Range of the function f(x) = frac{x^2}{x^2 + | vedclass

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

Question

(C) Let $y = \frac{x^2}{x^2 + 1}$.Since $x^2 \ge 0$ for all real $x$,the numerator is always non-negative and the denominator is always greater than t

Vedclass · Accepted Answer

$[0, 1)$

Vedclass · Answer

(C) Let $y = \frac{x^2}{x^2 + 1}$.Since $x^2 \ge 0$ for all real $x$,the numerator is always non-negative and the denominator is always greater than the numerator.As $x 	o \pm \infty$,$y 	o 1$,but $y$ never reaches $1$ because $x^2 At $x = 0$,$y = \frac{0}{0+1} = 0$.Since $x^2 \ge 0$ and $x^2 + 1 > 0$,the value of $y$ is always non-negative.Thus,the range of the function is $[0, 1)$.

Range of the function $f(x) = \frac{x^2}{x^2 + 1}$ is

Similar Questions

The function $f: R \rightarrow R$ is defined by $f(x) = \cos^2 x + \sin^4 x$ for $x \in R$. Then $f(R)$ is equal to:

For the function $f(x) = (1 + \frac{1}{x})^x$,the domain of $f(x)$ is:

Let $f(x) = \frac{\tan^n x}{\sum_{r=0}^{2n} \tan^r x}$,$n \in N$,where $x \in [0, \frac{\pi}{2})$.

The function is defined as $f(x) = (1 + \frac{1}{x})^x$. What is the range of the function $f(x)$?

The domain of $f(x)=\cos ^{-1}\left(\frac{x-3}{2}\right)-\log _{10}(4-x)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module