1
IIT-JEE 2002 Screening
MCQ (Single Correct Answer)
+2
-0.5
The point(s) in the curve $${y^3} + 3{x^2} = 12y$$ where the tangent is vertical, is (are)
A
$$\left( { \pm {4 \over {\sqrt 3 }}, - 2} \right)$$
B
$$\left( { \pm \sqrt {{{11} \over 3}} ,1} \right)$$
C
$$(0,0)$$
D
$$\left( { \pm {4 \over {\sqrt 3 }}, 2} \right)$$
2
IIT-JEE 2002 Screening
MCQ (Single Correct Answer)
+3
-0.75
The area bounded by the curves $$y = \left| x \right| - 1$$ and $$y = - \left| x \right| + 1$$ is
A
$$1$$
B
$$2$$
C
$$2\sqrt 2 $$
D
$$4$$
3
IIT-JEE 2002 Screening
MCQ (Single Correct Answer)
+3
-0.75
The integral $$\int\limits_{ - 1/2}^{1/2} {\left( {\left[ x \right] + \ell n\left( {{{1 + x} \over {1 - x}}} \right)} \right)dx} $$ equal to
A
$$ - {1 \over 2}$$
B
$$0$$
C
$$1$$
D
$$2\ell n\left( {{1 \over 2}} \right)$$
4
IIT-JEE 2002 Screening
MCQ (Single Correct Answer)
+3
-0.75
Let $$f\left( x \right) = \int\limits_1^x {\sqrt {2 - {t^2}} \,dt.} $$ Then the real roots of the equation
$${x^2} - f'\left( x \right) = 0$$ are
A
$$ \pm 1$$
B
$$ \pm {1 \over {\sqrt 2 }}$$
C
$$ \pm {1 \over 2}$$
D
$$0$$ and $$1$$
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