1

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

The sum of the real values of x for which the middle term in the binomial expansion of ${\left( {{{{x^3}} \over 3} + {3 \over x}} \right)^8}$ equals 5670 is :
A
0
B
8
C
6
D
4

## Explanation

${T_5} = {}^8{C_4}{{{x^{12}}} \over {81}} \times {{81} \over {{x^4}}} = 5670$

$\Rightarrow 70{x^8} = 5670$

$\Rightarrow x = \pm \sqrt 3$
2

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

Let (x + 10)50 + (x $-$ 10)50 = a0 + a1x + a2x2 + . . . . + a50x50, for all x $\in$ R; then ${{{a_2}} \over {{a_0}}}$ is equal to
A
12.25
B
12.75
C
12.00
D
12.50

## Explanation

(10 + x)50 + (10 $-$ x)50

$\Rightarrow$  a2 = 2.50C2 1048, a0 = 2.1050

${{{a_2}} \over {{a_0}}} = {{^{50}{C_2}} \over {{{10}^2}}} = 12.25$
3

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

Let Sn = 1 + q + q2 + . . . . . + qn and Tn = 1 + $\left( {{{q + 1} \over 2}} \right) + {\left( {{{q + 1} \over 2}} \right)^2}$ + . . . . . .+ ${\left( {{{q + 1} \over 2}} \right)^n}$ where q is a real number and q $\ne$ 1. If   101C1 + 101C2 . S1 + .... + 101C101 . S100 = $\alpha$T100 then $\alpha$ is equal to
A
202
B
200
C
2100
D
299

## Explanation

101C1 + 101C2S1 + . . . . . . . + 101C101S100

$=$ $\alpha$T100

101C1 + 101C2(1 + q) + 101C3(1 + q + q2) +

. . . . . .+101C101(1 + q + . . . . . + q100)

$= 2\alpha {{\left( {1 - {{\left( {{{1 + q} \over 2}} \right)}^{101}}} \right)} \over {\left( {1 - q} \right)}}$

$\Rightarrow$  101C1(1 $-$ q) + 101C2(1 $-$ q2) +

. . . . . . + 101C101(1 $-$ q101)

$= 2\alpha \left( {1 - {{\left( {{{1 + q} \over 2}} \right)}^{101}}} \right)$

$\Rightarrow$  (2101 $-$ 1) $-$ ((1 + q)101 $-$ 1)

$= 2\alpha \left( {1 - {{\left( {{{1 + q} \over 2}} \right)}^{101}}} \right)$

$\Rightarrow$  ${2^{101}}\left( {1 - {{\left( {{{1 + q} \over 2}} \right)}^{101}}} \right) = 2\alpha \left( {1 - {{\left( {{{1 + q} \over 2}} \right)}^{101}}} \right)$

$\Rightarrow$  $\alpha = {2^{100}}$
4

### JEE Main 2019 (Online) 12th January Morning Slot

A ratio of the 5th term from the beginning to the 5th term from the end in the binomial expansion of ${\left( {{2^{1/3}} + {1 \over {2{{\left( 3 \right)}^{1/3}}}}} \right)^{10}}$ is :
A
1 : 2(6)1/3
B
1 : 4(6)1/3
C
2(36)1/3 : 1
D
4(36)1/3 : 1

## Explanation

${{{T_5}} \over {T_5^1}} = {{{}^{10}{C_4}{{\left( {{2^{1/3}}} \right)}^{10 - 4}}{{\left( {{1 \over {2{{\left( 3 \right)}^{1/3}}}}} \right)}^4}} \over {{}^{10}{C_4}{{\left( {{1 \over {2\left( {{3^{1/3}}} \right)}}} \right)}^{10 - 4}}{{\left( {{2^{1/3}}} \right)}^4}}} = 4.{\left( {36} \right)^{1/3}}$