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) 9th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
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
$${1 \over 2}$$
3
JEE Main 2016 (Online) 9th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
If the function

f(x) = $$\left\{ {\matrix{ { - x} & {x < 1} \cr {a + {{\cos }^{ - 1}}\left( {x + b} \right),} & {1 \le x \le 2} \cr } } \right.$$

is differentiable at x = 1, then $${a \over b}$$ is equal to :
A
$${{\pi - 2} \over 2}$$
B
$${{ - \pi - 2} \over 2}$$
C
$${{\pi + 2} \over 2}$$
D
$$ - 1 - {\cos ^{ - 1}}\left( 2 \right)$$
4
JEE Main 2016 (Offline)
MCQ (Single Correct Answer)
+4
-1
Change Language
For $$x \in \,R,\,\,f\left( x \right) = \left| {\log 2 - \sin x} \right|\,\,$$

and $$\,\,g\left( x \right) = f\left( {f\left( x \right)} \right),\,\,$$ then :
A
$$g$$ is not differentiable at $$x=0$$
B
$$g'\left( 0 \right) = \cos \left( {\log 2} \right)$$
C
$$g'\left( 0 \right) = - \cos \left( {\log 2} \right)$$
D
$$g$$ is differentiable at $$x=0$$ and $$g'\left( 0 \right) = - \sin \left( {\log 2} \right)$$
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