1

### JEE Main 2018 (Online) 16th April Morning Slot

$\mathop {\lim }\limits_{x \to 0} \,\,{{{{\left( {27 + x} \right)}^{{1 \over 3}}} - 3} \over {9 - {{\left( {27 + x} \right)}^{{2 \over 3}}}}}$ equals.
A
${1 \over 3}$
B
$-$ ${1 \over 3}$
C
$-$ ${1 \over 6}$
D
${1 \over 6}$

## Explanation

Given,

$\mathop {lim}\limits_{x \to 0} \,{{{{\left( {27 + x} \right)}^{{1 \over 3}}} - 3} \over {9 - {{\left( {27 + x} \right)}^{{2 \over 3}}}}}$

=   $\mathop {\lim }\limits_{x \to 0} \,{{3\left[ {{{\left( {1 + {x \over {27}}} \right)}^{{1 \over 3}}} - 1} \right]} \over {9\left[ {1 - {{\left( {1 + {x \over {27}}} \right)}^{{2 \over 3}}}} \right]}}$

=   $\mathop {\lim }\limits_{x \to 0} \,\,{{\left( {1 + {x \over {3 \times 27}}} \right) - 1} \over {3\left[ {1 - \left( {1 + {{2x} \over {3 \times 27}}} \right)} \right]}}$

=   $\mathop {\lim }\limits_{x \to 0} \,\,{{{x \over {81}}} \over {3\left( { - {{2x} \over {81}}} \right)}} = - {1 \over 6}$
2

### JEE Main 2018 (Online) 16th April Morning Slot

If $x = \sqrt {{2^{\cos e{c^{ - 1}}}}}$ and $y = \sqrt {{2^{se{c^{ - 1}}t}}} \,\,\left( {\left| t \right| \ge 1} \right),$ then ${{dy} \over {dx}}$ is equal to :
A
${y \over x}$
B
${x \over y}$
C
$-$ ${y \over x}$
D
$-$ ${x \over y}$

## Explanation

x = $\sqrt {{2^{\cos e{c^{ - 1}}t}}}$

$\therefore\,\,\,\,$ ${{dx} \over {dt}}$ = ${1 \over {2\sqrt {{2^{\cos e{c^{ - 1}}t}}} }}$ $\times$ (${2^{\cos e{c^{ - 1}}t}}\,.\,\log 2$) $\times$ ${{ - 1} \over {t\sqrt {{t^2} - 1} }}$

${{dy} \over {dt}}$ = ${1 \over {2\sqrt {{2^{{{\sec }^{ - 1}}t}}} }}$ $\times$ $\left( {{2^{{{\sec }^{ - 1}}t}}\log 2} \right)$ $\times$ ${1 \over {t\sqrt {{t^2} - 1} }}$

$\therefore\,\,\,\,$ ${{dy} \over {dx}}$

= ${{{{dy} \over {dt}}} \over {{{dx} \over {dt}}}}$

= ${{ - \sqrt {{2^{\cos e{c^{ - 1}}t}}} } \over {\sqrt {{2^{\cos e{c^{ - 1}}t}}} }}$ $\times$ ${{{2^{{{\sec }^{ - 1}}t}}} \over {{2^{\cos e{c^{ - 1}}t}}}}$

= $- \sqrt {{{{2^{{{\sec }^{ - 1}}t}}} \over {{2^{\cos e{c^{ - 1}}t}}}}}$

= $-$ ${y \over x}$
3

### JEE Main 2018 (Online) 16th April Morning Slot

If the function f defined as

$f\left( x \right) = {1 \over x} - {{k - 1} \over {{e^{2x}} - 1}},x \ne 0,$ is continuous at

x = 0, then the ordered pair (k, f(0)) is equal to :
A
(3, 2)
B
(3, 1)
C
(2, 1)
D
$\left( {{1 \over 3},\,2} \right)$

## Explanation

If the function is continuous at x = 0, then
$\mathop {\lim }\limits_{x \to 0}$ f(x) will exist and f(0) = $\mathop {\lim }\limits_{x \to 0}$ f(x)

Now, $\mathop {\lim }\limits_{x \to 0}$ f(x) = $\mathop {\lim }\limits_{x \to 0} \left( {{1 \over x} - {{k - 1} \over {{e^{2x}} - 1}}} \right)$

= $\mathop {\lim }\limits_{x \to 0} \left( {{{{e^{2x}} - 1 - kx + x} \over {\left( x \right)\left( {{e^{2x}} - 1} \right)}}} \right)$

= $\mathop {\lim }\limits_{x \to 0} \left[ {{{\left( {1 + 2x + {{{{\left( {2x} \right)}^2}} \over {2!}} + {{{{\left( {2x} \right)}^3}} \over {3!}} + ....} \right) - 1 - kx + x} \over {\left( x \right)\left( {\left( {1 + 2x + {{{{\left( {2x} \right)}^2}} \over {2!}} + {{{{\left( {2x} \right)}^3}} \over {3!}} + ...} \right) - 1} \right)}}} \right]$

= $\mathop {\lim }\limits_{x \to 0} \left[ {{{\left( {3 - k} \right)x + {{4{x^2}} \over {2!}} + {{8{x^3}} \over {3!}} + ...} \over {\left( {2{x^2} + {{4{x^3}} \over {2!}} + {{8{x^3}} \over {3!}} + ....} \right)}}} \right]$

For the limit to exist, power of x in the numerator should be greater than or equal to the power of x in the denominator. Therefore, coefficient of x in numerator is equal to zero

$\Rightarrow$  3 $-$ k = 0

$\Rightarrow$  k = 3

So the limit reduces to

$\mathop {\lim }\limits_{x \to 0} {{\left( {{x^2}} \right)\left( {{4 \over {2!}} + {{8x} \over {3!}} + ...} \right)} \over {\left( {{x^2}} \right)\left( {2 + {{4x} \over {2!}} + {{8{x^2}} \over {3!}} + ...} \right)}}$

= $\mathop {\lim }\limits_{x \to 0} {{{4 \over {2!}} + {{8x} \over {3!}} + ...} \over {2 + {{4x} \over {2!}} + {{8{x^2}} \over {3!}} + ...}}$ = 1

Hence, f(0) = 1
4

### JEE Main 2019 (Online) 9th January Morning Slot

Let f : R $\to$ R be a function defined as
$f(x) = \left\{ {\matrix{ 5 & ; & {x \le 1} \cr {a + bx} & ; & {1 < x < 3} \cr {b + 5x} & ; & {3 \le x < 5} \cr {30} & ; & {x \ge 5} \cr } } \right.$

Then, f is
A
continuous if a = 0 and b = 5
B
continuous if a = –5 and b = 10
C
continuous if a = 5 and b = 5
D
not continuous for any values of a and b

## Explanation

Checking

if f(x) is continuous at x = 1 :

f(1$-$) = 5

f(1) = 5

f(1+) = a + b

if f(x) is continuous at x = 1,

then

f(1$-$) = f(1) = f(1+)

$\Rightarrow$  5 = 5 = a + b

$\therefore$  a + b = 5 . . . . . . . . (1)

checking if f(x) is continuous at x = 3 :

f(3$-$) = a + 3b

f(3) = b + 15

f(3+) = b + 15

if   f(x) = is continuous at x = 3

then,

f(3$-$) = f(3) = f(3+)

$\Rightarrow$  a + 3b = b + 15 = b + 15

$\Rightarrow$  a + 2b = 15 . . . . . (2)

checking if f(x) is continuous at x = 5 :

f(5$-$) = b + 25

f(5) = 30

f(5+) = 30

if f(x) is continuous at x = 5 then,

f(5$-$) = f(5) = f(5+)

$\Rightarrow$   b + 25 = 30 = 30

$\Rightarrow$   b = 5

By putting this value in equation (2), we get,

a + 2(5) = 15

$\Rightarrow$  a = 5

when a = 5 and b = 5 then equation (1)

a + b = 5 does not satisfy.

$\therefore$  f is not continuous for any value of a and b.