MHT CET 2020 19th October Evening Shift | MHT CET Year Wise Previous Years Questions

MHT CET 2020 19th October Evening Shift

MCQ (Single Correct Answer)

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The equation of a plane containing the point $(1,-1,2)$ and perpendicular to the planes $2 x+3 y-2 z=5$ and $x+2 y-3 z=8$ is

$r(4 \hat{\mathbf{i}}-5 \hat{\mathbf{j}}+3 \hat{\mathbf{k}})=15$

$\mathbf{r}(5 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}+2 \hat{\mathbf{k}})=5$

$\mathbf{r}(5 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}-\hat{\mathbf{k}})=5$

$\mathbf{r}(5 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}-\hat{\mathbf{k}})=7$

MHT CET 2020 19th October Evening Shift

MCQ (Single Correct Answer)

-0

The general solution of the differential equation $\left(1+y^2\right)+\left(x-e^{\tan ^{-1} y}\right) \frac{d y}{d x}=0$ is

$x \cdot e^{\tan ^{-1} y}=\frac{\left(e^{\tan ^{-1} y}\right)^2}{2}+C$

$e^{\tan ^{-1} y}=\left(e^{\tan ^{-1} x}\right)^2+C$

$x \cdot e^{\tan ^{-1} y}=\frac{\left(e^{\tan ^{-1} x} x\right)^2}{2}+C$

$e^{\tan ^{-1} y}=\left(e^{\tan ^{-1} y}\right)^2+C$

MHT CET 2020 19th October Evening Shift

MCQ (Single Correct Answer)

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The equation of the line passing through $(1,2,3)$ and perpendicular to the lines $x-1=\frac{y+2}{2}=\frac{z+4}{4}$ and $\frac{x-1}{2}=\frac{y-2}{2}=z+3$ is

$x-1=\frac{y-2}{2}=\frac{z-3}{4}$

$\frac{x-1}{4}=\frac{2-y}{5}=\frac{z-3}{2}$

$\frac{x-1}{6}=\frac{y-2}{7}=\frac{z-3}{2}$

$\frac{x-1}{6}=\frac{2-y}{7}=\frac{z-3}{2}$

MHT CET 2020 19th October Evening Shift

MCQ (Single Correct Answer)

-0

The cofactors of the elements of the first column of the matrix $A=\left[\begin{array}{ccc}2 & 0 & -1 \\ 3 & 1 & 2 \\ -1 & 1 & 2\end{array}\right]$ are

$0,-7,2$

$-1,3,-2$

$0,-8,4$

$0,-1,1$

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