1
MHT CET 2024 4th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

The value of $\int \frac{\mathrm{d} x}{x^2\left(x^4+1\right)^{\frac{3}{4}}}$ is

A
$\frac{\left(x^4+1\right)^{\frac{1}{4}}}{x}+\mathrm{c}$, where c is a constant of integration.
B
$\left(x^4+1\right)^{\frac{1}{4}}+\mathrm{c}$, where c is a constant of integration.
C
$\frac{-\left(x^4+1\right)^{\frac{1}{4}}}{x}+\mathrm{c}$, where c is a constant of integration.
D
$-\left(x^4+1\right)^{\frac{1}{4}}+c$, where $c$ is a constant of integration.
2
MHT CET 2024 4th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

If $(a+b) \cos C+(b+c) \cos A+(c+a) \cos B=72$ and if $a=18, b=24$, then area of the triangle $A B C$ is

A
144 sq.units
B
216 sq.units
C
256 sq.units
D
296 sq. units
3
MHT CET 2024 4th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

If $\cot ^{-1}(7)+\cot ^{-1}(8)+\cot ^{-1}(18)=\cot ^{-1} x$, then the value of $x$ is

A
$\frac{1}{3}$
B
2
C
3
D
$\frac{1}{2}$
4
MHT CET 2024 4th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

If $\mathrm{P}(\mathrm{X}=2)=0.3, \mathrm{P}(\mathrm{X}=3)=0.4, \mathrm{P}(\mathrm{X}=4)=0.3$, then the variance of random variable X is

A
1.6
B
3.6
C
6.6
D
0.6
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