MHT CET 2024 15th May Evening Shift | MHT CET Year Wise Previous Years Questions

MHT CET 2024 15th May Evening Shift

MCQ (Single Correct Answer)

-0

The equation of a line passing through the point $(2,-1,1)$ and parallel to the line joining the points $\hat{i}+2 \hat{j}+2 \hat{k}$ and $-\hat{i}+4 \hat{j}+\hat{k}$ is

$\bar{r}=(2 \hat{i}-\hat{j}+\hat{k})+\lambda(-2 \hat{i}+2 \hat{j}-\hat{k})$

$\bar{r}=(2 \hat{i}-\hat{j}+\hat{k})+\lambda(2 \hat{i}+6 \hat{j}+3 \hat{k})$

$\bar{r}=(2 \hat{i}-\hat{j}+\hat{k})+\lambda(2 \hat{i}-2 \hat{j}-\hat{k})$

$\overline{\mathrm{r}}=(2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}})+\lambda(2 \hat{\mathrm{i}}-6 \hat{\mathrm{j}}-3 \hat{\mathrm{k}})$

MHT CET 2024 15th May Evening Shift

MCQ (Single Correct Answer)

-0

The foot of the perpendicular drawn from origin to a plane is $\mathrm{M}(2,1,-2)$, then vector equation of the plane is

$\overline{\mathrm{r}} \cdot(2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}})=9$

$\overline{\mathrm{r}} \cdot(-2 \hat{\mathrm{i}}-\hat{\mathrm{j}}-2 \hat{\mathrm{k}})=7$

$\bar{r} \cdot(2 \hat{i}-\hat{j}-2 \hat{k})=9$

$\overline{\mathrm{r}} \cdot(2 \hat{\mathrm{i}}-\hat{\mathrm{j}}-\hat{\mathrm{k}})=7$

MHT CET 2024 15th May Evening Shift

MCQ (Single Correct Answer)

-0

If $f(1)=1, f^{\prime}(1)=3$, then the derivative of $\mathrm{f}(\mathrm{f}(\mathrm{f}(x)))+(\mathrm{f}(x))^2$ at $x=1$ is

MHT CET 2024 15th May Evening Shift

MCQ (Single Correct Answer)

-0

In a triangle ABC , with usual notations, $\frac{\cos \mathrm{B}+\cos \mathrm{C}}{\mathrm{b}+\mathrm{c}}+\frac{\cos \mathrm{A}}{\mathrm{a}}$ has the value

$\frac{1}{\mathrm{~b}+\mathrm{c}}$

$\frac{1}{\mathrm{~b}}$

$\frac{1}{\mathrm{c}}$

$\frac{1}{\mathrm{a}}$

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