The range of the function f(x) = x^2 + frac{1 | vedclass

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

Question

(A) Let $y = x^2 + \frac{1}{x^2+1}$.Let $t = x^2$. Since $x \in \mathbb{R}$,$t \geq 0$.Then $y = t + \frac{1}{t+1} = (t+1) + \frac{1}{t+1} - 1$.Since

Vedclass · Accepted Answer

$[1, \infty)$

Vedclass · Answer

(A) Let $y = x^2 + \frac{1}{x^2+1}$.Let $t = x^2$. Since $x \in \mathbb{R}$,$t \geq 0$.Then $y = t + \frac{1}{t+1} = (t+1) + \frac{1}{t+1} - 1$.Since $t \geq 0$,$t+1 \geq 1$.Using the Arithmetic Mean-Geometric Mean inequality $(AM \geq GM)$ for $t+1$ and $\frac{1}{t+1}$:$\frac{(t+1) + \frac{1}{t+1}}{2} \geq \sqrt{(t+1) \cdot \frac{1}{t+1}} = 1$.So,$(t+1) + \frac{1}{t+1} \geq 2$.Therefore,$y = (t+1) + \frac{1}{t+1} - 1 \geq 2 - 1 = 1$.The minimum value is attained at $t=0$ (i.e.,$x=0$),where $f(0) = 0 + \frac{1}{0+1} = 1$.Thus,the range of the function is $[1, \infty)$.

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

Similar Questions

The domain of the function $f(x) = \sec^{-1}(3x - 4) + \tanh^{-1}\left(\frac{x + 3}{5}\right)$ is

If the function $f: R - \{ 1, - 1\} \to A$ defined by $f(x) = \frac{x^2}{1 - x^2}$ is surjective,then $A$ is equal to

For $f(x) = \frac{\sin \pi[x]}{1+[x]} + \frac{x}{2+3x}$,where $[x]$ denotes the greatest integer function,the domain and range in $R$ are respectively

If the domain of the function $f(x) = \log_e \left( \frac{2x+3}{4x^2+x-3} \right) + \cos^{-1} \left( \frac{2x-1}{x+2} \right)$ is $(\alpha, \beta]$,then the value of $5\beta - 4\alpha$ is equal to

$A$ real valued function $f(x) = |x^2 - 3x + 2| + 2x - 3$ is defined on $[-2, 1]$. If $m$ and $M$ are absolute minimum and absolute maximum values of $f$ respectively,then $M - 4m =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module