QBManager

Complex Numbers and Quadratic Equation

118126 If $a, b, c$ are in Arithmetic Progression (AP), then the roots of the equation
$a x^{2} - 2 b x + c = 0$ are

1

1, \frac{c}{a}

2

\frac{- 1}{a}, - c

3

- 1, \frac{- c}{a}

4

- 2, \frac{- c}{2 a}

Explanation:

A Given
$a, b, c are in A. P.$
$∴ b - a = c - b$
$or 2 b = a + c$
${ax}^{2} - 2 bx + c = 0$
$α + β = + \frac{2 b}{a} = \frac{a + c}{a} = (1 + \frac{c}{a})$
$α \cdot β = c / a$
$(α - β)^{2} = (α + β)^{2} - 4 α β$
$= {(1 + \frac{c}{a})}^{2} - 4 \frac{c}{a}$
$= {(1 - \frac{c}{a})}^{2}$
$∵ α - β = 1 - \frac{c}{a}$
$α + β = 1 + \frac{c}{a}$
$∴ 2 α = 2$
$α = 1$
$∴ β = c / a$
$∴ Roots are (1, c/a)$
$∴ option (a) is correct.$

AP EAMCET-18.09.2020

Complex Numbers and Quadratic Equation

118163 The number of the real solutions of the equation $x^{2} - 3 | x | + 2 = 0$ is

1 2

2 4

3 1

4 3

Explanation:

B Given the equation
$x^{2} - 3 | x | + 2 = 0$
Case 1 : If $x > 0 \Rightarrow | x | = x$
$∴ x^{2} - 3 x + 2 = 0$
$\Rightarrow x^{2} - 2 x - x + 2 = 0$
$\Rightarrow x (x - 2) - 1 (x - 2) = 0$
$\Rightarrow (x - 2) (x - 1) = 0$
$∴ x = 1, 2$
$∵$ Both the roots are $+ ve$ . If yields two solutions.
Case 2: If $x < 0 \Rightarrow | x | = - x$
$∴ x^{2} - 3 (- x) + 2 = 0$
$x^{2} + 3 x + 2 = 0$
$x^{2} + 2 x + x + 2 = 0$
$(x + 2) (x + 1) = 0$
$x = - 1, - 2$
It also yields two solution because $x$ is -ve.
$∴$ The given equation has 4 real solution
$∴$ Option (b) is correct.

AIEEE-2003

Complex Numbers and Quadratic Equation

118127 If $x$ is real, then the value of $\frac{x^{2} - 3 x + 4}{x^{2} + 3 x + 4}$ lies in the interval

1

[\frac{1}{3}, 3]

2

[\frac{1}{5}, 5]

3

[\frac{1}{6}, 6]

4

[\frac{1}{7}, 7]

Explanation:

D Let
$y = \frac{x^{2} - 3 x + 4}{x^{2} + 3 x + 4}$
$\Rightarrow y (x^{2} + 3 x + 4) = x^{2} - 3 x + 4$
or $x^{2} (y - 1) + x (3 y + 3) + 4 y - 4 = 0$
$∴ D \geq 0$ (Discriminant)
$= (3 y + 3)^{2} - 4 (y - 1) 4 (y - 1) \geq 0$
$\Rightarrow 9 [(y + 1)^{2}] - 16 (y - 1)^{2} \geq 0$
$\Rightarrow 9 (y + 1)^{2} \geq 16 (y - 1)^{2}$
$\Rightarrow 3 (y + 1) \geq \pm 4 (y - 1)$
or $3 y + 3 \geq 4 y - 4$
or $y \leq 7$
And $3 y + 3 \geq - 4 y + 4$
$7 y \geq 1$
$y \geq 1 / 7$
$∴ \frac{1}{7} \leq y \leq 7$
$∴ y \in [\frac{1}{7}, 7]$
$∴$ Option (d) is correct.

AP EAMCET-2011

Complex Numbers and Quadratic Equation

118129 If $α, β, γ$ are the roots of the equation $x^{3} - 6 x^{2} + 11 x - 6 = 0$ and if $a = α^{2} + β^{2} + γ^{2} . b = α β + β γ$ $+ γ α$ and $c = (α + β) (β + γ) (γ + α)$ , then the correct inequality among he following is.

1

a < b < C

2

b < a < c

3

b < c < a

4

c < a < b

Explanation:

B Given $α, β, γ$ are the roots of $x^{3} - 6 x^{2} + 11 x -$ $6 = 0$
The equation can be written as.
$(x - α) (x - β) (x - γ) = 0$
$\Rightarrow x^{3} - x^{2} (α + β + γ) + x (α β + β γ + α γ) - α β γ = 0$
$Comparing, We get$
$α + β + γ = 6$
$α β + β γ + α γ = 11$
$α β γ = 6$
$Now, b = 11$
$c = (α + β) (β + γ) (γ + α)$
$= (6 - γ) (6 - α) (6 - β)$
$= 216 + 6 (α β + β γ + α γ) - 36 (α + β + γ) - α β γ$
$= 216 + 6 \times 11 - 36 \times 6 - 6$
$= 6 \times 10 = 60$
$a = α^{2} + β^{2} + γ^{2} = (α + β + γ)^{2} - 2 (α β + β γ + α γ)$
$= 36 - 2 \times 11 = 14$
$∴ b < a < c$
$∴ Option (b) is correct.$
Now, $b = 11$
$∴$ Option (b) is correct.

AP EAMCET-2010

Complex Numbers and Quadratic Equation

118126 If $a, b, c$ are in Arithmetic Progression (AP), then the roots of the equation
$a x^{2} - 2 b x + c = 0$ are

1

1, \frac{c}{a}

2

\frac{- 1}{a}, - c

3

- 1, \frac{- c}{a}

4

- 2, \frac{- c}{2 a}

Explanation:

A Given
$a, b, c are in A. P.$
$∴ b - a = c - b$
$or 2 b = a + c$
${ax}^{2} - 2 bx + c = 0$
$α + β = + \frac{2 b}{a} = \frac{a + c}{a} = (1 + \frac{c}{a})$
$α \cdot β = c / a$
$(α - β)^{2} = (α + β)^{2} - 4 α β$
$= {(1 + \frac{c}{a})}^{2} - 4 \frac{c}{a}$
$= {(1 - \frac{c}{a})}^{2}$
$∵ α - β = 1 - \frac{c}{a}$
$α + β = 1 + \frac{c}{a}$
$∴ 2 α = 2$
$α = 1$
$∴ β = c / a$
$∴ Roots are (1, c/a)$
$∴ option (a) is correct.$

AP EAMCET-18.09.2020

Complex Numbers and Quadratic Equation

118163 The number of the real solutions of the equation $x^{2} - 3 | x | + 2 = 0$ is

1 2

2 4

3 1

4 3

Explanation:

B Given the equation
$x^{2} - 3 | x | + 2 = 0$
Case 1 : If $x > 0 \Rightarrow | x | = x$
$∴ x^{2} - 3 x + 2 = 0$
$\Rightarrow x^{2} - 2 x - x + 2 = 0$
$\Rightarrow x (x - 2) - 1 (x - 2) = 0$
$\Rightarrow (x - 2) (x - 1) = 0$
$∴ x = 1, 2$
$∵$ Both the roots are $+ ve$ . If yields two solutions.
Case 2: If $x < 0 \Rightarrow | x | = - x$
$∴ x^{2} - 3 (- x) + 2 = 0$
$x^{2} + 3 x + 2 = 0$
$x^{2} + 2 x + x + 2 = 0$
$(x + 2) (x + 1) = 0$
$x = - 1, - 2$
It also yields two solution because $x$ is -ve.
$∴$ The given equation has 4 real solution
$∴$ Option (b) is correct.

AIEEE-2003

Complex Numbers and Quadratic Equation

118127 If $x$ is real, then the value of $\frac{x^{2} - 3 x + 4}{x^{2} + 3 x + 4}$ lies in the interval

1

[\frac{1}{3}, 3]

2

[\frac{1}{5}, 5]

3

[\frac{1}{6}, 6]

4

[\frac{1}{7}, 7]

Explanation:

D Let
$y = \frac{x^{2} - 3 x + 4}{x^{2} + 3 x + 4}$
$\Rightarrow y (x^{2} + 3 x + 4) = x^{2} - 3 x + 4$
or $x^{2} (y - 1) + x (3 y + 3) + 4 y - 4 = 0$
$∴ D \geq 0$ (Discriminant)
$= (3 y + 3)^{2} - 4 (y - 1) 4 (y - 1) \geq 0$
$\Rightarrow 9 [(y + 1)^{2}] - 16 (y - 1)^{2} \geq 0$
$\Rightarrow 9 (y + 1)^{2} \geq 16 (y - 1)^{2}$
$\Rightarrow 3 (y + 1) \geq \pm 4 (y - 1)$
or $3 y + 3 \geq 4 y - 4$
or $y \leq 7$
And $3 y + 3 \geq - 4 y + 4$
$7 y \geq 1$
$y \geq 1 / 7$
$∴ \frac{1}{7} \leq y \leq 7$
$∴ y \in [\frac{1}{7}, 7]$
$∴$ Option (d) is correct.

AP EAMCET-2011

Complex Numbers and Quadratic Equation

118129 If $α, β, γ$ are the roots of the equation $x^{3} - 6 x^{2} + 11 x - 6 = 0$ and if $a = α^{2} + β^{2} + γ^{2} . b = α β + β γ$ $+ γ α$ and $c = (α + β) (β + γ) (γ + α)$ , then the correct inequality among he following is.

1

a < b < C

2

b < a < c

3

b < c < a

4

c < a < b

Explanation:

B Given $α, β, γ$ are the roots of $x^{3} - 6 x^{2} + 11 x -$ $6 = 0$
The equation can be written as.
$(x - α) (x - β) (x - γ) = 0$
$\Rightarrow x^{3} - x^{2} (α + β + γ) + x (α β + β γ + α γ) - α β γ = 0$
$Comparing, We get$
$α + β + γ = 6$
$α β + β γ + α γ = 11$
$α β γ = 6$
$Now, b = 11$
$c = (α + β) (β + γ) (γ + α)$
$= (6 - γ) (6 - α) (6 - β)$
$= 216 + 6 (α β + β γ + α γ) - 36 (α + β + γ) - α β γ$
$= 216 + 6 \times 11 - 36 \times 6 - 6$
$= 6 \times 10 = 60$
$a = α^{2} + β^{2} + γ^{2} = (α + β + γ)^{2} - 2 (α β + β γ + α γ)$
$= 36 - 2 \times 11 = 14$
$∴ b < a < c$
$∴ Option (b) is correct.$
Now, $b = 11$
$∴$ Option (b) is correct.

AP EAMCET-2010

Complex Numbers and Quadratic Equation

118126 If $a, b, c$ are in Arithmetic Progression (AP), then the roots of the equation
$a x^{2} - 2 b x + c = 0$ are

1

1, \frac{c}{a}

2

\frac{- 1}{a}, - c

3

- 1, \frac{- c}{a}

4

- 2, \frac{- c}{2 a}

Explanation:

A Given
$a, b, c are in A. P.$
$∴ b - a = c - b$
$or 2 b = a + c$
${ax}^{2} - 2 bx + c = 0$
$α + β = + \frac{2 b}{a} = \frac{a + c}{a} = (1 + \frac{c}{a})$
$α \cdot β = c / a$
$(α - β)^{2} = (α + β)^{2} - 4 α β$
$= {(1 + \frac{c}{a})}^{2} - 4 \frac{c}{a}$
$= {(1 - \frac{c}{a})}^{2}$
$∵ α - β = 1 - \frac{c}{a}$
$α + β = 1 + \frac{c}{a}$
$∴ 2 α = 2$
$α = 1$
$∴ β = c / a$
$∴ Roots are (1, c/a)$
$∴ option (a) is correct.$

AP EAMCET-18.09.2020

Complex Numbers and Quadratic Equation

118163 The number of the real solutions of the equation $x^{2} - 3 | x | + 2 = 0$ is

1 2

2 4

3 1

4 3

Explanation:

B Given the equation
$x^{2} - 3 | x | + 2 = 0$
Case 1 : If $x > 0 \Rightarrow | x | = x$
$∴ x^{2} - 3 x + 2 = 0$
$\Rightarrow x^{2} - 2 x - x + 2 = 0$
$\Rightarrow x (x - 2) - 1 (x - 2) = 0$
$\Rightarrow (x - 2) (x - 1) = 0$
$∴ x = 1, 2$
$∵$ Both the roots are $+ ve$ . If yields two solutions.
Case 2: If $x < 0 \Rightarrow | x | = - x$
$∴ x^{2} - 3 (- x) + 2 = 0$
$x^{2} + 3 x + 2 = 0$
$x^{2} + 2 x + x + 2 = 0$
$(x + 2) (x + 1) = 0$
$x = - 1, - 2$
It also yields two solution because $x$ is -ve.
$∴$ The given equation has 4 real solution
$∴$ Option (b) is correct.

AIEEE-2003

Complex Numbers and Quadratic Equation

118127 If $x$ is real, then the value of $\frac{x^{2} - 3 x + 4}{x^{2} + 3 x + 4}$ lies in the interval

1

[\frac{1}{3}, 3]

2

[\frac{1}{5}, 5]

3

[\frac{1}{6}, 6]

4

[\frac{1}{7}, 7]

Explanation:

D Let
$y = \frac{x^{2} - 3 x + 4}{x^{2} + 3 x + 4}$
$\Rightarrow y (x^{2} + 3 x + 4) = x^{2} - 3 x + 4$
or $x^{2} (y - 1) + x (3 y + 3) + 4 y - 4 = 0$
$∴ D \geq 0$ (Discriminant)
$= (3 y + 3)^{2} - 4 (y - 1) 4 (y - 1) \geq 0$
$\Rightarrow 9 [(y + 1)^{2}] - 16 (y - 1)^{2} \geq 0$
$\Rightarrow 9 (y + 1)^{2} \geq 16 (y - 1)^{2}$
$\Rightarrow 3 (y + 1) \geq \pm 4 (y - 1)$
or $3 y + 3 \geq 4 y - 4$
or $y \leq 7$
And $3 y + 3 \geq - 4 y + 4$
$7 y \geq 1$
$y \geq 1 / 7$
$∴ \frac{1}{7} \leq y \leq 7$
$∴ y \in [\frac{1}{7}, 7]$
$∴$ Option (d) is correct.

AP EAMCET-2011

Complex Numbers and Quadratic Equation

118129 If $α, β, γ$ are the roots of the equation $x^{3} - 6 x^{2} + 11 x - 6 = 0$ and if $a = α^{2} + β^{2} + γ^{2} . b = α β + β γ$ $+ γ α$ and $c = (α + β) (β + γ) (γ + α)$ , then the correct inequality among he following is.

1

a < b < C

2

b < a < c

3

b < c < a

4

c < a < b

Explanation:

B Given $α, β, γ$ are the roots of $x^{3} - 6 x^{2} + 11 x -$ $6 = 0$
The equation can be written as.
$(x - α) (x - β) (x - γ) = 0$
$\Rightarrow x^{3} - x^{2} (α + β + γ) + x (α β + β γ + α γ) - α β γ = 0$
$Comparing, We get$
$α + β + γ = 6$
$α β + β γ + α γ = 11$
$α β γ = 6$
$Now, b = 11$
$c = (α + β) (β + γ) (γ + α)$
$= (6 - γ) (6 - α) (6 - β)$
$= 216 + 6 (α β + β γ + α γ) - 36 (α + β + γ) - α β γ$
$= 216 + 6 \times 11 - 36 \times 6 - 6$
$= 6 \times 10 = 60$
$a = α^{2} + β^{2} + γ^{2} = (α + β + γ)^{2} - 2 (α β + β γ + α γ)$
$= 36 - 2 \times 11 = 14$
$∴ b < a < c$
$∴ Option (b) is correct.$
Now, $b = 11$
$∴$ Option (b) is correct.

AP EAMCET-2010

Complex Numbers and Quadratic Equation

118126 If $a, b, c$ are in Arithmetic Progression (AP), then the roots of the equation
$a x^{2} - 2 b x + c = 0$ are

1

1, \frac{c}{a}

2

\frac{- 1}{a}, - c

3

- 1, \frac{- c}{a}

4

- 2, \frac{- c}{2 a}

Explanation:

A Given
$a, b, c are in A. P.$
$∴ b - a = c - b$
$or 2 b = a + c$
${ax}^{2} - 2 bx + c = 0$
$α + β = + \frac{2 b}{a} = \frac{a + c}{a} = (1 + \frac{c}{a})$
$α \cdot β = c / a$
$(α - β)^{2} = (α + β)^{2} - 4 α β$
$= {(1 + \frac{c}{a})}^{2} - 4 \frac{c}{a}$
$= {(1 - \frac{c}{a})}^{2}$
$∵ α - β = 1 - \frac{c}{a}$
$α + β = 1 + \frac{c}{a}$
$∴ 2 α = 2$
$α = 1$
$∴ β = c / a$
$∴ Roots are (1, c/a)$
$∴ option (a) is correct.$

AP EAMCET-18.09.2020

Complex Numbers and Quadratic Equation

118163 The number of the real solutions of the equation $x^{2} - 3 | x | + 2 = 0$ is

1 2

2 4

3 1

4 3

Explanation:

B Given the equation
$x^{2} - 3 | x | + 2 = 0$
Case 1 : If $x > 0 \Rightarrow | x | = x$
$∴ x^{2} - 3 x + 2 = 0$
$\Rightarrow x^{2} - 2 x - x + 2 = 0$
$\Rightarrow x (x - 2) - 1 (x - 2) = 0$
$\Rightarrow (x - 2) (x - 1) = 0$
$∴ x = 1, 2$
$∵$ Both the roots are $+ ve$ . If yields two solutions.
Case 2: If $x < 0 \Rightarrow | x | = - x$
$∴ x^{2} - 3 (- x) + 2 = 0$
$x^{2} + 3 x + 2 = 0$
$x^{2} + 2 x + x + 2 = 0$
$(x + 2) (x + 1) = 0$
$x = - 1, - 2$
It also yields two solution because $x$ is -ve.
$∴$ The given equation has 4 real solution
$∴$ Option (b) is correct.

AIEEE-2003

Complex Numbers and Quadratic Equation

118127 If $x$ is real, then the value of $\frac{x^{2} - 3 x + 4}{x^{2} + 3 x + 4}$ lies in the interval

1

[\frac{1}{3}, 3]

2

[\frac{1}{5}, 5]

3

[\frac{1}{6}, 6]

4

[\frac{1}{7}, 7]

Explanation:

D Let
$y = \frac{x^{2} - 3 x + 4}{x^{2} + 3 x + 4}$
$\Rightarrow y (x^{2} + 3 x + 4) = x^{2} - 3 x + 4$
or $x^{2} (y - 1) + x (3 y + 3) + 4 y - 4 = 0$
$∴ D \geq 0$ (Discriminant)
$= (3 y + 3)^{2} - 4 (y - 1) 4 (y - 1) \geq 0$
$\Rightarrow 9 [(y + 1)^{2}] - 16 (y - 1)^{2} \geq 0$
$\Rightarrow 9 (y + 1)^{2} \geq 16 (y - 1)^{2}$
$\Rightarrow 3 (y + 1) \geq \pm 4 (y - 1)$
or $3 y + 3 \geq 4 y - 4$
or $y \leq 7$
And $3 y + 3 \geq - 4 y + 4$
$7 y \geq 1$
$y \geq 1 / 7$
$∴ \frac{1}{7} \leq y \leq 7$
$∴ y \in [\frac{1}{7}, 7]$
$∴$ Option (d) is correct.

AP EAMCET-2011

Complex Numbers and Quadratic Equation

118129 If $α, β, γ$ are the roots of the equation $x^{3} - 6 x^{2} + 11 x - 6 = 0$ and if $a = α^{2} + β^{2} + γ^{2} . b = α β + β γ$ $+ γ α$ and $c = (α + β) (β + γ) (γ + α)$ , then the correct inequality among he following is.

1

a < b < C

2

b < a < c

3

b < c < a

4

c < a < b

Explanation:

B Given $α, β, γ$ are the roots of $x^{3} - 6 x^{2} + 11 x -$ $6 = 0$
The equation can be written as.
$(x - α) (x - β) (x - γ) = 0$
$\Rightarrow x^{3} - x^{2} (α + β + γ) + x (α β + β γ + α γ) - α β γ = 0$
$Comparing, We get$
$α + β + γ = 6$
$α β + β γ + α γ = 11$
$α β γ = 6$
$Now, b = 11$
$c = (α + β) (β + γ) (γ + α)$
$= (6 - γ) (6 - α) (6 - β)$
$= 216 + 6 (α β + β γ + α γ) - 36 (α + β + γ) - α β γ$
$= 216 + 6 \times 11 - 36 \times 6 - 6$
$= 6 \times 10 = 60$
$a = α^{2} + β^{2} + γ^{2} = (α + β + γ)^{2} - 2 (α β + β γ + α γ)$
$= 36 - 2 \times 11 = 14$
$∴ b < a < c$
$∴ Option (b) is correct.$
Now, $b = 11$
$∴$ Option (b) is correct.

AP EAMCET-2010