1
MHT CET 2023 12th May Morning Shift
MCQ (Single Correct Answer)
+2
-0

If general solution of $$\cos ^2 \theta-2 \sin \theta+\frac{1}{4}=0$$ is $$\theta=\frac{\mathrm{n} \pi}{\mathrm{A}}+(-1)^{\mathrm{n}} \frac{\pi}{\mathrm{B}}, \mathrm{n} \in \mathrm{Z}$$, then $$\mathrm{A}+\mathrm{B}$$ has the

A
7
B
6
C
1
D
$$-$$7
2
MHT CET 2023 12th May Morning Shift
MCQ (Single Correct Answer)
+2
-0

$$\overline{\mathrm{u}}, \overline{\mathrm{v}}, \overline{\mathrm{w}}$$ are three vectors such that $$|\overline{\mathrm{u}}|=1, |\bar{v}|=2,|\bar{w}|=3$$. If the projection of $$\bar{v}$$ along $$\bar{u}$$ is equal to projection of $$\bar{w}$$ along $$\bar{u}$$ and $$\bar{v}, \bar{w}$$ are perpendicular to each other, then $$|\bar{u}-\bar{v}+\bar{w}|=$$

A
4
B
$$\sqrt{7}$$
C
$$\sqrt{14}$$
D
2
3
MHT CET 2023 12th May Morning Shift
MCQ (Single Correct Answer)
+2
-0

The distance of the point $$\mathrm{P}(-2,4,-5)$$ from the line $$\frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6}$$ is

A
$$\frac{\sqrt{37}}{10}$$
B
$$\sqrt{\frac{37}{10}}$$
C
$$\frac{37}{\sqrt{10}}$$
D
$$\frac{37}{10}$$
4
MHT CET 2023 12th May Morning Shift
MCQ (Single Correct Answer)
+2
-0

$$\mathrm{A}$$ and $$\mathrm{B}$$ are independent events with $$\mathrm{P}(\mathrm{A})=\frac{1}{4}$$ and $$\mathrm{P}(\mathrm{A} \cup \mathrm{B})=2 \mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A})$$, then $$\mathrm{P}(\mathrm{B})$$ is

A
$$\frac{1}{4}$$
B
$$\frac{3}{5}$$
C
$$\frac{2}{3}$$
D
$$\frac{2}{5}$$
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