1
JEE Main 2017 (Offline)
+4
-1
$$\mathop {\lim }\limits_{x \to {\pi \over 2}} {{\cot x - \cos x} \over {{{\left( {\pi - 2x} \right)}^3}}}$$ equals
A
$${1 \over {16}}$$
B
$${1 \over 8}$$
C
$${1 \over {4}}$$
D
$${1 \over {24}}$$
2
JEE Main 2016 (Online) 10th April Morning Slot
+4
-1
$$\mathop {\lim }\limits_{x \to 0} \,{{{{\left( {1 - \cos 2x} \right)}^2}} \over {2x\,\tan x\, - x\tan 2x}}$$ is :
A
$$-$$ 2
B
$$-$$ $${1 \over 2}$$
C
$${1 \over 2}$$
D
2
3
JEE Main 2016 (Online) 10th April Morning Slot
+4
-1
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
$$\left( { - \sqrt 2 ,1 - \sqrt 3 } \right)$$
4
JEE Main 2016 (Online) 9th April Morning Slot
+4
-1
If    $$\mathop {\lim }\limits_{x \to \infty } {\left( {1 + {a \over x} - {4 \over {{x^2}}}} \right)^{2x}} = {e^3},$$ then 'a' is equal to :
A
2
B
$${3 \over 2}$$
C
$${2 \over 3}$$
D
$${1 \over 2}$$
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