1
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
$$\bar{\mathrm{r}}=\left(\frac{-1}{3} \hat{\mathrm{i}}+\frac{1}{3} \hat{\mathrm{j}}+\hat{\mathrm{k}}\right)+\lambda(2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-6 \hat{\mathrm{k}})$$
2
MHT CET 2021 22th September Morning Shift
MCQ (Single Correct Answer)
+2
-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

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

$$\int\limits_{ - \pi }^\pi {{{x\sin x} \over {1 + {{\cos }^2}x}}dx = } $$

A
$${{{\pi ^2}} \over 2}$$
B
$${\pi ^2}$$
C
$${{{\pi ^2}} \over 4}$$
D
$$3\pi $$
4
MHT CET 2021 22th September Morning Shift
MCQ (Single Correct Answer)
+2
-0

$$\int {{e^x}\left( {{{x - 1} \over {{x^2}}}} \right)dx = } $$

A
$${{ - {e^x}} \over {{x^2}}} + c$$
B
$${{ - {e^x}} \over x} + c$$
C
$${{{e^x}} \over {{x^2}}} + c$$
D
$${{{e^x}} \over x} + c$$
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