Differentiability of fg at x=1
Left-hand derivative :
(fg)′(1−)=limh→0−hfg(1−h)−fg(1)
=limh→0−h{(1−h)3−(1−h)2−hsin(1−h)}g(1−h)−0
=limh→0{(1−h)2+sin(1−h)}g(1−h).......(i)
Right-hand derivative :
(fg)′(1+)=limh→0hfg(1+h)−fg(1)
=limh→0h{(1+h)3−(1+h)2+hsin(1+h)}g(1+h)−0
=limh→0{(1+h)2+sin(1+h)}g(1+h).......(ii)
If g is continuous at x=1, then
limh→0g(1+h)=limh→0g(1−h)=g(1).........(iii)
From equations \left(i\right), \left(ii\right) \text{ and } \left(iii\right),weget</p><p>\lim_{h \to 0}{\left(fg\right)}^{'}\left({1}^{+}\right) = \lim_{h \to 0}{\left(fg\right)}^{'}\left({1}^{-}\right) = \left(1 + \sin 1\right)g\left(1\right)</p><p>∴ fgisdifferentiableatx = 1.</p><p>So,option(A) is correct.
Now, from equations \left(i\right) \text{ and } \left(ii\right),wecansaythatforfgtobedifferentiable,weneedonlyg\left(1 + h\right) = g\left(1 - h\right).</p><p>Butforgtobecontinuous,weneedg\left(1 + h\right) = g\left(1 - h\right) = g\left(1\right).</p><p>So,option(B) is incorrect.
Now, if g is differentiable at x=1 and f is already differentiable at x=1 as f′(1−)=f′(1+)=1+sin1, so product of two differentiable functions is also differentiable.
∴fg is differentiable at x=1.
So, option \left(C\right)iscorrect.</p><p>Now,fromoption(B), if fg is differentiable at x=1, we cannot guarantee g to be continuous at x=1.
So, we also cannot guarantee g to be differentiable at x=1.
So, option \left(D\right) is incorrect.
