MHT CET 2021 22th September Morning Shift | MHT CET Year Wise Previous Years Questions

MHT CET 2021 22th September Morning Shift

MCQ (Single Correct Answer)

-0

If $$A=\left[\begin{array}{lll}1 & 0 & 1 \\ 0 & 2 & 3 \\ 1 & 2 & 1\end{array}\right]$$, then the value of determinant of $$A^{-1}$$ is

$$-6$$

$$\frac{-1}{6}$$

$$\frac{1}{36}$$

$$36$$

MHT CET 2021 22th September Morning Shift

MCQ (Single Correct Answer)

-0

Solution of the differential equation $$\mathrm{y'=\frac{(x^2+y^2)}{xy}}$$, where y(1) = $$-$$2 is given by

$$y^2=4 x^2 \log x^2+x^2$$

$$y^2=x^2 \log x-x^2$$

$$y^2=x \log x^2+4 x^2$$

$$\mathrm{y}^2=\mathrm{x}^2 \log \mathrm{x}^2+4 \mathrm{x}^2$$

MHT CET 2021 22th September Morning Shift

MCQ (Single Correct Answer)

-0

The Cartesian equation of a line is $$3 x+1=6 y-2=1-z$$, then its vector equation is

$$\bar{\mathrm{r}}=\left(\frac{-1}{3} \hat{\mathrm{i}}+\frac{1}{3} \hat{\mathrm{j}}+\hat{\mathrm{k}}\right)+\lambda(2 \hat{\mathrm{i}}-\hat{\mathrm{j}}-6 \hat{\mathrm{k}})$$

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

$$\bar{r}=\left(\frac{-1}{3} \hat{i}+\frac{1}{3} \hat{j}+\hat{k}\right)+\lambda(2 \hat{i}-\hat{j}+6 \hat{k})$$

$$\bar{\mathrm{r}}=\left(\frac{-1}{3} \hat{\mathrm{i}}+\frac{1}{3} \hat{\mathrm{j}}+\hat{\mathrm{k}}\right)+\lambda(2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-6 \hat{\mathrm{k}})$$

MHT CET 2021 22th September Morning Shift

MCQ (Single Correct Answer)

-0

The position vector of the point of inersection of the medians of a triangle, whose vertices are $$A(1,2,3), B(1,0,3)$$ and $$C(4,1,-3)$$ is

$$6 \hat{i}+3 \hat{j}+3 \hat{k}$$

$$2 \hat{i}+\hat{j}+\hat{k}$$

$$3 \hat{i}+\hat{j}+\hat{k}$$

$$\hat{i}+\hat{j}+\hat{k}$$

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