1

### JEE Main 2016 (Online) 9th April Morning Slot

Let x, y, z be positive real numbers such that x + y + z = 12 and x3y4z5 = (0.1) (600)3. Then x3 + y3 + z3is equal to :
A
270
B
258
C
342
D
216

## Explanation

As we know

AM  $\ge$  GM

$\Rightarrow$   ${{3\left( {{x \over 3}} \right) + 4\left( {{y \over 4}} \right) + 5\left( {{z \over 5}} \right)} \over {12}}$  $\ge$  ${\left[ {{{\left( {{x \over 3}} \right)}^3}{{\left( {{y \over 4}} \right)}^4}{{\left( {{z \over 5}} \right)}^5}} \right]^{{1 \over {12}}}}$

$\Rightarrow$   1  $\ge$  ${{{x^3}{y^4}{z^5}} \over {{3^3}{4^4}{5^5}}}$

$\Rightarrow$   x3 y4 z5  $\le$  33 . 44 . 55

$\Rightarrow$   x3 y4 z5  $\le$  (0.1)(600)3

but given that,

x3 y4 z5 = (0.1) (600)3

$\therefore$   AM  $=$  GM

$\Rightarrow$   All the number are equal.

$\therefore$   ${x \over 3} = {y \over 4} = {z \over 5} = k$

$\Rightarrow$  x $=$ 3k, y = 4k, z = 5k

given that,

x + y + z $=$ 12

$\Rightarrow$   3k + 4k + 5k $=$ 12

$\Rightarrow$   12k $=$ 12

$\Rightarrow$   k = 1

$\therefore$   x $=$   3,  y $=$ 4,   z $=$ 5

So, x3 + y3 + z3

$=$ 33 + 43 + 53

$=$ 216
2

### JEE Main 2016 (Online) 9th April Morning Slot

For x $\in$ R, x $\ne$ -1,

if (1 + x)2016 + x(1 + x)2015 + x2(1 + x)2014 + . . . . + x2016 =

$\sum\limits_{i = 0}^{2016} {{a_i}} \,{x^i},\,\,$ then a17 is equal to :
A
${{2017!} \over {17!\,\,\,2000!}}$
B
${{2016!} \over {17!\,\,\,1999!}}$
C
${{2017!} \over {2000!}}$
D
${{2016!} \over {16!}}$

## Explanation

Assume,

P = (1 + x)2016 + x(1 + x)2015 + . . . . .+ x2015 . (1 + x) + x2016    . . . . .(1)

Multiply this with $\left( {{x \over {1 + x}}} \right),$

$\left( {{x \over {1 + x}}} \right)P =$ x(1 + x)2015 + x2(1 + x)2014 +

. . . . . . + x2016 + ${{{x^{2017}}} \over {1 + x}}$ . . . . . (2)

Performing (1) $-$ (2), we get

${P \over {1 + x}} =$ (1 + x)2016 $-$ ${{{x^{2017}}} \over {1 + x}}$

$\Rightarrow$    P = (1 + x)2017 $-$ x2017

$\therefore$    a17 = coefficient of x17 $=$ 2017C17 $=$ ${{2017!} \over {17!\,\,2000!}}$
3

### JEE Main 2016 (Online) 10th April Morning Slot

Let z = 1 + ai be a complex number, a > 0, such that z3 is a real number.

Then the sum 1 + z + z2 + . . . . .+ z11 is equal to :
A
$- 1250\,\sqrt 3 \,i$
B
$1250\,\sqrt 3 \,i$
C
$1365\,\sqrt 3 i$
D
$-$ $1365\,\sqrt 3 i$

## Explanation

z = 1 + ai

z2 = 1 $-$ a2 + 2ai

z2 . z = {(1 $-$ a2) + 2ai}   {1 + ai}

= (1 $-$ a2) + 2ai + (1 $-$ a2)    ai $-$ 2a2

$\because$    z3 is real  $\Rightarrow$  2a + (1 $-$ a2) a = 0

a (3 $-$ a2) = 0   $\Rightarrow$  a = $\sqrt 3$ (a > 0)

1 + z + z2 . . . . . . . z11 = ${{{z^{12}} - 1} \over {z - 1}} = {{{{\left( {1 + \sqrt 3 i} \right)}^{12}} - 1} \over {1 + \sqrt 3 i - 1}}$

= ${{{{\left( {1 + \sqrt 3 i} \right)}^{12}} - 1} \over {\sqrt 3 i}}$

(1 + ${\sqrt 3 i}$)12 = 212 ${\left( {{1 \over 2} + {{\sqrt 3 } \over 2}i} \right)^{12}}$

= 212 (cos${\pi \over 3}$ + isin${\pi \over 3}$)12 = 212 (cos4$\pi$ + isin4$\pi$) = 212

$\Rightarrow$    ${{{2^{12}} - 1} \over {\sqrt 3 i}} = {{4095} \over {\sqrt 3 i}} = - {{4095} \over 3}\sqrt 3 i = - 1365\sqrt 3 i$
4

### JEE Main 2016 (Online) 10th April Morning Slot

Let a1, a2, a3, . . . . . . . , an, . . . . . be in A.P.

If a3 + a7 + a11 + a15 = 72,

then the sum of its first 17 terms is equal to :
A
306
B
153
C
612
D
204

## Explanation

As  a1 a2 . . . . . an . . . . . are in A.P.

$\therefore$   a3 + a15 = a7 + a11 = a1 + a17

Given,

a3 + a7 + a11 + a15 + a15 = 72

$\Rightarrow$   (a3 + a15) + (a7 + a11) = 72

$\Rightarrow$   2(a1 + a17) = 72

$\Rightarrow$   (a1 + a17) = 36

$\therefore$   Sum of first 17 terms

= ${{17} \over 2}$ (a1 + a17)

= ${{17} \over 2}$ $\times$ 36

= 306

NEET