1

### JEE Main 2017 (Online) 8th April Morning Slot

If (27)999 is divided by 7, then the remainder is :
A
1
B
2
C
3
D
6

## Explanation

We have,

${{{{\left( {27} \right)}^{999}}} \over 7}$

= ${{{{\left( {28 - 1} \right)}^{999}}} \over 7}$

= ${{28\,\lambda - 1} \over 7}$

= ${{28\,\lambda - 7 + 7 - 1} \over \lambda }$

= ${{7\left( {4\lambda - 1} \right) + 6} \over 7}$

$\therefore\,\,\,$ Remainder = 6
2

### JEE Main 2017 (Online) 9th April Morning Slot

The coefficient of x−5 in the binomial expansion of

${\left( {{{x + 1} \over {{x^{{2 \over 3}}} - {x^{{1 \over 3}}} + 1}} - {{x - 1} \over {x - {x^{{1 \over 2}}}}}} \right)^{10}},$ where x $\ne$ 0, 1, is :
A
1
B
4
C
$-$ 4
D
$-$ 1

## Explanation

${\left( {{{x + 1} \over {{x^{2/3}} - {x^{1/3}} + 1}} - {{x - 1} \over {x - {x^{1/2}}}}} \right)^{10}}$

= ${\left( {{{{{\left( {{x^{1/3}}} \right)}^3} + {{\left( {{1^{1/3}}} \right)}^3}} \over {{x^{2/3}} - {x^{1/3}} + 1}} - {{{{\left( {\sqrt x } \right)}^2} - {{\left( 1 \right)}^2}} \over {x - {x^{1/2}}}}} \right)^{10}}$

= ${\left( {{{\left( {{x^{1/3}} + 1} \right)\left( {{x^{2/3}} - {x^{1/3}} + 1} \right)} \over {{x^{2/3}} - {x^{1/3}} + 1}} - {{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)} \over {\sqrt x \left( {\sqrt x - 1} \right)}}} \right)^{10}}$

= ${\left( {\left( {{x^{1/3}} + 1} \right) - {{\left( {\sqrt x + 1} \right)} \over {\sqrt x }}} \right)^{10}}$

= ${\left( {\left( {{x^{1/3}} + 1} \right) - \left( {1 + {1 \over {\sqrt x }}} \right)} \right)^{10}}$

= ${\left( {{x^{1/3}} - {1 \over {{x^{1/2}}}}} \right)^{10}}$

[Note:

For ${\left( {{x^\alpha } \pm {1 \over {{x^\beta }}}} \right)^n}$ the $\left( {r + 1} \right)$th term with power m of x is

$r = {{n\alpha - m} \over {\alpha + \beta }}$]

Here $\alpha = {1 \over 3}$, $\beta = {1 \over 2}$ and m = -5

then $r = {{10 \times {1 \over 3} - (-5)} \over {{1 \over 3} + {1 \over 2}}}$ = ${{25} \over 3} \times {6 \over 5}$ = 10

$\therefore$ T11 is the term with x-5.

$\therefore$ T11 = ${}^{10}{C_{10}}$ = 1
3

### JEE Main 2018 (Offline)

The sum of the co-efficients of all odd degree terms in the expansion of

${\left( {x + \sqrt {{x^3} - 1} } \right)^5} + {\left( {x - \sqrt {{x^3} - 1} } \right)^5}$, $\left( {x > 1} \right)$ is
A
2
B
-1
C
0
D
1

## Explanation 4

### JEE Main 2018 (Online) 15th April Morning Slot

If n is the degree of the polynomial,

${\left[ {{2 \over {\sqrt {5{x^3} + 1} - \sqrt {5{x^3} - 1} }}} \right]^8} +$ ${\left[ {{2 \over {\sqrt {5{x^3} + 1} + \sqrt {5{x^3} - 1} }}} \right]^8}$

and m is the coefficient of xn in it, then the ordered pair (n, m) is equal to :
A
(24, (10)8)
B
(8, 5(10)4)
C
(12, (20)4)
D
(12, 8(10)4)

## Explanation

Given,
${\left[ {{2 \over {\sqrt {5{x^3} + 1} - \sqrt {5{x^3} - 1} }}} \right]^8}$ + ${\left[ {{2 \over {\sqrt {5{x^3} + 1} + \sqrt {5{x^3} - 1} }}} \right]^8}$

= ${\left[ {{2 \over {\sqrt {5{x^3} + 1} - \sqrt {5{x^3} - 1} }} \times {{\sqrt {5{x^3} + 1} + \sqrt {5{x^3} - 1} } \over {\sqrt {5{x^3} + 1} + \sqrt {5{x^3} - 1} }}} \right]^8}$

+ ${\left[ {{2 \over {\sqrt {5{x^3} + 1} + \sqrt {5{x^3} - 1} }} \times {{\sqrt {5{x^3} + 1} - \sqrt {5{x^3} - 1} } \over {\sqrt {5{x^3} + 1} - \sqrt {5{x^3} - 1} }}} \right]^8}$

= ${\left[ {{{2(\sqrt {5x^3 + 1} + \sqrt {5x^3 - 1} } \over 2}} \right]^8}$ + ${\left[ {{{2(\sqrt {5x^3 - 1} - \sqrt {5x^3 - 1} } \over 2}} \right]^8}$

= (${\sqrt {5{x^3} + 1} }$ + ${\sqrt {5{x^3} - 1} }$)8 + (${\sqrt {5{x^3} - 1} }$ - ${\sqrt {5{x^3} - 1} }$)8

= 2 $\left[ {{}^8{C_0}} \right.{(\sqrt {5{x^3} + 1} )^8}$ +

+ ${}^8{C_2}{(\sqrt {5{x^3} + 1} )^6}\sqrt {5{x^3} - 1}$

+ ${}^8{C_4}{(\sqrt {5{x^3} + 1} )^4}{(5{x^3} - 1)^2}$

+ ${}^8{C_6}{(\sqrt {5{x^3} + 1} )^2}{(5{x^3} - 1)^3}$

+ $\left. {{}^8{C_8}{{(5{x^3} - 1)}^4}} \right]$

= 2$[{}^8{C_0}{(5{x^3} + 1)^4}$

+ ${}^8{C_2}{(5{x^3} + 1)^3}(5{x^3} - 1)$

+ ${}^8{C_4}(5{x^3} + 1)^2{(5{x^3} - 1)^2}$

+ ${}^8{C_6}(5{x^3} + 1){(5{x^3} - 1)^3}$

+ ${}^8{C_8}{(5{x^3} - 1)^4}]$

Here maximum power of x is 12

$\therefore$ Degree of polynomial = 12

Coefficient of x12

= 2 [8C6 54 + 8C2 $\times$ 53 $\times$ 5 + 8C4 $\times$ 52 $\times$ 52 + 8C6 $\times$ 5 $\times$ 53 + 8C8 $\times$ 54]

= 160000 = (20)4