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1
JEE Main 2018 (Online) 15th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
If $$\overrightarrow a ,\,\,\overrightarrow b ,$$ and $$\overrightarrow C $$ are unit vectors such that $$\overrightarrow a + 2\overrightarrow b + 2\overrightarrow c = \overrightarrow 0 ,$$ then $$\left| {\overrightarrow a \times \overrightarrow c } \right|$$ is equal to :
A
$${{\sqrt {15} } \over 4}$$
B
$${{1} \over {4}}$$
C
$${{15} \over {16}}$$
D
$${{\sqrt {15} } \over 16}$$
2
JEE Main 2018 (Online) 15th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
If the tangents drawn to the hyperbola 4y2 = x2 + 1 intersect the co-ordinate axes at the distinct points A and B then the locus of the mid point of AB is :
A
x2 $$-$$ 4y2 + 16x2y2 = 0
B
x2 $$-$$ 4y2 $$-$$ 16x2y2 = 0
C
4x2 $$-$$ y2 + 16x2y2 = 0
D
4x2 $$-$$ y2 $$-$$ 16x2y2 = 0
3
JEE Main 2018 (Online) 15th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
If $$f\left( {{{x - 4} \over {x + 2}}} \right) = 2x + 1,$$ (x $$ \in $$ R $$-$${1, $$-$$ 2}), then $$\int f \left( x \right)dx$$ is equal to :
(where C is a constant of integration)
A
12 loge | 1 $$-$$ x | + 3x + C
B
$$-$$ 12 loge | 1 $$-$$ x | $$-$$ 3x + C
C
12 loge | 1 $$-$$ x | $$-$$ 3x + C
D
$$-$$ 12 loge | 1 $$-$$ x | + 3x + C
4
JEE Main 2018 (Online) 15th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
Let y = y(x) be the solution of the differential equation $${{dy} \over {dx}} + 2y = f\left( x \right),$$

where $$f\left( x \right) = \left\{ {\matrix{ {1,} & {x \in \left[ {0,1} \right]} \cr {0,} & {otherwise} \cr } } \right.$$

If y(0) = 0, then $$y\left( {{3 \over 2}} \right)$$ is :
A
$${{{e^2} + 1} \over {2{e^4}}}$$
B
$${1 \over {2e}}$$
C
$${{{e^2} - 1} \over {{e^3}}}$$
D
$${{{e^2} - 1} \over {2{e^3}}}$$

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