1
MHT CET 2023 13th May Morning Shift
MCQ (Single Correct Answer)
+2
-0

A line drawn from the point $$\mathrm{A}(1,3,2)$$ parallel to the line $$\frac{x}{2}=\frac{y}{4}=\frac{z}{1}$$, intersects the plane $$3 x+y+2 z=5$$ in point $$\mathrm{B}$$, then co-ordinates of point $$\mathrm{B}$$ are

A
$$\left(\frac{1}{6}, \frac{4}{3}, \frac{19}{12}\right)$$
B
$$\left(-\frac{1}{6},-\frac{4}{3}, \frac{19}{12}\right)$$
C
$$\left(\frac{1}{6}, \frac{4}{3},-\frac{19}{12}\right)$$
D
$$\left(-\frac{1}{6},-\frac{4}{3},-\frac{19}{12}\right)$$
2
MHT CET 2023 13th May Morning Shift
MCQ (Single Correct Answer)
+2
-0

The value of $$\frac{\mathrm{i}^{248}+\mathrm{i}^{246}+\mathrm{i}^{244}+\mathrm{i}^{242}+\mathrm{i}^{240}}{\mathrm{i}^{249}+\mathrm{i}^{247}+\mathrm{i}^{245}+\mathrm{i}^{243}+\mathrm{i}^{241}}, (\mathrm{i}=\sqrt{-1})$$ is

A
i
B
1
C
$$-1$$
D
$$-$$i
3
MHT CET 2023 13th May Morning Shift
MCQ (Single Correct Answer)
+2
-0

If $$\mathrm{f}(x)=\int \frac{x^2 \mathrm{~d} x}{\left(1+x^2\right)\left(1+\sqrt{1+x^2}\right)}$$ and $$\mathrm{f}(0)=0$$, then $$\mathrm{f}(1)$$ is

A
$$\log (1+\sqrt{2})$$
B
$$\log (1+\sqrt{2})-\frac{\pi}{4}$$
C
$$\log (1+\sqrt{2})+\frac{\pi}{4}$$
D
$$\log (1-\sqrt{2})$$
4
MHT CET 2023 13th May Morning Shift
MCQ (Single Correct Answer)
+2
-0

A line $$\mathrm{L}_1$$ passes through the point, whose p. v. (position vector) $$3 \hat{i}$$, is parallel to the vector $$-\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}$$. Another line $$\mathrm{L}_2$$ passes through the point having p.v. $$\hat{i}+\hat{j}$$ is parallel to vector $$\hat{i}+\hat{k}$$, then the point of intersection of lines $$L_1$$ and $$L_2$$ has p.v.

A
$$2 \hat{i}+2 \hat{j}+\hat{k}$$
B
$$2 \hat{i}+\hat{j}+\hat{k}$$
C
$$2 \hat{\mathrm{i}}-\hat{\mathrm{j}}-\hat{\mathrm{k}}$$
D
$$2 \hat{i}-2 \hat{j}+\hat{k}$$
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