ExamSIDE
Questions
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1
JEE Main 2017 (Online) 8th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
$$\mathop {\lim }\limits_{x \to 3} $$ $${{\sqrt {3x} - 3} \over {\sqrt {2x - 4} - \sqrt 2 }}$$ is equal to :
A
$$\sqrt 3 $$
B
$${1 \over {\sqrt 2 }}$$
C
$${{\sqrt 3 } \over 2}$$
D
$${1 \over {2\sqrt 2 }}$$
2
JEE Main 2017 (Offline)
MCQ (Single Correct Answer)
+4
-1
Change Language
$$\mathop {\lim }\limits_{x \to {\pi \over 2}} {{\cot x - \cos x} \over {{{\left( {\pi - 2x} \right)}^3}}}$$ equals
A
$${1 \over {16}}$$
B
$${1 \over 8}$$
C
$${1 \over {4}}$$
D
$${1 \over {24}}$$
3
JEE Main 2017 (Offline)
MCQ (Single Correct Answer)
+4
-1
Change Language
If for $$x \in \left( {0,{1 \over 4}} \right)$$, the derivatives of

$${\tan ^{ - 1}}\left( {{{6x\sqrt x } \over {1 - 9{x^3}}}} \right)$$ is $$\sqrt x .g\left( x \right)$$, then $$g\left( x \right)$$ equals
A
$${{{3x\sqrt x } \over {1 - 9{x^3}}}}$$
B
$${{{3x} \over {1 - 9{x^3}}}}$$
C
$${{3 \over {1 + 9{x^3}}}}$$
D
$${{9 \over {1 + 9{x^3}}}}$$
4
JEE Main 2016 (Online) 10th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
$$\mathop {\lim }\limits_{x \to 0} \,{{{{\left( {1 - \cos 2x} \right)}^2}} \over {2x\,\tan x\, - x\tan 2x}}$$ is :
A
$$-$$ 2
B
$$-$$ $${1 \over 2}$$
C
$${1 \over 2}$$
D
2
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