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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- Assertion (A): $f(x) = \begin{cases} x \sin \frac{1}{x}, & x \neq 0 \\ 0, & x = 0 \end{cases}$ is continuous at $x = 0$.
Reason (R): When $x \to 0$, $\sin \frac{1}{x}$ is a finite value between -1 and 1. - If $f(x) = \begin{cases} ax + b, & x \leq 3 \\ 7, & 5<x \end{cases}$ is continuous in $\mathbb{R}$, then the values of $a$ and $b$ are:
- If $f(x) = \left\{ \begin{array}{ll} \frac{1 - \sin^3 x}{3 \cos^2 x} & \text{for} \, x \neq \frac{\pi}{2}, \\ k & \text{for} \, x = \frac{\pi}{2}, \end{array} \right. $ is continuous at $x = \frac{\pi}{2}$, then the value of $k$ is:
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In the given circuit the sliding contact is pulled outwards such that the electric current in the circuit changes at the rate of 8 A/s. At an instant when R is 12 Ω, the value of the current in the circuit will be A.- JEE Main - 2025
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- Let \( \vec{a} = \hat{i} + 2\hat{j} + \hat{k} \) and \( \vec{b} = 2\hat{i} + 7\hat{j} + 3\hat{k} \). Let \( L_1 : \vec{r} = (-\hat{i} + 2\hat{j} + \hat{k}) + \lambda \vec{a}, \lambda \in {R} \) and \( L_2 : \vec{r} = (\hat{j} + \hat{k}) + \mu \vec{b}, \mu \in \mathbb{R} \) be two lines. If the line \( L_3 \) passes through the point of intersection of \( L_1 \) and \( L_2 \), and is parallel to \( \vec{a} + \vec{b} \), then \( L_3 \) passes through the point:
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Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is:
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For $ \alpha, \beta, \gamma \in \mathbb{R} $, if $$ \lim_{x \to 0} \frac{x^2 \sin \alpha x + (\gamma - 1)e^{x^2} - 3}{\sin 2x - \beta x} = 3, $$ then $ \beta + \gamma - \alpha $ is equal to:
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.