Let `f(x) = {{:(x^(2)|"cos"(pi)/(x)|",",x ne 0","x in R),(0",",x = 0):}`, then f is

Options

A. differentiable both at x=0 and x=2
B. differentiable at x=0 but not differentiable at x=2
C. not differentiable at x=0 but differentiable at x=2
D. differentiable neither at x=0 nor at x=2

Answer

B

Explanation

We observe that
`underset(x to 0)lim (f(x)-f(0))/(x-0)=underset(x to 0)lim (x^(2)|"cos "(pi)/(x)|-0)/(x-0)=underset(x to 0)lim x|"cos "(pi)/(2)|=0`
So, f(x) is differentiable at x=0
Now,
`underset(x to 2^(-))lim (f(x)-f(2))/(x-2)`
`underset(h to 0 )lim (f(x)-f(2))/((2-h)-2)`
`underset(h to 0 )lim ((2-h)^(2)|cos((pi)/(2-h)))/(-h)`
`underset(h to 0 )lim ((2-h)^(2)sin((pi)/(2)-(pi)/(2-h)))/(h)a`
`underset(h to 0 )lim ((2-h)^(2))/(h)sin{(-pih)/(2(2-h))}`
`underset(h to 0 )lim (sin{(pih)/(2(2-h))})/((pih)/(2(2-h)))=pi` and `underset(x to 2^(+))lim=(f(x)-f(2))/(x-2)`
`underset(x to 2^(+))lim=(f(2+h)-f(2))/((2+h)-2)`
`underset(x to 2^(+))lim ((2+h)^(2)cos((pi)/(2+h))-0)/(h)`
`underset(x to 2^(+))lim ((2+h)^(2))/(h)sin((pi)/(2)-(pi)/(2+h))`
`underset(x to 2^(+))lim (2+h)^(2)/(h) sin {(pih)/(2(2+h))}`
`=underset(h to 0)lim (sin{(pih)/(2(2+h))})/((pi)/(2(2+h)))xx(pi)/(2)(2+h)=(pi)/(2)xx2=pi`
`therefore underset(x to 2^(-))lim (f(x)-f(2))/(x-2) ne underset(x to 2^(+))lim (f(x)-f(2))/(x-2)`
So, f(x) is not differentiable at x=2