1
JEE Main 2013 (Offline)
MCQ (Single Correct Answer)
+4
-1
If $$\int {f\left( x \right)dx = \psi \left( x \right),} $$ then $$\int {{x^5}f\left( {{x^3}} \right)dx} $$ is equal to
A
$${1 \over 3}\left[ {{x^3}\psi \left( {{x^3}} \right) - \int {{x^2}\psi \left( {{x^3}} \right)dx} } \right] + C$$
B
$${1 \over 3}{x^3}\psi \left( {{x^3}} \right) - 3\int {{x^3}\psi \left( {{x^3}} \right)dx} + C$$
C
$${1 \over 3}{x^3}\psi \left( {{x^3}} \right) - \int {{x^2}\psi \left( {{x^3}} \right)dx} + C$$
D
$${1 \over 3}\left[ {{x^3}\psi \left( {{x^3}} \right) - \int {{x^3}\psi \left( {{x^3}} \right)dx} } \right] + C$$
2
JEE Main 2013 (Offline)
MCQ (Single Correct Answer)
+4
-1
If $$x, y, z$$ are in A.P. and $${\tan ^{ - 1}}x,{\tan ^{ - 1}}y$$ and $${\tan ^{ - 1}}z$$ are also in A.P., then :
A
$$x=y=z$$
B
$$2x=3y=6z$$
C
$$6x=3y=2z$$
D
$$6x=4y=3z$$
3
JEE Main 2013 (Offline)
MCQ (Single Correct Answer)
+4
-1
If $$y = \sec \left( {{{\tan }^{ - 1}}x} \right),$$ then $${{{dy} \over {dx}}}$$ at $$x=1$$ is equal to :
A
$${1 \over {\sqrt 2 }}$$
B
$${1 \over 2}$$
C
$$1$$
D
$$\sqrt 2 $$
4
JEE Main 2013 (Offline)
MCQ (Single Correct Answer)
+4
-1
The equation of the circle passing through the foci of the ellipse $${{{x^2}} \over {16}} + {{{y^2}} \over 9} = 1$$, and having centre at $$(0,3)$$ is :
A
$${x^2} + {y^2} - 6y - 7 = 0$$
B
$${x^2} + {y^2} - 6y + 7 = 0$$
C
$${x^2} + {y^2} - 6y - 5 = 0$$
D
$${x^2} + {y^2} - 6y + 5 = 0$$
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