1

### JEE Main 2016 (Online) 9th April Morning Slot

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
${1 \over 2}$

## Explanation

Given,

$\mathop {\lim }\limits_{x \to \propto } {\left( {1 + {a \over x} - {4 \over {{x^2}}}} \right)^{2x}} = {e^3}$

So,   $\mathop {\lim }\limits_{x \to \propto } {\left( {1 + {a \over x} - {4 \over {{x^2}}}} \right)^{2x}}\,\left[ {{1^ \propto }\,\,\,form} \right]$

$= {e^{\mathop {\lim }\limits_{x \to \propto } \left[ {\left( {1 + {a \over x} - {4 \over {{x^2}}} - 1} \right)2x} \right]}}$

$= {e^{\mathop {\lim }\limits_{x \to \propto } \left( {2a - {8 \over x}} \right)}}$

$= {e^{2a}}$

$\therefore$   e2a = e3

$\therefore$    2a = 3

$\Rightarrow$   a $=$ ${3 \over 2}$
2

### JEE Main 2016 (Online) 9th April Morning Slot

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
$- 1 - {\cos ^{ - 1}}\left( 2 \right)$

## Explanation

As   f(x) is differentiable at x = 1

$\therefore$   $\mathop {\lim }\limits_{x \to {1^ - }} \left( { - x} \right) = \mathop {\lim }\limits_{x \to {1^ + }} \left( {a + {{\cos }^{ - 1}}\left( {x + b} \right)} \right) = f(1)$

$\Rightarrow$   $-$1 $=$ a + cos$-$1(1 + b)

$\Rightarrow$   cos$-$1 (1 + b) $=$ $-$1 $-$ a . . . . . .(1)

As   f(x) is differentiable, so,

2 . H . D $=$ R . H . D

Here, L . H . D $=$ $\mathop {\lim }\limits_{h \to 0}$ ${{f\left( {1 - h} \right) - f\left( 1 \right)} \over { - h}}$

$= \mathop {\lim }\limits_{h \to 0} {{ - \left( {1 - h} \right) - \left( { - 1} \right)} \over { - h}}$

$= \mathop {\lim }\limits_{x \to 0} {{ - 1 + h + 1} \over { - h}}$

$=$ $= \mathop {\lim }\limits_{x \to 0} {h \over { - h}}$

$=$ $-$ 1

R. H. D $= \mathop {\lim }\limits_{x \to 0} {{f\left( {1 + h} \right) - f\left( 1 \right)} \over h}$

$= \mathop {\lim }\limits_{h \to 0} {{a + {{\cos }^{ - 1}}\left( {1 + h + b} \right) - \left[ {a + {{\cos }^{ - 1}}\left( {1 + b} \right)} \right]} \over h}$

$= \mathop {\lim }\limits_{h \to 0} {{{{\cos }^{ - 1}}\left( {1 + h + b} \right) - {{\cos }^{ - 1}}\left( {1 + b} \right)} \over h}\left[ {{0 \over 0}\,\,form\left. \, \right]} \right.$

$= \mathop {\lim }\limits_{h \to 0} {{ - 1} \over {\sqrt {1 - {{\left( {1 + h + b} \right)}^2}} }}$ [ Using L' Hospital Rule]

$= {{ - 1} \over {\sqrt {1 - {{\left( {1 + b} \right)}^2}} }}$

$\therefore$   $- 1 = {{ - 1} \over {\sqrt {1 - {{\left( {1 + b} \right)}^2}} }}$

$\Rightarrow 1 - {\left( {1 + b} \right)^2} = 1$

$\Rightarrow$  ${\left( {1 + b} \right)^2} = 0$

$\Rightarrow$   $b = - 1$

putting value of b in equation (1), we get,

${\cos ^{ - 1}}\left( {1 - 1} \right) = - 1 - a$

$\Rightarrow$   ${\pi \over 2} = - 1 - a$

$\Rightarrow$   $a = - 1 - {\pi \over 2}$

$\therefore$   ${a \over b} = {{ - 1 - {\pi \over 2}} \over { - 1}} = 1 + {\pi \over 2} = {{\pi + 2} \over 2}$
3

### 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
4

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