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1
JEE Main 2016 (Online) 10th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
Let a, b $$ \in $$ R, (a $$ \ne $$ 0). If the function f defined as

$$f\left( x \right) = \left\{ {\matrix{ {{{2{x^2}} \over a}\,\,,} & {0 \le x < 1} \cr {a\,\,\,,} & {1 \le x < \sqrt 2 } \cr {{{2{b^2} - 4b} \over {{x^3}}},} & {\sqrt 2 \le x < \infty } \cr } } \right.$$

is continuous in the interval [0, $$\infty $$), then an ordered pair ( a, b) is :
A
$$\left( {\sqrt 2 ,1 - \sqrt 3 } \right)$$
B
$$\left( { - \sqrt 2 ,1 + \sqrt 3 } \right)$$
C
$$\left( {\sqrt 2 , - 1 + \sqrt 3 } \right)$$
D
$$\left( { - \sqrt 2 ,1 - \sqrt 3 } \right)$$
2
JEE Main 2016 (Online) 10th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
A ray of light is incident along a line which meets another line, 7x − y + 1 = 0, at the point (0, 1). The ray is then reflected from this point along the line, y + 2x = 1. Then the equation of the line of incidence of the ray of light is :
A
41x − 38y + 38 = 0
B
41x + 25y − 25 = 0
C
41x + 38y − 38 = 0
D
41x − 25y + 25 = 0
3
JEE Main 2016 (Online) 10th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
Let ABC be a triangle whose circumcentre is at P. If the position vectors of A, B, C and P are $$\overrightarrow a ,\overrightarrow b ,\overrightarrow c $$ and $${{\overrightarrow a + \overrightarrow b + \overrightarrow c } \over 4}$$ respectively, then the position vector of the orthocentre of this triangle, is :
A
$${\overrightarrow a + \overrightarrow b + \overrightarrow c }$$
B
$$ - \left( {{{\overrightarrow a + \overrightarrow b + \overrightarrow c } \over 2}} \right)$$
C
$$\overrightarrow 0 $$
D
$$\left( {{{\overrightarrow a + \overrightarrow b + \overrightarrow c } \over 2}} \right)$$
4
JEE Main 2016 (Online) 10th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
The solution of the differential equation

$${{dy} \over {dx}}\, + \,{y \over 2}\,\sec x = {{\tan x} \over {2y}},\,\,$$

where 0 $$ \le $$ x < $${\pi \over 2}$$, and y (0) = 1, is given by :
A
y = 1 $$-$$ $${x \over {\sec x + \tan x}}$$
B
y2 = 1 + $${x \over {\sec x + \tan x}}$$
C
y2 = 1 $$-$$ $${x \over {\sec x + \tan x}}$$
D
y = 1 + $${x \over {\sec x + \tan x}}$$

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