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

Let $A=\left[\begin{array}{ccc}1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1\end{array}\right]$ and $B=\left[\begin{array}{l}4 \\ 0 \\ 2\end{array}\right]$ such that $\mathrm{AX}=\mathrm{B}$, then $\mathrm{X}=$

A
$\left[\begin{array}{c}-1 \\ 2 \\ 1\end{array}\right]$
B
$\left[\begin{array}{c}2 \\ -1 \\ 1\end{array}\right]$
C
$\left[\begin{array}{c}-1 \\ 1 \\ 2\end{array}\right]$
D
$\left[\begin{array}{c}-2 \\ 1 \\ -1\end{array}\right]$
2
MHT CET 2024 11th May Morning Shift
MCQ (Single Correct Answer)
+2
-0

If $\bar{x}=\frac{\bar{b} \times \bar{c}}{[\overline{\mathrm{a}} \overline{\mathrm{b}} \overline{\mathrm{c}}]}, \bar{y}=\frac{\overline{\mathrm{c}} \times \overline{\mathrm{a}}}{[\overline{\mathrm{a}} \overline{\mathrm{b}} \overline{\mathrm{c}}]}$ and $\overline{\mathrm{z}}=\frac{\overline{\mathrm{a}} \times \overline{\mathrm{b}}}{[\overline{\mathrm{a}} \overline{\mathrm{b}} \overline{\mathrm{c}}]}$ where $\overline{\mathrm{a}}, \overline{\mathrm{b}}, \overline{\mathrm{c}}$ are non-coplanar vectors, then value of $\bar{x} \cdot(\overline{\mathrm{a}}+\overline{\mathrm{b}})+\bar{y} \cdot(\overline{\mathrm{~b}}+\overline{\mathrm{c}})+\overline{\mathrm{z}} \cdot(\overline{\mathrm{c}}+\overline{\mathrm{a}})$ is

A
3
B
1
C
$-$1
D
0
3
MHT CET 2024 11th May Morning Shift
MCQ (Single Correct Answer)
+2
-0

If in a triangle $A B C$, with usual notations, the angles are in A.P. and $b: c=\sqrt{3}: \sqrt{2}$, then angle $\mathrm{A}=$

A
$30^{\circ}$
B
$60^{\circ}$
C
$75^{\circ}$
D
$45^{\circ}$
4
MHT CET 2024 11th May Morning Shift
MCQ (Single Correct Answer)
+2
-0

The rate of change of the volume of a sphere with respect to its surface area, when its radius is 2 cm , is

A
$0.1 \mathrm{~cm}^3 / \mathrm{cm}^2$
B
$\frac{1}{2} \mathrm{~cm}^3 / \mathrm{cm}^2$
C
$1 \mathrm{~cm}^3 / \mathrm{cm}^2$
D
$2 \mathrm{~cm}^3 / \mathrm{cm}^2$
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