1
JEE Main 2023 (Online) 31st January Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language
Let $f: \mathbb{R}-\{2,6\} \rightarrow \mathbb{R}$ be real valued function

defined as $f(x)=\frac{x^2+2 x+1}{x^2-8 x+12}$.

Then range of $f$ is
A
$ \left(-\infty,-\frac{21}{4}\right] \cup[1, \infty) $
B
$\left(-\infty,-\frac{21}{4}\right) \cup(0, \infty) $
C
$\left(-\infty,-\frac{21}{4}\right] \cup[0, \infty) $
D
$\left(-\infty,-\frac{21}{4}\right] \cup\left[\frac{21}{4}, \infty\right)$
2
JEE Main 2023 (Online) 31st January Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language
The absolute minimum value, of the function

$f(x)=\left|x^{2}-x+1\right|+\left[x^{2}-x+1\right]$,

where $[t]$ denotes the greatest integer function, in the interval $[-1,2]$, is :
A
$\frac{3}{4}$
B
$\frac{3}{2}$
C
$\frac{1}{4}$
D
$\frac{5}{4}$
3
JEE Main 2023 (Online) 31st January Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language
If the domain of the function $$f(x)=\frac{[x]}{1+x^{2}}$$, where $$[x]$$ is greatest integer $$\leq x$$, is $$[2,6)$$, then its range is
A
$$\left(\frac{5}{37}, \frac{2}{5}\right]-\left\{\frac{9}{29}, \frac{27}{109}, \frac{18}{89}, \frac{9}{53}\right\}$$
B
$$\left(\frac{5}{37}, \frac{2}{5}\right]$$
C
$$\left(\frac{5}{26}, \frac{2}{5}\right]$$
D
$$\left(\frac{5}{26}, \frac{2}{5}\right]-\left\{\frac{9}{29}, \frac{27}{109}, \frac{18}{89}, \frac{9}{53}\right\}$$
4
JEE Main 2023 (Online) 30th January Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language
The range of the function $f(x)=\sqrt{3-x}+\sqrt{2+x}$ is :
A
$[2 \sqrt{2}, \sqrt{11}]$
B
$[\sqrt{5}, \sqrt{13}]$
C
$[\sqrt{2}, \sqrt{7}]$
D
$[\sqrt{5}, \sqrt{10}]$
JEE Main Subjects
EXAM MAP
Medical
NEET
Graduate Aptitude Test in Engineering
GATE CSEGATE ECEGATE EEGATE MEGATE CEGATE PIGATE IN
CBSE
Class 12