1
MHT CET 2023 9th May Morning Shift
MCQ (Single Correct Answer)
+2
-0

If two vertices of a triangle are $$\mathrm{A}(3,1,4)$$ and $$\mathrm{B}(-4,5,-3)$$ and the centroid of the triangle is $$G(-1,2,1)$$, then the third vertex $$C$$ of the triangle is

A
$$(2,0,2)$$
B
$$(-2,0,2)$$
C
$$(0,-2,2)$$
D
$$(2,-2,0)$$
2
MHT CET 2023 9th May Morning Shift
MCQ (Single Correct Answer)
+2
-0

In a Binomial distribution with $$\mathrm{n}=4$$, if $$2 \mathrm{P}(\mathrm{X}=3)=3 \mathrm{P}(\mathrm{X}=2)$$, then the variance is

A
$$\frac{36}{169}$$
B
$$\frac{144}{169}$$
C
$$\frac{9}{169}$$
D
$$\frac{16}{169}$$
3
MHT CET 2023 9th May Morning Shift
MCQ (Single Correct Answer)
+2
-0

Let two non-collinear vectors $$\hat{a}$$ and $$\hat{b}$$ form an acute angle. A point $$\mathrm{P}$$ moves, so that at any time $$t$$ the position vector $$\overline{\mathrm{OP}}$$, where $$\mathrm{O}$$ is origin, is given by $$\hat{a} \sin t+\hat{b} \cos t$$, when $$P$$ is farthest from origin $$O$$, let $$M$$ be the length of $$\overline{\mathrm{OP}}$$ and $$\hat{\mathrm{u}}$$ be the unit vector along $$\overline{\mathrm{OP}}$$, then

A
$$\hat{\mathrm{u}}=\frac{\hat{\mathrm{a}}+\hat{\mathrm{b}}}{|\hat{\mathrm{a}}+\hat{\mathrm{b}}|}$$ and $$\mathrm{M}=(1+\hat{\mathrm{a}} \cdot \hat{\mathrm{b}})^{\frac{1}{2}}$$
B
$$\hat{\mathrm{u}}=\frac{\hat{\mathrm{a}}-\hat{\mathrm{b}}}{|\hat{\mathrm{a}}-\hat{\mathrm{b}}|}$$ and $$\mathrm{M}=(1+\hat{\mathrm{a}} \cdot \hat{\mathrm{b}})^{\frac{1}{2}}$$
C
$$\hat{u}=\frac{\hat{a}+\hat{b}}{|\hat{a}+\hat{b}|}$$ and $$M=(1+2 \hat{a} \cdot \hat{b})^{\frac{1}{2}}$$
D
$$\hat{u}=\frac{\hat{a}-\hat{b}}{|\hat{a}-\hat{b}|}$$ and $$M=(1-2 \hat{a} \cdot \hat{b})^{\frac{1}{2}}$$
4
MHT CET 2023 9th May Morning Shift
MCQ (Single Correct Answer)
+2
-0

The number of solutions in $$[0,2 \pi]$$ of the equation $$16^{\sin ^2 x}+16^{\cos ^2 x}=10$$ is

A
2
B
4
C
6
D
8
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