1
JEE Main 2019 (Online) 11th January Evening Slot
+4
-1
If   $$\int {{{x + 1} \over {\sqrt {2x - 1} }}} \,dx$$ = f(x) $$\sqrt {2x - 1}$$ + C, where C is a constant of integration, then f(x) is equal to :
A
$${2 \over 3}$$ (x $$-$$ 4)
B
$${1 \over 3}$$ (x + 4)
C
$${1 \over 3}$$ (x + 1)
D
$${2 \over 3}$$ (x + 2)
2
JEE Main 2019 (Online) 11th January Morning Slot
+4
-1
If  $$\int {{{\sqrt {1 - {x^2}} } \over {{x^4}}}}$$ dx = A(x)$${\left( {\sqrt {1 - {x^2}} } \right)^m}$$ + C, for a suitable chosen integer m and a function A(x), where C is a constant of integration, then (A(x))m equals :
A
$${1 \over {27{x^6}}}$$
B
$${{ - 1} \over {27{x^9}}}$$
C
$${1 \over {9{x^4}}}$$
D
$${1 \over {3{x^3}}}$$
3
JEE Main 2019 (Online) 10th January Evening Slot
+4
-1
If  $$\int \,$$x5.e$$-$$4x3 dx = $${1 \over {48}}$$e$$-$$4x3 f(x) + C, where C is a constant of inegration, then f(x) is equal to -
A
$$-$$2x3 $$-$$ 1
B
$$-$$ 2x3 + 1
C
4x3 + 1
D
$$-$$4x3 $$-$$ 1
4
JEE Main 2019 (Online) 10th January Morning Slot
+4
-1
Let n $$\ge$$ 2 be a natural number and $$0 < \theta < {\pi \over 2}.$$ Then $$\int {{{{{\left( {{{\sin }^n}\theta - \sin \theta } \right)}^{1/n}}\cos \theta } \over {{{\sin }^{n + 1}}\theta }}} \,d\theta$$ is equal to - (where C is a constant of integration)
A
$${n \over {{n^2} - 1}}{\left( {1 + {1 \over {{{\sin }^{n - 1}}\theta }}} \right)^{{{n + 1} \over n}}} + C$$
B
$${n \over {{n^2} - 1}}{\left( {1 - {1 \over {{{\sin }^{n + 1}}\theta }}} \right)^{{{n + 1} \over n}}} + C$$
C
$${n \over {{n^2} - 1}}{\left( {1 - {1 \over {{{\sin }^{n - 1}}\theta }}} \right)^{{{n + 1} \over n}}} + C$$
D
$${n \over {{n^2} + 1}}{\left( {1 - {1 \over {{{\sin }^{n - 1}}\theta }}} \right)^{{{n + 1} \over n}}} + C$$
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