1
MHT CET 2021 22th September Morning Shift
MCQ (Single Correct Answer)
+2
-0

The general solution of $$\sin ^{-1}\left(\frac{d y}{d x}\right)=x+y$$ is

A
$$\tan (x+y)-\sec (x+y)=x^2+c$$
B
$$\tan (x+y)+\sec (x+y)=x^2+c$$
C
$$\tan (x+y)+\sec (x+y)=x+c$$
D
$$\tan (x+y)-\sec (x+y)=x+c$$
2
MHT CET 2021 22th September Morning Shift
MCQ (Single Correct Answer)
+2
-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

A
$$-6$$
B
$$\frac{-1}{6}$$
C
$$\frac{1}{36}$$
D
$$36$$
3
MHT CET 2021 22th September Morning Shift
MCQ (Single Correct Answer)
+2
-0

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

A
$$y^2=4 x^2 \log x^2+x^2$$
B
$$y^2=x^2 \log x-x^2$$
C
$$y^2=x \log x^2+4 x^2$$
D
$$\mathrm{y}^2=\mathrm{x}^2 \log \mathrm{x}^2+4 \mathrm{x}^2$$
4
MHT CET 2021 22th September Morning Shift
MCQ (Single Correct Answer)
+2
-0

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

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