JEE Main 2017 (Offline) | Limits, Continuity and Differentiability Question 166 | Mathematics

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JEE Main 2017 (Offline)

MCQ (Single Correct Answer)

-1

$$\mathop {\lim }\limits_{x \to {\pi \over 2}} {{\cot x - \cos x} \over {{{\left( {\pi - 2x} \right)}^3}}}$$ equals

$${1 \over {16}}$$

$${1 \over 8}$$

$${1 \over {4}}$$

$${1 \over {24}}$$

JEE Main 2017 (Offline)

MCQ (Single Correct Answer)

-1

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

$${{{3x\sqrt x } \over {1 - 9{x^3}}}}$$

$${{{3x} \over {1 - 9{x^3}}}}$$

$${{3 \over {1 + 9{x^3}}}}$$

$${{9 \over {1 + 9{x^3}}}}$$

JEE Main 2016 (Online) 10th April Morning Slot

MCQ (Single Correct Answer)

-1

$$\mathop {\lim }\limits_{x \to 0} \,{{{{\left( {1 - \cos 2x} \right)}^2}} \over {2x\,\tan x\, - x\tan 2x}}$$ is :

$$-$$ 2

$$-$$ $${1 \over 2}$$

$${1 \over 2}$$

JEE Main 2016 (Online) 10th April Morning Slot

MCQ (Single Correct Answer)

-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 :

$$\left( {\sqrt 2 ,1 - \sqrt 3 } \right)$$

$$\left( { - \sqrt 2 ,1 + \sqrt 3 } \right)$$

$$\left( {\sqrt 2 , - 1 + \sqrt 3 } \right)$$

$$\left( { - \sqrt 2 ,1 - \sqrt 3 } \right)$$

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