2 g(0)
• D. none of these.

Answer 1

Answer: (B)

$f^{\prime }\left(x\right)=L{t}_{h\to 0}\frac{f\left(x+h\right)-f\left(x\right)}{h}$
$=L{t}_{h\to 0}\frac{f\left(x\right).f\left(h\right)-f\left(x\right)}{h}$
$\left(\because f\left(x+y\right)=f\left(x\right)f\left(y\right)\right)$
$=f\left(x\right).L{t}_{h\to 0}\frac{f\left(h\right)-1}{h}$
$=f\left(x\right)L{t}_{h\to 0}\frac{1+\left(\sin 2h\right)g\left(h\right)-1}{h}$
$=2f\left(x\right)L{t}_{h\to 0}\frac{\sin 2h}{2h}.L{t}_{h\to 0}g\left(h\right)$
$=2f\left(x\right).g\left(0\right)$
$$\begin{gathered} \left[\left(\because g\left(x\right)\text{is}\text{continuous}\text{at}x=0\right) \\ \therefore L{t}_{x\to 0}g\left(x\right)=g\left(0\right)i.e.,L{t}_{h\to 0}g\left(h\right)=g\left(0\right)\right] \end{gathered}$$