1

### JEE Main 2016 (Online) 10th April Morning Slot

If the coefficients of x−2 and x−4 in the expansion of ${\left( {{x^{{1 \over 3}}} + {1 \over {2{x^{{1 \over 3}}}}}} \right)^{18}},\left( {x > 0} \right),$ are m and n respectively, then ${m \over n}$ is equal to :
A
182
B
${4 \over 5}$
C
${5 \over 4}$
D
27

## Explanation

Tr+1  =  18Cr  ${\left( {{x^{{1 \over 3}}}} \right)^{18 - r}}$  .  ${\left( {{1 \over {2{x^{{1 \over 3}}}}}} \right)^r}$

=  18Cr  ${\left( {{1 \over 2}} \right)^r}\,\,.\,\,{x^{{{18 - 2r} \over 3}}}$

For coefficient of x$-$2,

${{18 - 2r} \over 3}$   =   $-$2

$\Rightarrow$   r  =  12

$\therefore$   Coefficient of   x$-$2     is  (m) = 18C12  ${\left( {{1 \over 2}} \right)^{12}}$

For coefficient of x$-$4,

${{18 - 2r} \over 3}$ = $-$ 4

$\Rightarrow$   r = 15

$\therefore$   Coefficient of x$-$4 is (n) = 18C15 $\left( {{1 \over {2}}} \right)$15

$\therefore$   ${m \over n} = {{^{18}{C_{12}}{{\left( {{1 \over 2}} \right)}^{12}}} \over {^{18}{C_{15}}{{\left( {{1 \over 2}} \right)}^{15}}}}$

= ${{{}^{18}{C_6} \times {{\left( 2 \right)}^3}} \over {{}^{18}{C_3}}}$

= 182
2

### JEE Main 2017 (Offline)

The value of $\left( {{}^{21}{C_1} - {}^{10}{C_1}} \right) + \left( {{}^{21}{C_2} - {}^{10}{C_2}} \right) + \left( {{}^{21}{C_3} - {}^{10}{C_3}} \right)$
$\left( {{}^{21}{C_4} - {}^{10}{C_4}} \right)$$+ .... + \left( {{}^{21}{C_{10}} - {}^{10}{C_{10}}} \right)$ is
A
${2^{21}} - {2^{10}}$
B
${2^{20}} - {2^{9}}$
C
${2^{20}} - {2^{10}}$
D
${2^{21}} - {2^{11}}$

## Explanation 3

### JEE Main 2017 (Online) 8th April Morning Slot

If (27)999 is divided by 7, then the remainder is :
A
1
B
2
C
3
D
6

## Explanation

We have,

${{{{\left( {27} \right)}^{999}}} \over 7}$

= ${{{{\left( {28 - 1} \right)}^{999}}} \over 7}$

= ${{28\,\lambda - 1} \over 7}$

= ${{28\,\lambda - 7 + 7 - 1} \over \lambda }$

= ${{7\left( {4\lambda - 1} \right) + 6} \over 7}$

$\therefore\,\,\,$ Remainder = 6
4

### JEE Main 2017 (Online) 9th April Morning Slot

The coefficient of x−5 in the binomial expansion of

${\left( {{{x + 1} \over {{x^{{2 \over 3}}} - {x^{{1 \over 3}}} + 1}} - {{x - 1} \over {x - {x^{{1 \over 2}}}}}} \right)^{10}},$ where x $\ne$ 0, 1, is :
A
1
B
4
C
$-$ 4
D
$-$ 1

## Explanation

${\left( {{{x + 1} \over {{x^{2/3}} - {x^{1/3}} + 1}} - {{x - 1} \over {x - {x^{1/2}}}}} \right)^{10}}$

= ${\left( {{{{{\left( {{x^{1/3}}} \right)}^3} + {{\left( {{1^{1/3}}} \right)}^3}} \over {{x^{2/3}} - {x^{1/3}} + 1}} - {{{{\left( {\sqrt x } \right)}^2} - {{\left( 1 \right)}^2}} \over {x - {x^{1/2}}}}} \right)^{10}}$

= ${\left( {{{\left( {{x^{1/3}} + 1} \right)\left( {{x^{2/3}} - {x^{1/3}} + 1} \right)} \over {{x^{2/3}} - {x^{1/3}} + 1}} - {{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)} \over {\sqrt x \left( {\sqrt x - 1} \right)}}} \right)^{10}}$

= ${\left( {\left( {{x^{1/3}} + 1} \right) - {{\left( {\sqrt x + 1} \right)} \over {\sqrt x }}} \right)^{10}}$

= ${\left( {\left( {{x^{1/3}} + 1} \right) - \left( {1 + {1 \over {\sqrt x }}} \right)} \right)^{10}}$

= ${\left( {{x^{1/3}} - {1 \over {{x^{1/2}}}}} \right)^{10}}$

[Note:

For ${\left( {{x^\alpha } \pm {1 \over {{x^\beta }}}} \right)^n}$ the $\left( {r + 1} \right)$th term with power m of x is

$r = {{n\alpha - m} \over {\alpha + \beta }}$]

Here $\alpha = {1 \over 3}$, $\beta = {1 \over 2}$ and m = -5

then $r = {{10 \times {1 \over 3} - (-5)} \over {{1 \over 3} + {1 \over 2}}}$ = ${{25} \over 3} \times {6 \over 5}$ = 10

$\therefore$ T11 is the term with x-5.

$\therefore$ T11 = ${}^{10}{C_{10}}$ = 1