1
MHT CET 2024 11th May Morning Shift
MCQ (Single Correct Answer)
+2
-0

If angle $\theta$ in $[0,2 \pi]$ satisfies both the equations $\cot \theta=\sqrt{3}$ and $\sqrt{3} \sec \theta+2=0$, then $\theta$ is equal to

A
$\frac{\pi}{6}$
B
  $\frac{7 \pi}{6}$
C
$\frac{5 \pi}{6}$
D
$\frac{11 \pi}{6}$
2
MHT CET 2024 11th May Morning Shift
MCQ (Single Correct Answer)
+2
-0

The area of the triangle with vertices $(1,2,0)$, $(1,0,2)$ and $(0,3,1)$ is

A
$\sqrt{3}$ sq. units
B
$\sqrt{6}$ sq. units
C
$\sqrt{5}$ sq. units
D
$\sqrt{7}$ sq. units
3
MHT CET 2024 11th May Morning Shift
MCQ (Single Correct Answer)
+2
-0

The particular solution of the differential equation, $x y \frac{\mathrm{~d} y}{\mathrm{~d} x}=x^2+2 y^2$ when $y(1)=0$ is

A
$\frac{x^2+y^2}{x^3}=1$
B
$x^2+y^2=x$
C
$x^2+y^2=x^4$
D
$x^2+2 y^2=x^4$
4
MHT CET 2024 11th May Morning Shift
MCQ (Single Correct Answer)
+2
-0

The equation of the tangent to the circle, given by $x=5 \cos \theta, y=5 \sin \theta$ at the point $\theta=\frac{\pi}{3}$ on it , is

A
$x-\sqrt{3} y=-5$
B
$x+\sqrt{3} y=10$
C
$\sqrt{3} x+y=5 \sqrt{3}$
D
$\sqrt{3} x-y=0$
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