1

### 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

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

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

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

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

$\Rightarrow 100 = k$
2

### 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  ${1 \over 4}$
3

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

The positive value of $\lambda$ for which the co-efficient of x2 in the expression x2 ${\left( {\sqrt x + {\lambda \over {{x^2}}}} \right)^{10}}$ is 720, is -
A
4
B
$2\sqrt 2$
C
3
D
$\sqrt 5$

## Explanation

${x^2}\left( {{}^{10}{C_r}{{\left( {\sqrt x } \right)}^{10 - r}}{{\left( {{\lambda \over {{x^2}}}} \right)}^r}} \right)$

${x^2}\left[ {{}^{10}{C_r}{{\left( x \right)}^{{{10 - r} \over 2}}}{{\left( \lambda \right)}^r}{{\left( x \right)}^{ - 2r}}} \right]$

${x^2}\left[ {{}^{10}{C_r}{\lambda ^r}{x^{{{10 - r} \over 2}}}} \right]$

$\therefore$  r = 2

Hence, ${}^{10}{C_2}{\lambda ^2} = 720$

${\lambda ^2} = 16$

$\lambda = \pm 4$
4

### JEE Main 2019 (Online) 11th January Morning Slot

The value of r for which 20Cr 20C0 + 20Cr$-$1 20C1 + 20Cr$-$2 20C2 + . . . . .+ 20C0 20Cr  is maximum, is
A
20
B
15
C
10
D
11

## Explanation

Given sum = coefficient of xr in the expansion of

(1 + x)20(1 + x)20,

Which is equal to 40Cr

It is maximum when r = 20