The function f(x) = cases ( 2x)/x + x &, if \ x ≠ 0 \\ K &, if \ x = 0 cases is continuous at x = 0, then the value of K is: Option 1: 1 Option 2: 2 Option 3: 0 Option 4: 3

Answer: 3. Solution: For the function to be continuous at x = 0: _(x 0) f(x) = f(0) = K Finding the limit as x approaches 0: _(x 0) (( 2x)/x + x) = _(x 0) ( 2x)/x + _(x 0) x For _(x 0) ( 2x)/x: ( 2x)/x = (2 2x)/2x = 2 * ( 2x)/2x Using _(u 0) ( u)/u = 1 where u = 2x: _(x 0) ( 2x)/x = 2 x 1 = 2 For _(x 0) x: _(x 0) x = (0) = 1 Combining the results: _(x 0) f(x) = 2 + 1 = 3 Since the function is continuous at x = 0: K = 3

Calculus

Continuity & Differentiability

Medium

core

The function $f(x) = \begin{cases} \frac{(\sin 2x)}{x} + \cos x & , if \ x \neq 0 \\ K & , if \ x = 0 \end{cases}$ is continuous at $x = 0$ , then the value of K is:

✅ Correct Option: 4

For the function to be continuous at $x = 0$ :

$\lim_{x \to 0} f(x) = f(0) = K$

Finding the limit as $x$ approaches $0$ :

$\lim_{x \to 0} \left(\frac{\sin 2x}{x} + \cos x\right)$

$= \lim_{x \to 0} \frac{\sin 2x}{x} + \lim_{x \to 0} \cos x$

For $\lim_{x \to 0} \frac{\sin 2x}{x}$ :

$\frac{\sin 2x}{x} = \frac{2 \sin 2x}{2x}$

$= 2 \cdot \frac{\sin 2x}{2x}$

Using $\lim_{u \to 0} \frac{\sin u}{u} = 1$ where $u = 2x$ :

$\lim_{x \to 0} \frac{\sin 2x}{x} = 2 \times 1$

$= 2$

For $\lim_{x \to 0} \cos x$ :

$\lim_{x \to 0} \cos x = \cos(0)$

$= 1$

Combining the results:

$\lim_{x \to 0} f(x) = 2 + 1$

$= 3$

Since the function is continuous at $x = 0$ :

$K = 3$

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The function $f(x) = e^{|x|}$ is

(a) continuous everywhere on $R$

(b) not continuous at $x = 0$

(d) not differentiable at $x = 0$

(e) continuous and differentiable on $R$

Choose the most appropriate answer from the options given below :

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Match List-I with List-II

List-I	List-II
Function f(x)	Points of Non-Differentiability
---	---
(A) $f(x) = \vert x\vert + 1$	(I) Not differentiable at $x = 3$ only
(B) $f(x) = \vert x - 3\vert$	(II) Not differentiable at $x = -3$ only
(C) $f(x) = \vert x + 3\vert$	(III) Not differentiable at $x = 3, -3$ only
(D) $f(x) = \vert x^2 - 9\vert$	(IV) Not differentiable at $x = 0$ only

Choose the correct answer from the options given below:

The function f(x)={(sin⁡2x)x+cos⁡x,if x≠0K,if x=0f(x) = \begin{cases} \frac{(\sin 2x)}{x} + \cos x & , if \ x \neq 0 \\ K & , if \ x = 0 \end{cases}f(x)={x(sin2x)​+cosxK​,if x=0,if x=0​ is continuous at x=0x = 0x=0, then the value of K is:

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CUET Mathematics Past Year Questions (Topic-Wise)

The function $f(x) = \begin{cases} \frac{(\sin 2x)}{x} + \cos x & , if \ x \neq 0 \\ K & , if \ x = 0 \end{cases}$ is continuous at $x = 0$ , then the value of K is: