1
BITSAT 2021
MCQ (Single Correct Answer)
+3
-1

Solution of $$\left( {{{x + y - 1} \over {x + y - 2}}} \right){{dy} \over {dx}} = \left( {{{x + y + 1} \over {x + y + 2}}} \right)$$, given that y = 1 when x = 1 is

A
$$\ln \left| {{{{{(x - y)}^2} - 2} \over 2}} \right| = 2(x + y)$$
B
$$\ln \left| {{{{{(x + y)}^2} - 2} \over 2}} \right| = 2(x - y)$$
C
$$\ln \left| {{{{{(x - y)}^2} + 2} \over 2}} \right| = 2(x + y)$$
D
$$\ln \left| {{{{{(x + y)}^2} + 2} \over 2}} \right| = 2(x + y)$$
2
BITSAT 2021
MCQ (Single Correct Answer)
+3
-1

If $$y = \sin \left( {2{{\tan }^{ - 1}}\sqrt {{{1 - x} \over {1 + x}}} } \right)$$, then $${{dy} \over {dx}}$$ is :

A
1
B
$$-$$1
C
$${x \over {\sqrt {{x^2} - 1} }}$$
D
$${{ - x} \over {\sqrt {1 - {x^2}} }}$$
3
BITSAT 2021
MCQ (Single Correct Answer)
+3
-1

The maximum value of the function y = x(x $$-$$ 1)2, is

A
0
B
$${4 \over {27}}$$
C
$$-$$4
D
None of these
4
BITSAT 2021
MCQ (Single Correct Answer)
+3
-1

The solution of $${x^3}{{dy} \over {dx}} + 4{x^2}\tan y = {e^x}\sec y$$ satisfying y (1) = 0, is

A
$$\tan y = (x - 2){e^x}\log x$$
B
$$\sin y = {e^x}(x - 1){x^{ - 4}}$$
C
$$\tan y = (x - 1){e^x}{x^{ - 3}}$$
D
$$\sin y = {e^x}(x - 1){x^3}$$
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