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1

### 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)

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

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

$$\Rightarrow$$  $$\alpha = {2^{100}}$$
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 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

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

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

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

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