Let `f(x)={x^2|(cos)pi/x|, x!=0 and 0,x=0,x in RR,` then `f` isA. differentiable both at x=0 and x=2B. differentiable at x=0 but not differentiable at x=2C. not differentiable at x=0 but differentiable at x=2D. differentiable neither at x=0 nor at x=2

Let `f(x)={x^2|(cos)pi/x|, x!=0 and 0,x=0,x in RR,` then `f` isA. diff

1.	Let `f(x)={x^2\|(cos)pi/x\|, x!=0 and 0,x=0,x in RR,` then `f` isA. differentiable both at x=0 and x=2B. differentiable at x=0 but not differentiable at x=2C. not differentiable at x=0 but differentiable at x=2D. differentiable neither at x=0 nor at x=2
Answer» Correct Answer - B 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