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

$$\int {{1 \over {1 - 2\sin x}}dx} $$ is equal to

A
$${1 \over {2\sqrt 3 }}\log \left| {{{\tan {x \over 2} - 2 - \sqrt 3 } \over {\tan {x \over 2} - 2 + \sqrt 3 }}} \right| + c$$
B
$${{\sqrt 3 } \over 2}\log \left| {{{\tan {x \over 2} - 2 - \sqrt 3 } \over {\tan {x \over 2} - 2 + \sqrt 3 }}} \right| + c$$
C
$${1 \over {\sqrt 3 }}\log \left| {{{\tan {x \over 2} - 2 - \sqrt 3 } \over {\tan {x \over 2} - 2 + \sqrt 3 }}} \right| + c$$
D
None of the above
2
BITSAT 2021
MCQ (Single Correct Answer)
+3
-1

$$\int\limits_0^1 {{{\log (1 + x)} \over {1 + {x^2}}}dx} $$ is equal to :

A
$${\pi \over 8}\log 2$$
B
$${\pi \over 8}\log {1 \over 2}$$
C
$${\pi \over 4}\log 2$$
D
None of these
3
BITSAT 2021
MCQ (Single Correct Answer)
+3
-1

The area of one curvilinear triangle formed by curves y = sin x, y = cos x and X-axis, is

A
2 sq units
B
(2 + $$\sqrt2$$) sq units
C
(2 $$-$$ $$\sqrt2$$) sq units
D
None of these
4
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
$$\ln \left| {{{{{(x + y)}^2} + 2} \over 2}} \right| = 2(x + y)$$
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