1

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

If the equations x2 + bx−1 = 0 and x2 + x + b = 0 have a common root different from −1, then $\left| b \right|$ is equal to :
A
$\sqrt 2$
B
2
C
3
D
$\sqrt 3$

## Explanation

Given,

x2 + bx $-$ 1 = 0 . . . . .(1)

and x2 + x + b = 0 . . . . . (2)

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

bx $-$ 1 $-$ x $-$ b = 0

$\Rightarrow$    x(b $-$ 1) = b + 1

$\Rightarrow$   x = ${{b + 1} \over {b - 1}}$

putting value of x in equation (2),

${\left( {{{b + 1} \over {b - 1}}} \right)^2} + \left( {{{b + 1} \over {b - 1}}} \right) + b = 0$

$\Rightarrow$   (b + 1)2 + (b + 1) (b $-$ 1) + b (b $-$ 1)2 = 0

$\Rightarrow$    b2 + 2b + 1 + b2 $-$ 1 + b (b2 $-$ 2b + 1) = 0

$\Rightarrow$    2b3 + 2b + b3 $-$ 2b2 + b = 0

$\Rightarrow$     b3 + 3b = 0

$\Rightarrow$    b(b2 + 3) = 0

b2 = $-$ 3, b = 0

$\therefore$    b = $\pm \sqrt 3 i$

$\Rightarrow$    $\left| b \right|$ = $\sqrt 3$
2

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

If x is a solution of the equation, $\sqrt {2x + 1}$ $- \sqrt {2x - 1} = 1,$ $\,\,\left( {x \ge {1 \over 2}} \right),$ then $\sqrt {4{x^2} - 1}$ is equal to :
A
${3 \over 4}$
B
${1 \over 2}$
C
2
D
$2\sqrt 2$

## Explanation

Given,

$\sqrt {2x + 1} - \sqrt {2x - 1} = 1$

$\Rightarrow$   $\sqrt {2x + 1} = 1 + \sqrt {2x - 1}$

Squaring both sides, we get

2x + 1 $=$ 1 + 2x $-$ 1 + 2$\sqrt {2x - 1}$

$\Rightarrow$   1 $=$ 2$\sqrt {2x - 1}$

$\Rightarrow$   1 $=$ 4(2x $-$ 1)

$\Rightarrow$   8x $-$ 4 $=$ 1

$\Rightarrow$   x $=$ ${5 \over 8}$

So,    $\sqrt {4{x^2} - 1}$

$= \sqrt {4\left( {{{25} \over {64}}} \right) - 1}$

$= \sqrt {{{36} \over {64}}}$

$= {6 \over 8}$

$= {3 \over 4}$
3

### JEE Main 2017 (Offline)

If for a positive integer n, the quadratic equation

$x\left( {x + 1} \right) + \left( {x + 1} \right)\left( {x + 2} \right)$$+ .... + \left( {x + \overline {n - 1} } \right)\left( {x + n} \right)$$ = 10n$

has two consecutive integral solutions, then n is equal to :
A
9
B
10
C
11
D
12

## Explanation

$\sum\limits_{r = 1}^n {\left( {x + r - 1} \right)\left( {x + r} \right)} = 10n$

$\Rightarrow$ $\sum\limits_{r = 1}^n {\left( {{x^2} + xr + \left( {r - 1} \right)x + {r^2} - r} \right)} = 10n$

$\Rightarrow$ $\sum\limits_{r = 1}^n {\left( {{x^2} + \left( {2r - 1} \right)x + r\left( {r - 1} \right)} \right)} = 10n$

$\Rightarrow$ $n{x^2} + \left\{ {1 + 3 + 5 + .... + \left( {2n - 1} \right)} \right\}x$

$+ \left\{ {1.2 + 2.3 + ... + \left( {n - 1} \right)n} \right\}$ = 10n

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

$\Rightarrow$ ${x^2} + nx + {{\left( {{n^2} - 31} \right)} \over 3} = 0$

Let $\alpha$ and $\alpha$ + 1 be its two solutions

$\therefore$ $\alpha$ + ($\alpha$ + 1) = -n

$\Rightarrow$ $\alpha$ = ${{ - n - 1} \over 2}$ ....(1)

Also $\alpha$($\alpha$ + 1) = ${{\left( {{n^2} - 31} \right)} \over 3}$ ......(2)

Putting value of (1) in (2), we get

$- \left( {{{n + 1} \over 2}} \right)\left( {{{1 - n} \over 2}} \right) = {{\left( {{n^2} - 31} \right)} \over 3}$

$\Rightarrow$ n2 = 121

$\Rightarrow$ n = 11
4

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

Let p(x) be a quadratic polynomial such that p(0)=1. If p(x) leaves remainder 4 when divided by x$-$ 1 and it leaves remainder 6 when divided by x + 1; then :
A
p(2) = 11
B
p(2) = 19
C
p($-$ 2) = 19
D
p($-$ 2) = 11

## Explanation

Let, P(x) = ax2 + bx + c

As, P(0) = 1,

$\therefore\,\,\,$ a(0)2 + b(0) + c = 1

$\Rightarrow $$\,\,\, c = 1 \therefore\,\,\, P(x) = ax2 + bx + 1 If P(x) is divided by x - 1, remainder = 4 \Rightarrow$$\,\,\,$ P$\left( 1 \right) = 4$

$\therefore\,\,\,$ a + b + 1 = 4 . . . . . (1)

If    P(x) is divided by x + 1, remainder = 6

$\Rightarrow$$\,\,\,$ P($-$ 1) = 6

$\therefore\,\,\,$ a $-$ b + 1 = 6 . . . .(2)

By solving (1) and (2) we get,

a = 4, and b = $-$1

$\therefore\,\,\,$ P(x) = 4x2 $-$ x + 1

P(2) = 4(2)2 $-$ 2 + 1 = 15

P($-$ 2) = 4 ($-$2)2 $-$ ($-$ 2) + 1 = 19