1
MCQ (Single Correct Answer)

JEE Main 2016 (Online) 10th April Morning Slot

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

Explanation

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

$$ = \mathop {\lim }\limits_{x \to 0} {{{{\left( {2{{\sin }^2}x} \right)}^2}} \over {2x\left( {x + {{{x^3}} \over 3} + {{2{x^5}} \over {15}} + ....} \right) - x\left( {2x + {{{2^3}{x^3}} \over 3} + 2.{{{2^5}{x^5}} \over {15}}+...} \right)}}$$

$$ = \mathop {\lim }\limits_{x \to 0} {{4{{\left( {x - {{{x^3}} \over {3!}} + {{{x^5}} \over {5!}} - ....} \right)}^4}} \over {{x^4}\left( {{2 \over 3} - {8 \over 3}} \right) + {x^6}\left( {{4 \over {15}} - {{64} \over {15}}} \right) + ....}}$$

$$ = \mathop {\lim }\limits_{x \to 0} {{4{{\left( {1 - {{{x^2}} \over {3!}} + {{{3^4}} \over {5!}} - ....} \right)}^4}} \over { - 2 + {x^2}\left( { - {{60} \over {15}}} \right) + ....}}$$

(By dividing numerator and denominator by x4)

$$ = {{4{{\left( {1 - 0} \right)}^4}} \over { - 2 + 0}}$$

$$ = {4 \over { - 2}}$$

$$=$$ $$-$$ 2
2
MCQ (Single Correct Answer)

JEE Main 2016 (Online) 10th April Morning Slot

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

Explanation

f(x) is continuous at x = 1

$$ \therefore $$    $${{2{{\left( 1 \right)}^2}} \over a} = a$$

$$ \Rightarrow $$   a2 = 2

$$ \Rightarrow $$   a = $$ \pm $$ $$\sqrt 2 $$

Also f(x) is continuous at x = $$\sqrt 2 $$

$$ \therefore $$   a = $${{2{b^2} - 4b} \over {2\sqrt 2 }}$$

Now, when a = $$\sqrt 2 ,$$ then

4 = 2b2 $$-$$ 4b

$$ \Rightarrow $$   b2 $$-$$ 2b = 2

$$ \Rightarrow $$   b2 $$-$$ 2b $$-$$ 2 = 0

$$ \Rightarrow $$   b = $${{2 \mp \sqrt {4 + 4.2} } \over 2}$$

= 1 $$ \pm $$ $$\sqrt 3 $$

$$ \therefore $$   (a, b) = $$\left( {\sqrt 2 ,1 \pm \sqrt 3 } \right)$$

When a = $$-$$ $$\sqrt 2 $$, then

$$-$$ $$\sqrt 2 $$ = $${{2{b^2} - 4b} \over {2\sqrt 2 }}$$

$$ \Rightarrow $$   $$-$$ 4 = 2b2 $$-$$ 4b

$$ \Rightarrow $$   b2 $$-$$ 2b + 2 = 0

$$ \therefore $$    b = $${{2 \pm \sqrt {4 - 8} } \over 2}$$ = 1 $$ \pm $$ i

As  b $$ \in $$ Real number so,

b = 1 $$ \pm $$ i  is not accepted.

$$ \therefore $$   (a, b) = $$\left( {\sqrt 2 ,1 + \sqrt 3 } \right)$$   or  $$\left( {\sqrt 2 ,1 - \sqrt 3 } \right)$$
3
MCQ (Single Correct Answer)

JEE Main 2017 (Offline)

$$\mathop {\lim }\limits_{x \to {\pi \over 2}} {{\cot x - \cos x} \over {{{\left( {\pi - 2x} \right)}^3}}}$$ equals
A
$${1 \over {16}}$$
B
$${1 \over 8}$$
C
$${1 \over {4}}$$
D
$${1 \over {24}}$$

Explanation

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

= $$\mathop {\lim }\limits_{x \to {\pi \over 2}} {1 \over 8}.{{\cos x\left( {1 - \sin x} \right)} \over {\sin x{{\left( {{\pi \over 2} - x} \right)}^3}}}$$

Put $${{\pi \over 2} - x}$$ = t

$$ \Rightarrow $$ x $$ \to $$ $${{\pi \over 2}}$$

t $$ \to $$ 0

= $$\mathop {\lim }\limits_{t \to 0} {1 \over 8}.{{\cos \left( {{\pi \over 2} - t} \right)\left( {1 - \sin \left( {{\pi \over 2} - t} \right)} \right)} \over {\sin \left( {{\pi \over 2} - t} \right){{\left( {{\pi \over 2} - {\pi \over 2} + t} \right)}^3}}}$$

= $$\mathop {{1 \over 8}\lim }\limits_{t \to 0} {{\sin t\left( {1 - \cos t} \right)} \over {\cos t{{\left( t \right)}^3}}}$$

= $$\mathop {{1 \over 8}\lim }\limits_{t \to 0} {{\sin t\left( {2{{\sin }^2}{t \over 2}} \right)} \over {\cos t{{\left( t \right)}^3}}}$$

= $$\mathop {{1 \over 4}\lim }\limits_{t \to 0} {{\sin t\left( {{{\sin }^2}{t \over 2}} \right)} \over {\cos t{{\left( t \right)}^3}}}$$

= $$\mathop {\lim }\limits_{t \to 0} \left( {{{\sin t} \over t}} \right){\left( {{{\sin {t \over 2}} \over {{t \over 2}}}} \right)^2}{1 \over {\cos t}}{1 \over 4}$$

= $${1 \over 4} \times {1 \over 4}$$

= $${1 \over {16}}$$
4
MCQ (Single Correct Answer)

JEE Main 2017 (Offline)

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

Explanation

Let y = $${\tan ^{ - 1}}\left( {{{6x\sqrt x } \over {1 - 9{x^3}}}} \right)$$

= $${\tan ^{ - 1}}\left[ {{{2.\left( {3{x^{{3 \over 2}}}} \right)} \over {1 - {{\left( {3{x^{{3 \over 2}}}} \right)}^2}}}} \right]$$

= 2$${\tan ^{ - 1}}\left( {3{x^{{3 \over 2}}}} \right)$$

$$ \therefore $$ $${{dy} \over {dx}} = 2.{1 \over {1 + {{\left( {3{x^{{3 \over 2}}}} \right)}^2}}}.3 \times {3 \over 2}{\left( x \right)^{{1 \over 2}}}$$

= $${9 \over {1 + 9{x^3}}}.\sqrt x $$

$$ \therefore $$ g(x) = $${9 \over {1 + 9{x^3}}}$$

Questions Asked from Limits, Continuity and Differentiability

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