For differentiability,
L.H.D. = R.H.D.
Thus at
x=0 L.H.D. L.H.D. = R.H.D. =h→0lim−hf(0−h)−f(0)=h→0lim−hf(−h)−f(0)=h→0lim−h(−h+π)−(πcos0)=11=h→0limhf(0+h)−f(0)=h→0limhf(h)−f(0)=h→0limhπcosh−πcos0=h→0limhπ[cosh−1]$\begin{aligned} & =\lim _{h \rightarrow 0} \frac{\pi\left[1-\frac{h^2}{2!}+\frac{h^4}{4!} \ldots \ldots-1\right]}{h} \\ & =\lim _{h \rightarrow 0} \frac{\pi\left[1-\frac{1}{2} h^2+\frac{1}{24} h^4 \ldots..-1\right]}{h} \\ & =\lim _{h \rightarrow 0} \frac{\pi\left[-\frac{1}{2} h^2+\frac{1}{24} h^4 \ldots \ldots\right]}{h}=0\end{aligned}\text{ L.H.D. } \neq \text{ R.H.D. }
Soatx=0
functionisnotdifferentiable.Statement(1)isnotcorrect.Atx=\frac{\pi}{2}
f(2π) R.H.D. =πcos2π=0=h→0lim−hf(2π+h)−f(2π)=h→0limh(2π+h−2π)2−0=h→0limhh2=0 L.H.D. =h→0lim−hf(2π−h)−f(2π)=h→0lim−hπcos(2π−h)−0=h→0limπ(−hsinh)=−πh→0lim(hsinh)=−π(1)=−π L.H.D. = R.H.D.
Hence function is not differentiable at
x=2π. Statement (2) is not correct.
