1
MHT CET 2025 23rd April Morning Shift
MCQ (Single Correct Answer)
+2
-0

A particle moves along a curve $y=\frac{2 x^3-1}{3}$. The points on the curve at which the $y$ co-ordinate is changing 18 times the $x$ co-ordinate are

A
$\left(-3,-\frac{55}{3}\right),\left(3,-\frac{53}{3}\right)$
B
$\left(-3, \frac{53}{3}\right),\left(3, \frac{55}{3}\right)$
C
$\left(-3,-\frac{53}{3}\right),\left(3, \frac{55}{3}\right)$
D
$\left(-3,-\frac{55}{3}\right),\left(3, \frac{53}{3}\right)$
2
MHT CET 2025 23rd April Morning Shift
MCQ (Single Correct Answer)
+2
-0

The equation of motion of the particle is $\mathrm{s}=\mathrm{at}^2+\mathrm{bt}+\mathrm{c}$. If the displacement after 1 second is 20 m , velocity after 2 seconds is $30 \mathrm{~m} /$ seconds and the acceleration is $10 \mathrm{~m} /$ seconds $^2$, then

A
$a+c=2 b$
B
$\mathrm{a}+\mathrm{c}=\mathrm{b}$
C
$\mathrm{a}-\mathrm{c}=\mathrm{b}$
D
$a+c=3 b$
3
MHT CET 2025 23rd April Morning Shift
MCQ (Single Correct Answer)
+2
-0

If $y=\tan ^{-1}\left[\frac{4 \sin 2 x}{\cos 2 x-6 \sin ^2 x}\right]$, then $\frac{\mathrm{d} y}{\mathrm{~d} x}$ at $x=0$ is

A
$\frac{1}{8}$
B
-8
C
8
D
$-\frac{1}{8}$
4
MHT CET 2025 23rd April Morning Shift
MCQ (Single Correct Answer)
+2
-0

If $x=a \sin 2 t(1+\cos 2 t), y=b \cos 2 t(1-\cos 2 t)$ then $\frac{\mathrm{d} y}{\mathrm{~d} x}$ is equal to

A
$\frac{\mathrm{b}}{\mathrm{a}} \tan t$
B
$\frac{a}{b} \tan t$
C
$\frac{b}{\operatorname{atan} t}$
D
$\frac{\mathrm{a}}{\mathrm{b} \tan \mathrm{t}}$
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