1
JEE Main 2023 (Online) 1st February Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Let $$f:\mathbb{R}-{0,1}\to \mathbb{R}$$ be a function such that $$f(x)+f\left(\frac{1}{1-x}\right)=1+x$$. Then $$f(2)$$ is equal to

A
$$\frac{9}{4}$$
B
$$\frac{7}{4}$$
C
$$\frac{7}{3}$$
D
$$\frac{9}{2}$$
2
JEE Main 2023 (Online) 1st February Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Let $$f(x) = \left| {\matrix{ {1 + {{\sin }^2}x} & {{{\cos }^2}x} & {\sin 2x} \cr {{{\sin }^2}x} & {1 + {{\cos }^2}x} & {\sin 2x} \cr {{{\sin }^2}x} & {{{\cos }^2}x} & {1 + \sin 2x} \cr } } \right|,\,x \in \left[ {{\pi \over 6},{\pi \over 3}} \right]$$. If $$\alpha$$ and $$\beta$$ respectively are the maximum and the minimum values of $$f$$, then

A
$${\alpha ^2} - {\beta ^2} = 4\sqrt 3 $$
B
$${\beta ^2} - 2\sqrt \alpha = {{19} \over 4}$$
C
$${\beta ^2} + 2\sqrt \alpha = {{19} \over 4}$$
D
$${\alpha ^2} + {\beta ^2} = {9 \over 2}$$
3
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
$\left(-\infty,-\frac{21}{4}\right] \cup\left[\frac{21}{4}, \infty\right)$
4
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
$\frac{5}{4}$
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