Let $f(x)=\begin{cases}x^2 \sin \left(\frac{1}{x}\right) & , x \neq 0 \\ 0 & , x=0\end{cases}$Then at $x=0$
- $f$ is continuous but not differentiable
- $f$ is continuous but $f^{\prime}$ is not continuous
- $f^{\prime}$ is continuous but not differentiable
- $f$ and $f^{\prime}$ both are continuous
The Correct Option is B
Solution and Explanation
is continuous
is differentiable.
is not continuous (as is highly oscillating at )
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Concepts Used:
Continuity
A function is said to be continuous at a point x = a, if
limx→a
f(x) Exists, and
limx→a
f(x) = f(a)
It implies that if the left hand limit (L.H.L), right hand limit (R.H.L) and the value of the function at x=a exists and these parameters are equal to each other, then the function f is said to be continuous at x=a.
If the function is undefined or does not exist, then we say that the function is discontinuous.
Conditions for continuity of a function: For any function to be continuous, it must meet the following conditions:
- The function f(x) specified at x = a, is continuous only if f(a) belongs to real number.
- The limit of the function as x approaches a, exists.
- The limit of the function as x approaches a, must be equal to the function value at x = a.