1
MHT CET 2024 10th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

$\overline{\mathrm{a}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}, \overline{\mathrm{b}}=4 \hat{\mathrm{i}}-2 \hat{j}+3 \hat{k}, \overline{\mathrm{c}}=\hat{i}-2 \hat{j}+\hat{k}$, then $a$ vector of magnitude 6 units, which is parallel to the vector $2 \bar{a}-\bar{b}+3 c$, is

A
$2 \hat{i}-4 \hat{j}+4 \hat{k}$
B
$\hat{i}-\hat{j}+2 \hat{k}$
C
$4 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}-2 \hat{\mathrm{k}}$
D
$2 \hat{i}+4 \hat{j}+4 \hat{k}$
2
MHT CET 2024 10th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

If $y=\sin ^2\left(\cot ^{-1} \sqrt{\frac{1+x}{1-x}}\right)$, then $\frac{\mathrm{d} y}{\mathrm{~d} x}$ has the value

A
$\frac{-1}{2}$
B
$\frac{1}{2}$
C
$-1$
D
$1$
3
MHT CET 2024 10th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

If the line $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-\mathrm{k}}{2}=\frac{\mathrm{z}}{1}$ intersect, then the value of k is

A
$\frac{3}{2}$
B
$\frac{9}{2}$
C
$-\frac{2}{9}$
D
$-\frac{3}{2}$
4
MHT CET 2024 10th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

Inverse of the matrix $\left[\begin{array}{cc}0.8 & -0.6 \\ 0.6 & 0.8\end{array}\right]$ is

A
$\left[\begin{array}{cc}0.8 & 0.6 \\ -0.6 & 0.8\end{array}\right]$
B
$\left[\begin{array}{ll}-0.8 & 0.6 \\ -0.6 & 0.8\end{array}\right]$
C
$\left[\begin{array}{cc}-0.8 & -0.6 \\ 0.6 & 0.8\end{array}\right]$
D
$\left[\begin{array}{cc}8 & -6 \\ 6 & 8\end{array}\right]$
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