1

### 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

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

### 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
3

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

If   A > 0, B > 0   and    A + B = ${\pi \over 6}$,

then the minimum value of tanA + tanB is :
A
$\sqrt 3 - \sqrt 2$
B
$2 - \sqrt 3$
C
$4 - 2\sqrt 3$
D
${2 \over {\sqrt 3 }}$

## Explanation

Given,

A + B = ${\pi \over 6}$

$\therefore$   tan(A + B) = tan$\left( {{\pi \over 6}} \right)$ = ${1 \over {\sqrt 3 }}$

We know,

tan(A + B) = ${{\tan A + \tan B} \over {1 - \tan A\tan B}}$

$\Rightarrow$  ${1 \over {\sqrt 3 }}$ = ${y \over {1 - \tan A\tan B}}$

where y = tan A + tan B

$\Rightarrow$    tanA tanB = 1 $-$ $\sqrt 3$ y

Also AM   $\ge$  GM

$\Rightarrow$    ${{\tan A + \tan B} \over 2} \ge \sqrt {\tan A\tan B}$

$\Rightarrow$   y $\ge$ 2$\sqrt {1 - \sqrt 3 y}$

$\Rightarrow$   y2 $\ge$ 4 $-$ 4${\sqrt 3 y}$

$\Rightarrow$    y2 + 4${\sqrt 3 y}$ $-$ 4 $\ge$ 0

$\Rightarrow$   y  $\le$ $-$ 2$\sqrt 3$ $-$ 4

or   y $\ge$ $-$ 2$\sqrt 3$ + 4

(y $\le$ $-$ 2$\sqrt 3$ $-$ 4 is not possible as tan B > 0)
4

### JEE Main 2017 (Offline)

Let $a$, b, c $\in R$. If $f$(x) = ax2 + bx + c is such that
$a$ + b + c = 3 and $f$(x + y) = $f$(x) + $f$(y) + xy, $\forall x,y \in R,$

then $\sum\limits_{n = 1}^{10} {f(n)}$ is equal to
A
165
B
190
C
255
D
330

## Explanation

f(x) = ax2 + bx + c

f(1) = a + b + c = 3 $\Rightarrow$ f (1) = 3

Now f(x + y) = f(x) + f(y) + xy ...(1)

Put x = y = 1 in eqn (1)

f(2) = f(1) + f(1) + 1

= 2f(1) + 1

$\Rightarrow$ f(2) = 7

Similarly f(3) = 12

f(4) = 18

$\sum\limits_{n = 1}^{10} {f(n)}$ = 3 + 7 + 12 + 18 + 25 + 33 + 42 + 52 + 63 + 75 = 330