1
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
$\phi$
2
MHT CET 2024 15th May Morning Shift
MCQ (Single Correct Answer)
+2
-0

The probability, that a year selected at random will have 53 Mondays, is

A
$\frac{1}{4}$
B
$\frac{3}{28}$
C
$\frac{5}{28}$
D
$\frac{3}{4}$
3
MHT CET 2024 15th May Morning Shift
MCQ (Single Correct Answer)
+2
-0

Let $\mathrm{L}_1: \frac{x+2}{5}=\frac{y-3}{2}=\frac{\mathrm{z}-6}{1}$ and $\mathrm{L}_2: \frac{x-3}{4}=\frac{y+2}{3}=\frac{z-3}{5}$ be the given lines. Then the unit vector perpendicular to both $\mathrm{L}_1$ and $\mathrm{L}_2$ is

A
$\frac{-\hat{i}-3 \hat{j}+\hat{k}}{11}$
B
$\frac{\hat{i}-3 \hat{j}+\hat{k}}{11}$
C
$\frac{\hat{i}+3 \hat{j}-\hat{k}}{11}$
D
$\frac{\hat{i}+3 \hat{j}+\hat{k}}{11}$
4
MHT CET 2024 15th May Morning Shift
MCQ (Single Correct Answer)
+2
-0

A multiple choice examination has 5 questions. Each question has three alternative answers of which exactly one is correct. The probability that a student will get 4 or more correct answers just by guessing is

A
$\frac{17}{243}$
B
$\frac{13}{243}$
C
$\frac{11}{243}$
D
$\frac{10}{243}$
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