1
MHT CET 2021 21th September Morning Shift
MCQ (Single Correct Answer)
+2
-0

$$\mathrm{A}^{-1}=\frac{-1}{2}\left[\begin{array}{cc}1 & -4 \\ -1 & 2\end{array}\right]$$, then $$2 A+I_2=\quad$$

where $$I_2$$ is a unit matrix of order 2

A
$$\left[\begin{array}{ll}5 & 8 \\ 1 & 2\end{array}\right]$$
B
$$\left[\begin{array}{ll}5 & 8 \\ 2 & 2\end{array}\right]$$
C
$$\left[\begin{array}{ll}2 & 4 \\ 1 & 1\end{array}\right]$$
D
$$\left[\begin{array}{ll}5 & 8 \\ 2 & 3\end{array}\right]$$
2
MHT CET 2021 21th September Morning Shift
MCQ (Single Correct Answer)
+2
-0

If $$y=\operatorname{cosec}^{-1}\left[\frac{\sqrt{x}+1}{\sqrt{x}-1}\right]+\cos ^{-1}\left[\frac{\sqrt{x}-1}{\sqrt{x}+1}\right]$$, then $$\frac{d y}{d x}=$$

A
0
B
1
C
$$\frac{2}{\sqrt{x}+1}$$
D
$$\frac{1}{2(\sqrt{x}-1)}$$
3
MHT CET 2021 21th September Morning Shift
MCQ (Single Correct Answer)
+2
-0

If $$f(x)=|x-1|+|x-2|+|x-3|, \forall x \in[1,4]$$, then $$\int_\limits1^4 f(x) d x=$$

A
$$\frac{1}{2}$$
B
7
C
$$\frac{9}{2}$$
D
$$\frac{19}{2}$$
4
MHT CET 2021 21th September Morning Shift
MCQ (Single Correct Answer)
+2
-0

The general solution of the differential equation $$\frac{d y}{d x}=\frac{x+2 y-1}{x+2 y+1}$$ is

A
$$3(x+y)+4 \log |3 x+6 y-1|=K$$
B
$$3(x-y)+4 \log |3 x+6 y-1|=K$$
C
$$6(-x+y)+4 \log |3 x+6 y-1|=K$$
D
$$6(x+y)+4 \log |3 x+6 y-1|=K$$
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