1

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

If the fractional part of the number $\left\{ {{{{2^{403}}} \over {15}}} \right\} is \, {k \over {15}}$, then k is equal to :
A
8
B
14
C
6
D
1

## Explanation

${{{2^{403}}} \over {15}}$

$= {{{2^3}\, \cdot \,{2^{400}}} \over {15}}$

$= {8 \over {15}}{\left( {16} \right)^{100}}$

$= {8 \over {15}}{\left( {15 + 1} \right)^{100}}$

$= {8 \over {15}}\left( {{}^{100}{C_0} + {}^{100}{C_1}\,15 + {}^{100}{C_2}{{\left( {15} \right)}^2} + .....{{\left( {15} \right)}^{100}}} \right)$

$= {8 \over {15}} + 8\left( {{}^{100}{C_1} + {}^{100}{C_2}\,\left( {15} \right) + ..... + {{\left( {15} \right)}^{99}}} \right)$

$= {8 \over {15}} + 8$ (integer)

$\therefore$  Fractional part $= {8 \over {15}}$

According to the question,

${k \over {15}} = {8 \over {15}}$

$\Rightarrow$   K $=$ 8
2

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

The coefficient of t4 in the expansion of ${\left( {{{1 - {t^6}} \over {1 - t}}} \right)^3}$ is :
A
14
B
15
C
10
D
12

## Explanation

${\left( {{{1 - {t^6}} \over {1 - t}}} \right)^3}$

= (1 $-$ t6)3 (1 $-$ t)$-$3

= (1 $-$ 3C1t6 + 3C2t12 $-$ 3C3t18) $\times$ (1 $-$ t)$-$3

coefficient of t4 is 1 $\times$ coefficient of t4 in (1 $-$ t)$-$3

= 1 $\times$ 3+4$-$1C4 (By multinomial theorem)

= 6C4 = 15
3

### JEE Main 2019 (Online) 10th January Morning Slot

If  ${\sum\limits_{i = 1}^{20} {\left( {{{{}^{20}{C_{i - 1}}} \over {{}^{20}{C_i} + {}^{20}{C_{i - 1}}}}} \right)} ^3} = {k \over {21}}$  then k is equal to
A
100
B
200
C
50
D
400

## Explanation

${\sum\limits_{i = 1}^{20} {\left( {{{^{20}{C_{i - 1}}} \over {^{20}{C_i}{ + ^{20}}{C_{i - 1}}}}} \right)} ^3} = {k \over {21}}$

$\Rightarrow \,\,\sum\limits_{i = 1}^{20} {{{\left( {{{{}^{20}{C_{i - 1}}} \over {{}^{21}{C_i}}}} \right)}^3}} = {k \over {21}}$

$\Rightarrow \,\,\sum\limits_{i = 1}^{20} {{{\left( {{i \over {21}}} \right)}^3}} = {k \over {21}}$

$\Rightarrow \,\,{1 \over {{{\left( {21} \right)}^3}}}{\left[ {{{20\left( {21} \right)} \over 2}} \right]^2} = {k \over {21}}$

$\Rightarrow 100 = k$
4

### JEE Main 2019 (Online) 10th January Morning Slot

If the third term in the binomial expansion
of ${\left( {1 + {x^{{{\log }_2}x}}} \right)^5}$ equals 2560, then a possible value of x is -
A
$2\sqrt 2$
B
$4\sqrt 2$
C
${1 \over 8}$
D
${1 \over 4}$

## Explanation

${\left( {1 + {x^{{{\log }_2}x}}} \right)^5}$

${T_3} = {}^5{C_2}.{\left( {{x^{{{\log }_2}x}}} \right)^2} = 2560$

$\Rightarrow \,\,10.{x^{2{{\log }_2}x}} = 2560$

$\Rightarrow \,\,{x^{2\log 2x}} = 256$

$\Rightarrow \,\,2{({\log _2}x)^2} = {\log _2}256$

$\Rightarrow 2{({\log _2}x)^2} = 8$

$\Rightarrow \,\,{({\log _2}x)^2} = 4$

$\Rightarrow \,\,{\log _2}x = 2$  or  $-$ 2

$x = 4$   or  ${1 \over 4}$