1
MHT CET 2025 21st April Evening Shift
MCQ (Single Correct Answer)
+2
-0

The differential equation of all circles having their centres on the line $y=5$ and touching ( X -axis) is $\qquad$

A
$\quad(5-y) \frac{\mathrm{d} y}{\mathrm{~d} x}+y^2-10 y=0$
B
$\quad(5-y)^2 \frac{\mathrm{~d}^2 y}{\mathrm{~d} x^2}+y^2-10 y=0$
C
$\quad(5-y) \frac{\mathrm{d} y}{\mathrm{~d} x}+y-10=0$
D
$\quad(5-y)^2\left(\frac{\mathrm{~d} y}{\mathrm{~d} x}\right)^2+y^2-10 y=0$
2
MHT CET 2025 21st April Evening Shift
MCQ (Single Correct Answer)
+2
-0

$$ \int_0^{\frac{\pi}{4}} \frac{\cos ^2 x \sin ^2 x}{\cos ^3 x+\sin ^3 x} d x= $$

A
$\frac{1}{3}$
B
$\frac{-1}{3}$
C
$\frac{1}{6}$
D
$\frac{-1}{6}$
3
MHT CET 2025 21st April Evening Shift
MCQ (Single Correct Answer)
+2
-0

If $\{(\mathrm{p} \wedge \sim \mathrm{q}) \wedge(\mathrm{p} \wedge \mathrm{r})\} \rightarrow \sim \mathrm{p} \vee \mathrm{q}$ has truth value false then truth values of the statements $p, q, r$ are respectively

A
$\mathrm{T}, \mathrm{T}, \mathrm{T}$
B
$\mathrm{F}, \mathrm{F}, \mathrm{F}$
C
$F, F, T$
D
$T, F, T$
4
MHT CET 2025 21st April Evening Shift
MCQ (Single Correct Answer)
+2
-0

In a triangle ABC , with usual notations, $\tan \left(\frac{\mathrm{A}}{2}\right)=\frac{5}{6}, \tan \left(\frac{\mathrm{C}}{2}\right)=\frac{2}{5}$, then

A
$\mathrm{a}, \mathrm{c}, \mathrm{b}$ are in A.P.
B
$\mathrm{b}, \mathrm{a}, \mathrm{c}$ are in A.P.
C
$\mathrm{a}, \mathrm{b}, \mathrm{c}$ are in A.P.
D
$\mathrm{a}, \mathrm{b}, \mathrm{c}$ are in G.P.
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