1
MHT CET 2024 4th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

The particular solution of differential equation $\left(1+y^2\right)(1+\log x) \mathrm{d} x+x \mathrm{~d} y=0$ at $x=1, y=1$ is

A
$\log x-\frac{1}{2}(\log x)^2-\tan ^{-1} y=-\frac{\pi}{4}$
B
$\log x+\frac{1}{2}(\log x)^2+\tan ^{-1} y=\frac{\pi}{4}$
C
$\log x-\frac{1}{2}(\log x)^2+\tan ^{-1} y=\frac{\pi}{4}$
D
$\log x+\frac{1}{2}(\log x)^2-\tan ^{-1} y=\frac{\pi}{4}$
2
MHT CET 2024 4th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

If $\lim _\limits{x \rightarrow 1} \frac{x^2-a x+b}{x-1}=7$, then $a+b$ is equal to

A
$-$1
B
1
C
$-$11
D
11
3
MHT CET 2024 4th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

A ladder 5 m long rests against a vertical wall. If its top slides downwards at the rate of $10 \mathrm{~cm} / \mathrm{sec}$., then the foot of the ladder is sliding at the rate of _________ $\mathrm{m} / \mathrm{sec}$., when it is 4 m away from the wall.

A
0.75
B
7.5
C
0.0075
D
0.075
4
MHT CET 2024 4th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

If $\mathrm{f}(x)=\frac{x}{2-x}, \mathrm{~g}(x)=\frac{x+1}{x+2}$, then (gogof) $(x)=$

A
$\frac{6+x}{10-2 x}$
B
$\frac{6-x}{10+2 x}$
C
$\frac{6+x}{10+2 x}$
D
$\frac{6-x}{10-2 x}$

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