1
KCET 2019
MCQ (Single Correct Answer)
+1
-0

If $$3 A+4 B^{\prime}=\left[\begin{array}{ccc}7 & -10 & 17 \\ 0 & 6 & 31\end{array}\right]$$ and $$2 B+3 A^{\prime}\left[\begin{array}{cc}-1 & 18 \\ 4 & 0 \\ -5 & -7\end{array}\right]$$ then $$B=$$

A
$$\left[\begin{array}{cc}-1 & -18 \\ 4 & -16 \\ -5 & -7\end{array}\right]$$
B
$$\left[\begin{array}{cc}1 & 3 \\ -1 & 1 \\ 2 & 4\end{array}\right]$$
C
$$\left[\begin{array}{cc}1 & 3 \\ -1 & 1 \\ 2 & -4\end{array}\right]$$
D
$$\left[\begin{array}{cc}1 & -3 \\ -1 & 1 \\ 2 & 4\end{array}\right]$$
2
KCET 2019
MCQ (Single Correct Answer)
+1
-0

If $$A=\left[\begin{array}{ll}1 & 3 \\ 4 & 2\end{array}\right], B=\left[\begin{array}{cc}2 & -1 \\ -1 & 2\end{array}\right]$$, Then $$\left|A B B^{\prime}\right|=$$

A
100
B
50
C
250
D
$$-$$250
3
KCET 2019
MCQ (Single Correct Answer)
+1
-0

If the value of a third order determinant is 16, then the value of the determinant formed by replacing each of its elements by its cofactor is

A
256
B
96
C
16
D
48
4
KCET 2019
MCQ (Single Correct Answer)
+1
-0

$$\int x^3 \sin 3 x d x=$$

A
$$-\frac{x^3 \cdot \cos 3 x}{3}+\frac{x^2 \sin 3 x}{3}+\frac{2 x \cos 3 x}{9} -\frac{2 \sin 3 x}{27}+C$$
B
$$-\frac{x^3 \cdot \cos 3 x}{3}-\frac{x^2 \sin 3 x}{3}+\frac{2 x \cos 3 x}{9} -\frac{2 \sin 3 x}{27}+C$$
C
$$-\frac{x^3 \cdot \cos 3 x}{3}+\frac{x^2 \sin 3 x}{3}-\frac{2 x \cos 3 x}{9} -\frac{2 \sin 3 x}{27}+C$$
D
$$\frac{x^3 \cdot \cos 3 x}{3}+\frac{x^2 \sin 3 x}{3}-\frac{2 x \cos 3 x}{9} -\frac{2 \sin 3 x}{27}+C$$
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