1
MCQ (Single Correct Answer)

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
MCQ (Single Correct Answer)

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
MCQ (Single Correct Answer)

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
MCQ (Single Correct Answer)

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

Questions Asked from Quadratic Equation and Inequalities

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