1
MCQ (Single Correct Answer)

### JEE Main 2018 (Online) 16th April Morning Slot

The coefficient of x2 in the expansion of the product
(2$-$x2) .((1 + 2x + 3x2)6 + (1 $-$ 4x2)6) is :
A
107
B
106
C
108
D
155

## Explanation

Given,

(2 $-$ x2) . (1 + 2x + 3x2) 6 + (1 $-$ 4x2)6)

Let, a = ((1 + 2x + 3x2)6 + (1 $-$ 4x2)6)

$\therefore\,\,\,\,$ Given statement becomes,

(2 $-$ x2) . (a)

=    2a $-$ x2 (a)

Here coefficients of x2 is

=    2 (coefficient of x2 in a ) $-$ 1 (constant in a)

(1 + 2x + 3x2) 6  =   6C0 + 6C1 (2x + 3x2) + 6C2 (2x + 3x2)2

+ . . . . . .+ (2x + 3x2)6

(1 $-$ 4x2)6 = 6C0 $-$ 6C1 (4x2) + 6C2 (4x2)2

+ . . . . .+ (4x2)6

Coefficient of x2 in (1 + 2x + 3x2)6

= 6C1 $\times$ 3 + 6C2 $\times$ 4

= 18 + 60

Coefficient of x2 in (1 $-$ 4x2)6

= $-$ 6C1 $\times$ 4

= $-$ 24

Coefficient of x2 in ((1 + 2x + 3x2)6 + (1 $-$ 4x2)6)

= 60 + 18 $-$ 24

= 54

Constant term in (1 + 2x + 3x2)6 = 6C0 = 1

Constant term in (1 $-$ 4x)6 = 6C0 = 1

$\therefore\,\,\,\,$ Constant term in ((1 + 2x + 3x2)6 + (1 $-$ 4x)6)

= 1 + 1 = 2

$\therefore\,\,\,\,$ Coefficient of x2 in (2 $-$ x2) ((1 + 2x + 3x2)6 + (1 $-$ 4x2)6)

= 2 (54) $-$ 1 (2)

= 108 $-$ 2

= 106
2
MCQ (Single Correct Answer)

### 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
3
MCQ (Single Correct Answer)

### 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
4
MCQ (Single Correct Answer)

### 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

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

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

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

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

$\Rightarrow 100 = k$

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