1
MHT CET 2021 21th September Morning Shift
MCQ (Single Correct Answer)
+2
-0

The vertices of triangle $$\mathrm{ABC}$$ are $$\mathrm{A} \equiv(3,0,0) ; \mathrm{B} \equiv(0,0,4) ; \mathrm{C} \equiv(0,5,4)$$. Find the position vector of the point in which the bisector of angle A meets B C is

A
$$5 \hat{\mathrm{i}}+12 \hat{\mathrm{j}}$$
B
$$\frac{5 \hat{\mathrm{j}}+12 \hat{\mathrm{k}}}{3}$$
C
$$\frac{5 \hat{\mathrm{i}}+12 \hat{\mathrm{j}}}{13}$$
D
$$\frac{5 \hat{\mathrm{i}}-12 \hat{\mathrm{j}}}{3}$$
2
MHT CET 2021 21th September Morning Shift
MCQ (Single Correct Answer)
+2
-0

If the lines $\frac{1-x}{3}=\frac{7 y-14}{2 \lambda}=\frac{z-3}{2}$ and $\frac{7-7 x}{3 \lambda}=\frac{y-5}{1}=\frac{6-z}{5}$ are at right angles, then $\lambda=$

A
$$\frac{-70}{11}$$
B
$$\frac{70}{11}$$
C
$$\frac{11}{70}$$
D
$$\frac{-11}{70}$$
3
MHT CET 2021 21th September Morning Shift
MCQ (Single Correct Answer)
+2
-0

A lot of 100 bulbs contains 10 defective bulbs. Five bulbs selected at random from the lot and sent to retain store, then the probability that the store will receive at most one defective bulb is

A
0.59049
B
0.91854
C
0.6561
D
0.32805
4
MHT CET 2021 21th September Morning Shift
MCQ (Single Correct Answer)
+2
-0

The differential equation of family of circles whose centres lie on $$\mathrm{X}$$-axis is

A
$$\frac{d^2 y}{d x^2}+\left(\frac{d y}{d x}\right)^2+1=0$$
B
$$y\left(\frac{d^2 y}{d x^2}\right)+\left(\frac{d y}{d x}\right)^2+1=0$$
C
$$y\left(\frac{d^2 y}{d x^2}\right)-\left(\frac{d y}{d x}\right)^2-1=0$$
D
$$y\left(\frac{d^2 y}{d x^2}\right)+\left(\frac{d y}{d x}\right)^2-1=0$$
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