1
MHT CET 2023 14th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

If $$A=\left[\begin{array}{ll}1 & -1 \\ 2 & -1\end{array}\right], B=\left[\begin{array}{cc}1 & 1 \\ 4 & -1\end{array}\right]$$, then $$(A+B)^{-1}$$ is

A
$$\left[\begin{array}{cc}\frac{-1}{2} & 0 \\ \frac{-3}{2} & \frac{1}{2}\end{array}\right]$$
B
$$\left[\begin{array}{cc}\frac{1}{2} & 0 \\ \frac{3}{2} & \frac{-1}{2}\end{array}\right]$$
C
$$\left[\begin{array}{cc}\frac{1}{2} & 0 \\ \frac{-3}{2} & \frac{1}{2}\end{array}\right]$$
D
$$\left[\begin{array}{ll}\frac{1}{2} & 0 \\ \frac{3}{2} & \frac{1}{2}\end{array}\right]$$
2
MHT CET 2023 14th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

The function $$\mathrm{f}(x)=x^3-6 x^2+9 x+2$$ has maximum value when $$x$$ is

A
1
B
2
C
3
D
6
3
MHT CET 2023 14th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

If $$I_n=\int_\limits0^{\frac{\pi}{4}} \tan ^n \theta d \theta$$, then $$I_{12}+I_{10}=$$

A
$$\frac{1}{8}$$
B
$$\frac{1}{12}$$
C
$$\frac{1}{11}$$
D
$$\frac{1}{10}$$
4
MHT CET 2023 14th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

The centre of the circle whose radius is 3 units and touching internally the circle $$x^2+y^2-4 x-6 y-12=0$$ at the point $$(-1,-1)$$ is

A
$$\left(\frac{4}{5}, \frac{7}{5}\right)$$
B
$$\left(\frac{4}{5}, \frac{-7}{5}\right)$$
C
$$\left(\frac{-4}{5}, \frac{-7}{5}\right)$$
D
$$\left(\frac{-4}{5}, \frac{7}{5}\right)$$
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