1
MHT CET 2024 16th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

The co-ordinates of the foot of perpendicular, drawn from the point $(-2,3)$ on the line $3 x-y-1=0$ are

A
$(-1,2)$
B
$(1,-2)$
C
$(-1,-2)$
D
$(1,2)$
2
MHT CET 2024 16th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

The value of $\tan ^{-1}(-\sqrt{3})-\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)+\cos ^{-1}\left(\frac{-1}{2}\right)$ is

A
$\frac{-\pi}{4}$
B
$\frac{4 \pi}{3}$
C
$\frac{\pi}{12}$
D
$\frac{7 \pi}{12}$
3
MHT CET 2024 16th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

The general solution of the differential equation $\frac{\mathrm{d} y}{\mathrm{~d} x}=y \tan x-y^2 \sec x$ is

A
$\tan x=(\mathrm{c}+\sec x) y$, where c is constant of integration.
B
$\sec y=(\mathrm{c}+\tan y) x$, where c is constant of integration.
C
$\sec x=(\mathrm{c}+\tan x) y$, where c is constant of integration.
D
$\cos y=(\mathrm{c}+\tan y)$, where c is constant of integration.
4
MHT CET 2024 16th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

If $\overline{\mathrm{a}}, \overline{\mathrm{b}}, \overline{\mathrm{c}}$ are three vectors such that $\overline{\mathrm{a}} \neq \overline{0}$ and $\overline{\mathrm{a}} \times \overline{\mathrm{b}}=2 \overline{\mathrm{a}} \times \overline{\mathrm{c}},|\overline{\mathrm{a}}|=|\overline{\mathrm{c}}|=1,|\overline{\mathrm{~b}}|=4$ and $|\overline{\mathrm{b}} \times \overline{\mathrm{c}}|=\sqrt{15}$. If $\overline{\mathrm{b}}-2 \overline{\mathrm{c}}=\lambda \overline{\mathrm{a}}$, then $\lambda$ is

A
1
B
$-$4
C
3
D
$-$2
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