Answer:
Option B
Explanation:
Concept Involved To check differentiability at a point we use RHD and LHD at a point and if RHD = LHD, then f(x) is differentiable at the point.
Description of Situation
As, R
{f′(x)}=limh→0f(x+h)−f(x)h
and L{f′(x)}=limh→0f(x−h)−f(x)−h
Here, students generaliy gets confused in defining modulus
Sol. To check differentiable at x=0
R{f′(0)}=limh→0f(0+h)−f(0)h
=limh→0h2|cosπh|−0h
=limh→0h.|cosπh|=0
L{f′(0)}=limh→0f(0−h)−f(0)−h
=\lim_{h \rightarrow 0}\frac{h^{2}|\cos \left(-\frac{\pi}{h}\right)|-0}{-h}}
=0
so, f(x) is differentiable at x=0
To check differentiability at x=2
R{f′(2)}=limh→0f(2+h)−f(2)h
=limh→0(2+h)2|cos(π2+h)|−0h
=limh→0(2+h)2sin(π2−π2+h)h
=limh→0(2+h)2sin(πh2(2+h))h.π2(2+h)
.π2(2+h)
⇒ R(f′(2))=π
L(f′(2))=limh→0f(2−h)−f(2)−h
=limh→0(2−h)2.|cosπ2−h|−22.|cosπ2|−h
=limh→0(2−h)2−(−cosπ2−h)−0−h
=limh→0−(2−h)2.sin(π2−π2−h)h
limh→0(2−h)2.sin(−πh2(2−h))h×−π2(2−h)
×−π2(2−h)
⇒ L.F′(2))=−π
Thus, f(x) is differentiable at x=0 but not at x=2
