1
MHT CET 2024 11th May Morning Shift
MCQ (Single Correct Answer)
+2
-0

If $x^2+y^2=\mathrm{t}+\frac{1}{\mathrm{t}}, x^4+y^4=\mathrm{t}^2+\frac{1}{\mathrm{t}^2}$, then $\frac{\mathrm{d} y}{\mathrm{~d} x}=$

A
$\frac{1}{x^3 y}$
B
$\frac{1}{x y^3}$
C
$-\frac{1}{x y^3}$
D
$-\frac{1}{x^3 y}$
2
MHT CET 2024 11th May Morning Shift
MCQ (Single Correct Answer)
+2
-0

The general solution of $\frac{\mathrm{d} y}{\mathrm{~d} x}+\sin \left(\frac{x+y}{2}\right)=\sin \left(\frac{x-y}{2}\right)$ is

A
$\log \tan \left(\frac{y}{2}\right)=\mathrm{C}-2 \sin x$
B
$\log \tan \left(\frac{y}{4}\right)=\mathrm{C}-2 \sin \left(\frac{x}{2}\right)$
C
$\log \tan \left(\frac{y}{2}+\frac{\pi}{4}\right)=\mathrm{C}-2 \sin x$
D
$\log \tan \left(\frac{y}{2}+\frac{\pi}{4}\right)=\mathrm{C}-2 \sin \left(\frac{x}{2}\right)$
3
MHT CET 2024 11th May Morning Shift
MCQ (Single Correct Answer)
+2
-0

If $A$ and $B$ are two independent events such that $\mathrm{P}\left(\mathrm{A}^{\prime}\right)=0.75, \mathrm{P}(\mathrm{A} \cup \mathrm{B})=0.65$ and $\mathrm{P}(\mathrm{B})=\mathrm{p}$, then value of $p$ is

A
$\frac{9}{14}$
B
$\frac{7}{15}$
C
$\frac{5}{14}$
D
$\frac{8}{15}$
4
MHT CET 2024 11th May Morning Shift
MCQ (Single Correct Answer)
+2
-0

If the lines $\frac{x+1}{-10}=\frac{y+k}{-1}=\frac{z-4}{1} \quad$ and $\frac{x+10}{-1}=\frac{y+1}{-3}=\frac{z-1}{4}$ intersect each other, then the value of $k$ is

A
$-$3
B
3
C
4
D
2
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