1
BITSAT 2022
MCQ (Single Correct Answer)
+3
-1

$$\left( {{{dy} \over {dx}}} \right)\tan x = y{\sec ^2}x + \sin x$$, find general solution

A
$$y = \tan x(\log |{\mathop{\rm cosec}\nolimits} x - \cot x| + \cos x + c)$$
B
$$y = {\sec ^2}x + \tan x + c$$
C
$$y = \log |\sec x + \tan x| + {\mathop{\rm cosec}\nolimits} \,x + c$$
D
$$y = {\tan ^2}x + \sin x + c$$
2
BITSAT 2022
MCQ (Single Correct Answer)
+3
-1

If the straight line $$y = mx + c$$ touches the parabola $${y^2} - 4ax + 4{a^3} = 0$$, then c is

A
$$am + {a \over m}$$
B
$$am - {a \over m}$$
C
$${a \over m} + {a^2}m$$
D
$${a \over m} - {a^2}m$$
3
BITSAT 2022
MCQ (Single Correct Answer)
+3
-1

A normal is drawn at the point P to the parabola $${y^2} = 8x$$, which is inclined at 60$$^\circ$$ with the straight line $$y = 8$$. Then the point P lies on the straight line

A
$$2x + y - 12 - 4\sqrt 3 = 0$$
B
$$2x - y - 12 + 4\sqrt 3 = 0$$
C
$$2x - y - 12 - 4\sqrt 3 = 0$$
D
None of these
4
BITSAT 2022
MCQ (Single Correct Answer)
+3
-1

The value of $$\int {{1 \over {{{[{{(x - 1)}^3}{{(x + 2)}^5}]}^{{1 \over 4}}}}}dx} $$, is

A
$${4 \over 3}{\left( {{{x + 1} \over {x - 2}}} \right)^{{1 \over 4}}} + C$$
B
$${3 \over 4}{\left( {{{x - 1} \over {x + 2}}} \right)^{{1 \over 4}}} + C$$
C
$${4 \over 3}{\left( {{{x - 1} \over {x + 2}}} \right)^{{1 \over 4}}} + C$$
D
$${1 \over 3}{\left( {{{2x - 1} \over {4x - 3}}} \right)^{{1 \over 4}}} + C$$
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