JEE Advanced 2012 · Paper 1 · Mathematics · Continuity and Differentiability
Let ∫(x)=\ array r x^ 2 | (π)/(x)|, x ≠ 0 0, x=0 array , x ∈ R....
Let \int(x)=\left\{x2cosxπ,x=00,x=0, x \in R\right. then ∫ is
Options
Adifferentiable both at x=0 and at x=2
Bdifferentiable at x=0 but not differentiable at x=2
Cnot differentiable at x=0 but differentiable at x=2
Ddifferentiable neither at x=0 nor at x=2
Correct answer
B. differentiable at x=0 but not differentiable at x=2
Step-by-step solution
f^{\prime}\left(0^{+}\right)=\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h}=h→0limhh2coshπ=h→0limhcoshπ=0× some finite value =0=0× some finite value =0 and f′(0−)=limh→0−hf(0−h)−f(0)=limh→0−hh2∣cos−hπ∣=limh→0−hcoshπ=0× some finite value =0∵f′(0+)=f′(0−)∴f is differentiable at x=0 Now f′(2+)=limh→0hf(2+h)−f(2)=limh→0h(2+h)2∣cos2+hπ∣−4∣cos2π∣=limh→0h(2+h)2(cos2+hπ)=limh→0h(2+h)2sin(2π−2+hπ)=\lim _{h \rightarrow 0} \frac{(2+h)^{2}}{h} \sin \left(\frac{\pi h}{2(2+h)}\right)=limh→0h(2+h)2×(2(2+h)πh)sin(2(2+h)πh)×2(2+h)πh=π
and f′(2−)=limh→0−hf(2−h)−f(2)=limh→0−h(2−h)2∣cos(2−hπ)∣−0=limh→0−h−(2−h)2cos(2−hπ)=limh→0h(2−h)2sin(2π−2−hπ)=limh→0h(2−h)2×(2(2−h)−πh)sin(2(2−h)−πh)×(2(2−h)−πh)=−π∵f′(2+)=f′(2−),∴f is not differentiable at $x=2 .
