1
MHT CET 2023 13th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

The abscissae of two points $$A$$ and $$B$$ are the roots of the equation $$x^2+2 a x-b^2=0$$ and their ordinates are roots of the equation $$y^2+2 p y-q^2=0$$. Then, the equation of the circle with $$A B$$ as diameter is given by

A
$$x^2+y^2-2 a x-2 p y+\left(b^2+q^2\right)=0$$
B
$$x^2+y^2-2 a x-2 p y-\left(b^2+q^2\right)=0$$
C
$$x^2+y^2+2 a x+2 p y+\left(b^2+q^2\right)=0$$
D
$$x^2+y^2+2 a x+2 p y-\left(b^2+q^2\right)=0$$
2
MHT CET 2023 13th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

The incentre of the $$\triangle A B C$$, whose vertices are $$A(0,2,1), B(-2,0,0)$$ and $$C(-2,0,2)$$, is

A
$$\left(-\frac{3}{2}, \frac{1}{2}, 1\right)$$
B
$$\left(\frac{3}{2}, \frac{1}{2}, 1\right)$$
C
$$\left(-\frac{3}{2},-\frac{1}{2},-1\right)$$
D
$$\left(\frac{3}{2},-\frac{1}{2},-1\right)$$
3
MHT CET 2023 13th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

The acute angle between the line joining the points $$(2,1,-3),(-3,1,7)$$ and a line parallel to $$\frac{x-1}{3}=\frac{y}{4}=\frac{z+3}{5}$$ through the point $$(-1,0,4)$$ is

A
$$\cos ^{-1}\left(\frac{1}{\sqrt{10}}\right)$$
B
$$\cos ^{-1}\left(\frac{5}{7 \sqrt{10}}\right)$$
C
$$\cos ^{-1}\left(\frac{7}{5 \sqrt{10}}\right)$$
D
$$\cos ^{-1}\left(\frac{3}{5 \sqrt{10}}\right)$$
4
MHT CET 2023 13th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

A random variable $$X$$ has the probability distribution

$$X=x$$ 1 2 3 4 5 6 7 8
$$P(X=x)$$ 0.15 0.23 0.12 0.20 0.08 0.10 0.05 0.07

For the events $$E=\{X$$ is a prime number $$\}$$ and $$F=\{x<5\}, P(E U F)$$ is

A
0.63
B
0.75
C
0.83
D
0.90
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