Question

Let $f\left(x+y\right) =f\left(x\right)f\left(y\right) f\left(x\right) = 1+\left(\sin2x\right)g\left(x\right)$ where $ g\left(x\right) $ is continuous. Then $f'\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)

$f'\left(x\right) =Lt_{h\to0} \frac{f\left(x+h\right)-f\left(x\right)}{h}$
$ = Lt_{h\to0} \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) .Lt_{h\to0} \frac{f\left(h\right)-1}{h} $
$= f\left(x\right) Lt_{h\to0} \frac{1+ \left(\sin2h\right)g\left(h\right)-1}{h}$
$ = 2f\left(x\right) Lt_{h \to0} \frac{\sin2h}{2h} .Lt_{h\to0} g\left(h\right)$
$ =2f\left(x\right).g\left(0\right)$
$ \left[ \left(\because g\left(x\right) \, \text{is} \, \, \, \text{continuous} \, \text{at} \, x = 0 \right) \\ \therefore Lt_{x\to0} g\left(x\right) = g\left(0\right) i.e., Lt_{h \to0} g\left(h\right) =g\left(0\right)\right] $