Let f: R → R be a function defined by \f(x) = \{ \table {\sin(x^2)}/x , \if x \neq0;0, \if x= 0 . Then, at x = 0, f is

Correct Answer is : differentiable and the derivative is continuous

continuous but not differentiable

differentiable and the derivative is not continuous

differentiable and the derivative is continuous

KVPY 2019 SB SX Exam Previous Papers

Question : 20 of 120

Marks: +1, -0

Let f: R → R be a function defined by

\f(x) = \{ \table {\sin(x^2)}/x , \if x ≠0;0, \if x= 0

. Then, at x = 0, f is

Solution:

\lim↙{h→0}{\f(h) – \f(0)}/h=\lim↙{h→0}{{\sinh^2}/h-0}/h

= 1 = \f '(0)

⇒ f(x) is differentiable at x = 0
for x ≠ 0, f'(x) =

{2x \cosx^2 -\sinx^2}/x^2

=2\cos(x^2) -{\sin(x^2)}/x^2

⇒

\f(x) = \{ \table 1{,},x=0;2\cos x^2-{\sin x^2 }/x^2 {,} , x ≠0

which is continuous

Go to Question:

Prev Question

Next Question