1
MHT CET 2021 20th September Evening Shift
MCQ (Single Correct Answer)
+2
-0

The joint equation of pair of lines through the origin and making an equilateral triangle with the line $$y=3$$ is

A
$$x^2+3 y^2=0$$
B
$$3 x^2-y^2=0$$
C
$$x^2-3 y^2=0$$
D
$$3 x^2+y^2=0$$
2
MHT CET 2021 20th September Evening Shift
MCQ (Single Correct Answer)
+2
-0

If $$\bar{r}=-4 \hat{i}-6 \hat{j}-2 \hat{k}$$ is a linear combination of the vectors $$\bar{a}=-\hat{i}+4 \hat{j}+3 \hat{k}$$ and $$\bar{b}=-8 \hat{i}-\hat{j}+3 \hat{k}$$, then

A
$$\overline{\mathrm{r}}=\frac{-4}{3} \overline{\mathrm{a}}+\frac{2}{3} \overline{\mathrm{b}}$$
B
$$\overline{\mathrm{r}}=\frac{4}{3} \overline{\mathrm{a}}+\frac{2}{3} \bar{b}$$
C
$$\overline{\mathrm{r}}=\frac{-1}{3} \overline{\mathrm{a}}+\frac{2}{3} \overline{\mathrm{b}}$$
D
$$\overline{\mathrm{r}}=\frac{1}{3} \overline{\mathrm{a}}-\frac{1}{3} \overline{\mathrm{b}}$$
3
MHT CET 2021 20th September Evening Shift
MCQ (Single Correct Answer)
+2
-0

If $$A^{-1}=\left[\begin{array}{cc}2 & -3 \\ -1 & 2\end{array}\right]$$ and $$B^{-1}=\left[\begin{array}{cc}1 & 0 \\ -3 & 1\end{array}\right]$$, then $$(A B)^{-1}=$$

A
$$\left[\begin{array}{cc}2 & 7 \\ 3 & -1\end{array}\right]$$
B
$$\left[\begin{array}{cc}2 & -7 \\ -3 & 11\end{array}\right]$$
C
$$\left[\begin{array}{cc}2 & -3 \\ -7 & 11\end{array}\right]$$
D
$$\left[\begin{array}{cc}2 & 3 \\ 7 & -11\end{array}\right]$$
4
MHT CET 2021 20th September Evening Shift
MCQ (Single Correct Answer)
+2
-0

Two unbiased dice are thrown. Then the probability that neither a doublet nor a total of 10 will appear is

A
$$\frac{1}{12}$$
B
$$\frac{1}{36}$$
C
$$\frac{2}{9}$$
D
$$\frac{7}{9}$$
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