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

An open tank with a square bottom, to contain 4000 cubic cm . of liquid, is to be constructed. The dimensions of the tank, so that the surface area of the tank is minimum, are

A
side of square bottom $=40 \mathrm{~cm}$, height $=10 \mathrm{~cm}$.
B
side of square bottom $=20 \mathrm{~cm}$, height $=10 \mathrm{~cm}$.
C
side of square bottom $=10 \mathrm{~cm}$, height $=40 \mathrm{~cm}$.
D
side of square bottom $=5 \mathrm{~cm}$, height $=160 \mathrm{~cm}$.
2
MHT CET 2024 15th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

$$\int \operatorname{cosec}(x-a) \cdot \operatorname{cosec} x d x=$$

A
$\frac{-1}{\operatorname{sina}} \log (\sin (x-\mathrm{a}) \sin x)+\mathrm{c}$, where c is a constant of integration.
B
$\frac{1}{\sin \mathrm{a}} \log (\sin (x-\mathrm{a}) \sin x)+\mathrm{c}$, where c is a constant of integration.
C
$\frac{1}{\operatorname{sina}} \log (\sin (x-a) \cdot \operatorname{cosec} x)+c$, where c is a constant of integration.
D
$\frac{-1}{\operatorname{sina}} \log (\operatorname{cosec}(x-\mathrm{a}) \cdot \sin x)+\mathrm{c}$, where c is a constant of integration.
3
MHT CET 2024 15th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

The contrapositive of the inverse of $\mathrm{p} \rightarrow(\mathrm{p} \rightarrow \mathrm{q})$ is

A
$(\sim p \wedge q) \rightarrow p$
B
$(\sim \mathrm{p} \vee \mathrm{q}) \rightarrow \mathrm{p}$
C
$\mathrm{p} \rightarrow(\sim \mathrm{p} \vee \mathrm{q})$
D
$(\mathrm{p} \vee \mathrm{q}) \rightarrow \mathrm{p}$
4
MHT CET 2024 15th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

The area enclosed between the parabola $y^2=4 x$ and the line $y=2 x-4$ is

A
$\frac{17}{3}$ sq. units
B
15 sq. units
C
$\frac{19}{3}$ sq. units
D
9 sq. units
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