Which of the following functions is an even f | vedclass

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

Question

(B) function $f(x)$ is even if $f(-x) = f(x)$ and odd if $f(-x) = -f(x)$.For option $(A)$: $f(-x) = \frac{a^{-x} + 1}{a^{-x} - 1} = \frac{\frac{1}{a^x

Vedclass · Accepted Answer

$f(x) = x \left( \frac{a^x - 1}{a^x + 1} \right)$

Vedclass · Answer

(B) function $f(x)$ is even if $f(-x) = f(x)$ and odd if $f(-x) = -f(x)$.For option $(A)$: $f(-x) = \frac{a^{-x} + 1}{a^{-x} - 1} = \frac{\frac{1}{a^x} + 1}{\frac{1}{a^x} - 1} = \frac{1 + a^x}{1 - a^x} = -\frac{a^x + 1}{a^x - 1} = -f(x)$. Thus,it is an odd function.For option $(B)$: $f(-x) = (-x) \left( \frac{a^{-x} - 1}{a^{-x} + 1} \right) = (-x) \left( \frac{\frac{1}{a^x} - 1}{\frac{1}{a^x} + 1} \right) = (-x) \left( \frac{1 - a^x}{1 + a^x} \right) = x \left( \frac{a^x - 1}{a^x + 1} \right) = f(x)$. Thus,it is an even function.For option $(C)$: $f(-x) = \frac{a^{-x} - a^x}{a^{-x} + a^x} = -\frac{a^x - a^{-x}}{a^x + a^{-x}} = -f(x)$. Thus,it is an odd function.For option $(D)$: $f(-x) = \sin(-x) = -\sin x = -f(x)$. Thus,it is an odd function.Therefore,the correct option is $(B)$.

Which of the following functions is an even function?

Similar Questions

For a real number $r$,we denote by $[r]$ the largest integer less than or equal to $r$. If $x, y$ are real numbers with $x, y \geq 1$,then which of the following statements is always true?

If $f(x) = \log \left( \frac{1 + x}{1 - x} \right)$,then $f(x)$ is

Number of solutions of the equation $2^x + x = 2^{\sin x} + \sin x$ in $[0, 10\pi]$ is -

The number of elements in the set $\{x \in R : (|x|-3)|x+4|=6\}$ is equal to

For a real number $x$,let $[x]$ denote the greatest integer less than or equal to $x$,and let $\{x\} = x - [x]$. The number of solutions $x$ to the equation $[x]\{x\} = 5$ with $0 \leq x \leq 2015$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module