1
COMEDK 2024 Afternoon Shift
MCQ (Single Correct Answer)
+1
-0

If $$A=\left[\begin{array}{lll}5 & 0 & 4 \\ 2 & 3 & 2 \\ 1 & 2 & 1\end{array}\right] \quad B^{-1}=\left[\begin{array}{lll}1 & 3 & 3 \\ 1 & 4 & 3 \\ 1 & 3 & 4\end{array}\right]$$ then $$(A B)^{-1}$$ is equal to

A
$$ \left[\begin{array}{ccc} -2 & 19 & -27 \\ -2 & 18 & -25 \\ 3 & 29 & 42 \end{array}\right] $$
B
$$ \left[\begin{array}{lll} -2 & 19 & -27 \\ -2 & 18 & -25 \\ -3 & 29 & -42 \end{array}\right] $$
C
$$ \left[\begin{array}{ccc} -2 & -2 & -3 \\ 19 & 18 & 29 \\ -27 & -25 & -42 \end{array}\right] $$
D
$$ \left[\begin{array}{lll} 2 & -19 & 27 \\ 2 & -18 & 25 \\ 3 & -29 & 42 \end{array}\right] $$
2
COMEDK 2024 Afternoon Shift
MCQ (Single Correct Answer)
+1
-0

Degree of the differential equation $$\log \left(\frac{d y}{d x}\right)^{\frac{1}{2}}=5 x+4 y$$ is

A
Not defined
B
4
C
1
D
2
3
COMEDK 2024 Afternoon Shift
MCQ (Single Correct Answer)
+1
-0

$$ \text { The area of the region (in square units) bounded by the line } y+3=x ; x=1 \text { and } x=5 \text { is } $$

A
2
B
32
C
24
D
4
4
COMEDK 2024 Afternoon Shift
MCQ (Single Correct Answer)
+1
-0

If an ellipse has an equation in the standard form and it passes through the points $$\left(\frac{5}{2}, \frac{\sqrt{6}}{4}\right)$$ and $$\left(-2, \frac{\sqrt{15}}{5}\right)$$ then the length of its latus rectum is

A
$$\frac{1}{10}$$
B
$$\frac{1}{\sqrt{10}}$$
C
$$\sqrt{\frac{10}{5}}$$
D
$$\frac{\sqrt{10}}{5}$$
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