1
MHT CET 2021 24th September Morning Shift
MCQ (Single Correct Answer)
+2
-0

The general solution of the differential equation $$(2 y-1) d x-(2 x+3) d y=0$$ is

A
$$(2 x+3)^2=c(2 y-1)$$
B
$$\frac{2 x+3}{2 y-1}=c$$
C
$$(2 x+3)(2 y-1)=c$$
D
$$(2 x+3)(2 y-1)^2=c$$
2
MHT CET 2021 24th September Morning Shift
MCQ (Single Correct Answer)
+2
-0

If $$y=\log _{10} x+\log _x 10+\log _x x+\log _{10} 10$$, then $$\frac{d y}{d x}=$$

A
$$\frac{1}{x \log _e 10}+\frac{1}{x \log _{10} e}$$
B
$$\frac{1}{x \log _e 10}+\frac{\log _e 10}{x\left(\log _{10} e\right)^2}$$
C
$$\frac{1}{x \log _e 10}-\frac{1}{x \log _{10} e}$$
D
$$\frac{1}{x \log _e 10}-\frac{\log _e 10}{x\left(\log _e x\right)^2}$$
3
MHT CET 2021 24th September Morning Shift
MCQ (Single Correct Answer)
+2
-0

If the vectors $$\vec{a}=2 \hat{i}+p \hat{j}+4 \hat{k}$$ and $$\vec{b}=6 \hat{i}-9 \hat{j}+q \hat{k}$$ are collinear, then $$p$$ and $$q$$ are

A
$$\mathrm{p=3, q=-12}$$
B
$$\mathrm{p}=3, \mathrm{q}=12$$
C
$$\mathrm{p}=-3, \mathrm{q}=12$$
D
$$\mathrm{p}=-3, \mathrm{q}=-12$$
4
MHT CET 2021 24th September Morning Shift
MCQ (Single Correct Answer)
+2
-0

If $$y=\tan ^{-1}\left[\frac{\log \left(\frac{e}{x^2}\right)}{\log \left(e x^2\right)}\right]+\tan ^{-1}\left[\frac{3+2 \log x}{1-6 \log x}\right]$$, then $$\frac{d^2 y}{d x^2}=$$

A
$$\frac{2}{1+x^2}$$
B
$$\frac{1}{1+x^2}$$
C
$$\frac{3}{1+x^2}$$
D
0
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