Match the following: List-IList-IIA. frac{x}{ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Step $1$: Analyze $A$. Let $f(x) = \frac{x}{e^x-1} + \frac{x}{2} + 4$. Check $f(-x) = \frac{-x}{e^{-x}-1} + \frac{-x}{2} + 4 = \frac{-x e^x}{1-e^x

Vedclass · Accepted Answer

$A-II, B-III, C-IV, D-I$

Vedclass · Answer

(C) Step $1$: Analyze $A$. Let $f(x) = \frac{x}{e^x-1} + \frac{x}{2} + 4$. Check $f(-x) = \frac{-x}{e^{-x}-1} + \frac{-x}{2} + 4 = \frac{-x e^x}{1-e^x} - \frac{x}{2} + 4 = \frac{x e^x}{e^x-1} - \frac{x}{2} + 4$. Since $\frac{e^x}{e^x-1} = 1 + \frac{1}{e^x-1}$,$f(-x) = x(1 + \frac{1}{e^x-1}) - \frac{x}{2} + 4 = x + \frac{x}{e^x-1} - \frac{x}{2} + 4 = \frac{x}{e^x-1} + \frac{x}{2} + 4 = f(x)$. Thus,$A$ is an even function $(II)$.Step $2$: Analyze $B$. Let $f(x) = 	an^{-1}(\log|x+\sqrt{x^2+1}|)$. Since $x > 0$,the domain is restricted. However,considering the nature of the expression,it is neither odd nor even $(I)$.Step $3$: Analyze $C$. For $3 Step $4$: Analyze $D$. $f(x) = \sin 2x + \sin^2 x + \cos 3x$. This is a combination of trigonometric functions and is neither odd nor even $(I)$.Wait,re-evaluating $A$: $f(x) = \frac{x}{e^x-1} + \frac{x}{2} + 4$. This is a known even function. $C$ is identity. $B$ is neither. $D$ is neither. Checking options,$A-II$ is correct. $C-IV$ is correct. Thus,option $C$ is the correct match.

List-$I$	List-$II$
$A$. $\frac{x}{e^x-1} + \frac{x}{2} + 4; x \neq 0$	$I$. is neither odd nor even function
$B$. $\tan^{-1}(\log\|x+\sqrt{x^2+1}\|), x > 0$	$II$. is an even function
$C$. For $3 < x < 5, \|x-2\|+\|x-3\|+\|x-5\|$	$III$. is an odd function
$D$. $\sin 2x + \sin^2 x + \cos 3x, \forall x \in \mathbb{R}$	$IV$. is the identity function
	$V$. is a constant function

Similar Questions

Which of the following pairs of functions are identical?

If $f(x)=\frac{2^{x}+2^{-x}}{2}$,then $f(x+y) \cdot f(x-y)$ is

Let $f(x)$ and $g(x)$ be two continuous functions defined from $R \rightarrow R$,such that $f(x_1) > f(x_2)$ and $g(x_1) < g(x_2)$ for all $x_1 > x_2$. Then the solution set of $f(g(\alpha^2 - 2\alpha)) > f(g(3\alpha - 4))$ is

Let $f(x)$ be a non-constant polynomial with real coefficients such that $f\left(\frac{1}{2}\right)=100$ and $f(x) \leq 100$ for all real $x$. Which of the following statements is $NOT$ necessarily true?

If $f:[0,2) \rightarrow R$ is defined by $f(x)=\begin{cases} 1+\frac{2x}{k} & \text{for } 0 \leq x < 1 \\ kx & \text{for } 1 \leq x < 2 \end{cases}$ where $k>0$,and $f$ is such that $\lim_{x \rightarrow 1^{-}} f(x)=\lim_{x \rightarrow 1^{+}} f(x)$,then find the value of $k^2$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module