1
MHT CET 2024 4th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

The incenter of the triangle ABC , whose vertices are $\mathrm{A}(0,2,1), \mathrm{B}(-2,0,0)$ and $\mathrm{C}(-2,0,2)$ 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
$\left(-\frac{3}{2},-\frac{1}{2},-1\right)$
2
MHT CET 2024 4th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

The combined equation of two lines through the origin and making an angle of $45^{\circ}$ with the line $3 x+y=0$, is

A
$2 x^2-3 x y-2 y^2=0$
B
$2 x^2+3 x y+4 y^2=0$
C
$2 x^2+3 x y-2 y^2=0$
D
$2 x^2-3 x y+2 y^2=0$
3
MHT CET 2024 4th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

The general solution of the differential equation $\frac{1}{x} \frac{\mathrm{~d} y}{\mathrm{~d} x}=\tan ^{-1}$ is

A
$y+\frac{x^2 \tan ^{-1} x}{2}+\mathrm{c}=0$, where c is a constant of integration.
B
$y+x \tan ^{-1} x+\mathrm{c}=0$, where c is a constant integration.
C
$y-x-\tan ^{-1} x+\mathrm{c}=0$, where is a constant of integration.
D
$y=\frac{x^2 \tan ^{-1} x}{2}-\frac{1}{2}\left(x-\tan ^{-1} x\right)+\mathrm{c}$, where c is constant of integration.
4
MHT CET 2024 4th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

The area (in square units) in the first quadrant bounded by the curve $y=x^2+2$ and the lines $y=x+1, x=0$ and $x=3$, is

A
$\frac{15}{4}$
B
$\frac{21}{2}$
C
$\frac{17}{4}$
D
$\frac{15}{2}$
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