1
MHT CET 2024 16th May Morning Shift
MCQ (Single Correct Answer)
+1
-0

Range of the function $\mathrm{f}(x)=\frac{x^2+x+2}{x^2+x+1}, x \in \mathbb{R}$ is

A
$\left(1, \frac{7}{3}\right)$
B
$\left[1, \frac{7}{3}\right)$
C
$\left(1, \frac{7}{3}\right]$
D
$\left[1, \frac{7}{3}\right]$
2
MHT CET 2024 15th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

If $[x]^2-5[x]+6=0$, where $[\cdot]$ denotes the greatest integer function, then

A
$x \in(2,4]$
B
$x \in[2,4]$
C
$x \in[2,4)$
D
$x \in(2,4)$
3
MHT CET 2024 15th May Morning Shift
MCQ (Single Correct Answer)
+2
-0

Let $\mathrm{f}(x)=(x+1)^2-1, x \geqslant-1$, then the set $\left\{x / f(x)=f^{-1}(x)\right\}$ is

A
$\{0,1,-1\}$
B
$\{0,-1\}$
C
$\left\{0,-1, \frac{-3+\mathrm{i} \sqrt{3}}{2}, \frac{-3-\mathrm{i} \sqrt{3}}{2}\right.$, where $\left.\mathrm{i}=\sqrt{-1}\right\}$
D
$\phi$
4
MHT CET 2024 11th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

If $\mathrm{f}:[1, \infty) \rightarrow[2, \infty)$ is given by $\mathrm{f}(x)=x+\frac{1}{x}$ then $\mathrm{f}^{-1}(x)$ equals

A
$\frac{x+\sqrt{x^2-4}}{2}$
B
$\frac{2}{1+x^2}$
C
$\frac{x-\sqrt{x^2-4}}{2}$
D
$1+\sqrt{x^2-4}$
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