1
MHT CET 2022 11th August Evening Shift
MCQ (Single Correct Answer)
+2
-0

The general solution of the differential equation $$\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{3 x+y}{x-y}$$ is (where $$C$$ is a constant of integration.)

A
$$\tan ^{-1}\left(\frac{y}{x}\right)+\log \left(\frac{y^2+3 x^2}{x^2}\right)=\log (x)+C$$
B
$$\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{y}{x \sqrt{3}}\right)+\log \left(\frac{y^2+3 x^2}{x^2}\right)^{\frac{1}{2}}=\log (x)+C$$
C
$$\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{y}{x \sqrt{3}}\right)-\log \left(\frac{y^2+3 x^2}{x^2}\right)^{\frac{1}{2}}=\log (x)+C$$
D
$$\tan ^{-1}\left(\frac{x}{y}\right)+\log \left(\frac{y^2+3 x^2}{x^2}\right)=\log (x)+C$$
2
MHT CET 2022 11th August Evening Shift
MCQ (Single Correct Answer)
+2
-0

If $$A=\left[\begin{array}{lll}1 & 2 & 1 \\ 3 & 1 & 3\end{array}\right]$$ and $$B=\left[\begin{array}{ll}2 & 3 \\ 1 & 2 \\ 1 & 2\end{array}\right]$$, then $$(A B)^{-1}=$$

A
$$\left[\begin{array}{ll}\frac{17}{5} & \frac{9}{5} \\ 2 & 1\end{array}\right]$$
B
$$\left[\begin{array}{cc}\frac{-17}{5} & \frac{9}{5} \\ 2 & -1\end{array}\right]$$
C
$$\left[\begin{array}{ll}\frac{17}{5} & 2 \\ \frac{9}{5} & 1\end{array}\right]$$
D
$$\left[\begin{array}{cc}\frac{-17}{5} & 2 \\ \frac{-9}{5} & -1\end{array}\right]$$
3
MHT CET 2022 11th August Evening Shift
MCQ (Single Correct Answer)
+2
-0

The distance between parallel lines

$$\frac{x-1}{2}=\frac{y-2}{-2}=\frac{z-3}{1}$$ and

$$\frac{x}{2}=\frac{y}{-2}=\frac{z}{1}$$ is :

A
$$\frac{2 \sqrt{5}}{3}$$ units
B
$$\frac{\sqrt{5}}{3}$$ units
C
$$\frac{5 \sqrt{5}}{3}$$ units
D
$$\frac{4 \sqrt{5}}{3}$$ units
4
MHT CET 2022 11th August Evening Shift
MCQ (Single Correct Answer)
+2
-0

Maximum value of $$Z=5 x+2 y$$, subject to $$2 x-y \geq 2, x+2 y \leq 8$$ and $$x, y \geq 0$$ is

A
40
B
17.6
C
28
D
25.6
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