1
MHT CET 2024 15th May Morning Shift
MCQ (Single Correct Answer)
+2
-0

Let $\overline{\mathrm{a}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}, \overline{\mathrm{b}}=\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}$ and $\overline{\mathrm{c}}=\hat{\mathrm{i}}-\hat{\mathrm{j}}-\hat{\mathrm{k}}$ be three vectors. A vector $\bar{v}$ in the plane of $\overline{\mathrm{a}}$ and $\overline{\mathrm{b}}$, whose projection on $\overline{\mathrm{c}}$ is $\frac{1}{\sqrt{3}}$, is given by

A
$\hat{i}-3 \hat{j}+3 \hat{k}$
B
$-3 \hat{i}-3 \hat{j}-\hat{k}$
C
$3 \hat{i}-\hat{j}+3 \hat{k}$
D
$\hat{\mathrm{i}}+3 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}$
2
MHT CET 2024 15th May Morning Shift
MCQ (Single Correct Answer)
+2
-0

The differential equation, having general solution as $A x^2+B y^2=1$, where $A$ and $B$ are arbitrary constants, is

A
$x y \frac{\mathrm{~d}^2 y}{\mathrm{~d} x^2}-x\left(\frac{\mathrm{~d} y}{\mathrm{~d} x}\right)^2-y \frac{\mathrm{~d} y}{\mathrm{~d} x}=0$
B
$x y \frac{\mathrm{~d}^2 y}{\mathrm{~d} x^2}-x\left(\frac{\mathrm{~d} y}{\mathrm{~d} x}\right)^2+y \frac{\mathrm{~d} y}{\mathrm{~d} x}=0$
C
$x y \frac{\mathrm{~d}^2 y}{\mathrm{~d} x^2}+x\left(\frac{\mathrm{~d} y}{\mathrm{~d} x}\right)^2+y \frac{\mathrm{~d} y}{\mathrm{~d} x}=0$
D
$x y \frac{\mathrm{~d}^2 y}{\mathrm{~d} x^2}+x\left(\frac{\mathrm{~d} y}{\mathrm{~d} x}\right)^2-y \frac{\mathrm{~d} y}{\mathrm{~d} x}=0$
3
MHT CET 2024 15th May Morning Shift
MCQ (Single Correct Answer)
+2
-0

Let $z$ be a complex number such that $|z|+z=2+i$, where $i=\sqrt{-1}$, then $|z|$ is equal to

A
$\frac{4}{5}$
B
$\frac{5}{4}$
C
$\frac{5}{3}$
D
$\frac{\sqrt{41}}{4}$
4
MHT CET 2024 15th May Morning Shift
MCQ (Single Correct Answer)
+2
-0

If $\overline{\mathrm{a}}$ and $\overline{\mathrm{b}}$ are two unit vectors such that $5 \overline{\mathrm{a}}+4 \overline{\mathrm{~b}}$ and $\overline{\mathrm{a}}-2 \overline{\mathrm{~b}}$ are perpendicular to each other, then the angle between $\bar{a}$ and $\bar{b}$ is

A
$\frac{\pi}{3}$
B
$\cos ^{-1}\left(\frac{2}{3}\right)$
C
$\frac{2 \pi}{3}$
D
$\cos ^{-1}\left(\frac{1}{3}\right)$
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