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

The solution of $$(1+x y) y d x+(1-x y) x d y=0$$ is

A
$$\log \left(\frac{x}{y}\right)+\frac{1}{x y}=k$$, where $$k$$ is constant of integration
B
$$\log \left(\frac{x}{y}\right)=\frac{1}{x y}+k$$, where $$k$$ is constant of integration
C
$$\log \left(\frac{x}{y}\right)+{x y}=k$$, where $$k$$ is constant of integration
D
$$\log \left(\frac{x}{y}\right)={x y}+k$$, where $$k$$ is constant of integration
2
MHT CET 2023 13th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

The integral $$\int_\limits{\pi / 6}^{\pi / 3} \sec ^{\frac{2}{3}} x \operatorname{cosec}^{\frac{4}{3}} x d x$$ is equal to

A
$$3^{\frac{5}{6}}-3^{\frac{2}{3}}$$
B
$$3^{\frac{7}{6}}-3^{\frac{5}{6}}$$
C
$$3^{\frac{5}{3}}-3^{\frac{1}{3}}$$
D
$$3^{\frac{4}{3}}-3^{\frac{1}{3}}$$
3
MHT CET 2023 13th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

$$\lim _\limits{x \rightarrow 0} \frac{\sqrt{1+x \sin x}-\sqrt{\cos x}}{\tan ^2 \frac{x}{2}}=$$

A
1
B
2
C
3
D
$$-$$1
4
MHT CET 2023 13th May Evening Shift
MCQ (Single Correct Answer)
+2
-0

$$A(1,-3), B(4,3)$$ are two points on the curve $$y=x-\frac{4}{x}$$. The points on the curve, the tangents at which are parallel to the chord $$A B$$, are

A
$$(1,2),(-1,-2)$$
B
$$(2,0),(-2,0)$$
C
$$(0,2),(1,-2)$$
D
$$(3,2),(-3,1)$$
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