QBManager

Complex Numbers and Quadratic Equation

118107 The minimum value of the sum of the squares of the roots of $x^{2} + (3 - a) x + 1 = 2 a$ is :

1 4

2 5

3 6

4 8

Explanation:

C Given,
$x^{2} + (3 - a) x + 1 = 2 a$
$x^{2} + (3 - a) x + 1 - 2 a = 0$
Let, $α, β$ are the roots of the above,
Then, $α + β = (a - 3)$
$α β = 1 - 2 a$
$α^{2} + β^{2} = (α + β)^{2} - 2 α β$
$= (a - 3)^{2} - 2 (1 - 2 a)$
$= a^{2} + 9 - 6 a - 2 + 4 a$
$= a^{2} - 2 a + 7$
$= (a - 1)^{2} + 7 - 1 = (a - 1)^{2} + 6$
$∴ Minimum value of α^{2} + β^{2} = 6 [∵ (a - 1)^{2}] = 0$

JEE Main-26.07.2022

Complex Numbers and Quadratic Equation

118108 For $x^{2} - (a + 3) | x | + 4 = 0$ to have real solutions, the range of a is

1

(- \infty, - 7] \cup [1, \infty)

2

(- 3, \infty)

3

(- \infty, - 7]

4

[1, \infty)

Explanation:

A Given, $x^{2} - (a + 3) | x | + 4 = 0$
Case I:- $x > 0$ , then $| x | = x$
$∴ x^{2} - (a + 3) x + 4 = 0$
But $x$ is real i.e. real solution.
$D = (a + 3)^{2} - 16 \geq 0$
$D = (a + 3 + 4) (a + 3 - 4) \geq 0$
$= (a + 7) (a - 1) \geq 0$
$= a \geq 1 or a \leq - 7 But x > 0$
$Case II : - x < 0 \Rightarrow | x | = - x$
$∴ x^{2} - (a + 3) (- x) + 4 = 0$
$\Rightarrow x^{2} + (a + 3) x + 4 = 0$
$Discriminant (D) = (a + 3)^{2} - 16 \geq 0$
$∴ We get, a \geq 1 or a \leq - 7$
$∴ a \in (- \infty, - 7) \cup (1, \infty)$

AMU-2019

Complex Numbers and Quadratic Equation

118109 The number of real solutions of equation $3 (x^{2} + \frac{1}{x^{2}}) - 2 (x + \frac{1}{x}) + 5 = 0$ is

1 4

2 3

3 2

4 0

Explanation:

D Given,
$3 (x^{2} + \frac{1}{x^{2}}) - 2 (x + \frac{1}{x}) + 5 = 0$
$3 \cdot [{(x + \frac{1}{x})}^{2} - 2] - 2 (x + \frac{1}{x}) + 5 = 0$
Let, $(x + \frac{1}{x}) = t$ we get-
or
$3 [t^{2} - 2] - 2 t + 5 = 0$
$3 t^{2} - 6 - 2 t + 5 = 0$
$3 t^{2} - 2 t - 1 = 0$
$3 t^{2} - 3 t + t - 1 = 0$
$3 t (t - 1) + 1 (t - 1) = 0$
$(3 t + 1) (t - 1) = 0$
So, we get,
$3 t + 1 = 0, t = - 1 / 3$
$x + \frac{1}{x} = 1$
$x^{2} + 1 = x$
or
$x^{2} - x + 1 = 0$
$p = (- 1)^{2} - 4 = - 3 < 0$
$\Rightarrow$ No real solution.
and $3 (x + \frac{1}{x}) + 1 = 0$
$\Rightarrow 3 (x^{2} + 1) + x = 0$
$\Rightarrow 3 x^{2} + 3 + x = 0$
$∴ D = (1)^{2} - 4 \times 3 \times 3 = - 35$
$\Rightarrow$ No real solution
$∴$ The number of real solutions is zero.

JEE Main-24.01.2023

Complex Numbers and Quadratic Equation

118110 Let $α$ be a root of the equation $1 + x^{2} + x^{4} = 0$ . Then the value of $α^{1011 +} α^{2022]} - α^{3033}$ is equal to :

1 1

2

α

3

1 + α

4

1 + 2 α

Explanation:

A Given,
$1 + x^{2} + x^{4} = 0 .$
Let, $ω$ is root ( $ω$ is cube root of unity)
Then, $ω^{1011} + ω^{2022} - ω^{3033}$
$= {(ω^{3})}^{337} + {(ω^{3})}^{674} - {(ω^{3})}^{1011}$
$= 1 + 1 - 1 = 1$

JEE Main-29.06.2022

Complex Numbers and Quadratic Equation

118107 The minimum value of the sum of the squares of the roots of $x^{2} + (3 - a) x + 1 = 2 a$ is :

1 4

2 5

3 6

4 8

Explanation:

C Given,
$x^{2} + (3 - a) x + 1 = 2 a$
$x^{2} + (3 - a) x + 1 - 2 a = 0$
Let, $α, β$ are the roots of the above,
Then, $α + β = (a - 3)$
$α β = 1 - 2 a$
$α^{2} + β^{2} = (α + β)^{2} - 2 α β$
$= (a - 3)^{2} - 2 (1 - 2 a)$
$= a^{2} + 9 - 6 a - 2 + 4 a$
$= a^{2} - 2 a + 7$
$= (a - 1)^{2} + 7 - 1 = (a - 1)^{2} + 6$
$∴ Minimum value of α^{2} + β^{2} = 6 [∵ (a - 1)^{2}] = 0$

JEE Main-26.07.2022

Complex Numbers and Quadratic Equation

118108 For $x^{2} - (a + 3) | x | + 4 = 0$ to have real solutions, the range of a is

1

(- \infty, - 7] \cup [1, \infty)

2

(- 3, \infty)

3

(- \infty, - 7]

4

[1, \infty)

Explanation:

A Given, $x^{2} - (a + 3) | x | + 4 = 0$
Case I:- $x > 0$ , then $| x | = x$
$∴ x^{2} - (a + 3) x + 4 = 0$
But $x$ is real i.e. real solution.
$D = (a + 3)^{2} - 16 \geq 0$
$D = (a + 3 + 4) (a + 3 - 4) \geq 0$
$= (a + 7) (a - 1) \geq 0$
$= a \geq 1 or a \leq - 7 But x > 0$
$Case II : - x < 0 \Rightarrow | x | = - x$
$∴ x^{2} - (a + 3) (- x) + 4 = 0$
$\Rightarrow x^{2} + (a + 3) x + 4 = 0$
$Discriminant (D) = (a + 3)^{2} - 16 \geq 0$
$∴ We get, a \geq 1 or a \leq - 7$
$∴ a \in (- \infty, - 7) \cup (1, \infty)$

AMU-2019

Complex Numbers and Quadratic Equation

118109 The number of real solutions of equation $3 (x^{2} + \frac{1}{x^{2}}) - 2 (x + \frac{1}{x}) + 5 = 0$ is

1 4

2 3

3 2

4 0

Explanation:

D Given,
$3 (x^{2} + \frac{1}{x^{2}}) - 2 (x + \frac{1}{x}) + 5 = 0$
$3 \cdot [{(x + \frac{1}{x})}^{2} - 2] - 2 (x + \frac{1}{x}) + 5 = 0$
Let, $(x + \frac{1}{x}) = t$ we get-
or
$3 [t^{2} - 2] - 2 t + 5 = 0$
$3 t^{2} - 6 - 2 t + 5 = 0$
$3 t^{2} - 2 t - 1 = 0$
$3 t^{2} - 3 t + t - 1 = 0$
$3 t (t - 1) + 1 (t - 1) = 0$
$(3 t + 1) (t - 1) = 0$
So, we get,
$3 t + 1 = 0, t = - 1 / 3$
$x + \frac{1}{x} = 1$
$x^{2} + 1 = x$
or
$x^{2} - x + 1 = 0$
$p = (- 1)^{2} - 4 = - 3 < 0$
$\Rightarrow$ No real solution.
and $3 (x + \frac{1}{x}) + 1 = 0$
$\Rightarrow 3 (x^{2} + 1) + x = 0$
$\Rightarrow 3 x^{2} + 3 + x = 0$
$∴ D = (1)^{2} - 4 \times 3 \times 3 = - 35$
$\Rightarrow$ No real solution
$∴$ The number of real solutions is zero.

JEE Main-24.01.2023

Complex Numbers and Quadratic Equation

118110 Let $α$ be a root of the equation $1 + x^{2} + x^{4} = 0$ . Then the value of $α^{1011 +} α^{2022]} - α^{3033}$ is equal to :

1 1

2

α

3

1 + α

4

1 + 2 α

Explanation:

A Given,
$1 + x^{2} + x^{4} = 0 .$
Let, $ω$ is root ( $ω$ is cube root of unity)
Then, $ω^{1011} + ω^{2022} - ω^{3033}$
$= {(ω^{3})}^{337} + {(ω^{3})}^{674} - {(ω^{3})}^{1011}$
$= 1 + 1 - 1 = 1$

JEE Main-29.06.2022

Complex Numbers and Quadratic Equation

118107 The minimum value of the sum of the squares of the roots of $x^{2} + (3 - a) x + 1 = 2 a$ is :

1 4

2 5

3 6

4 8

Explanation:

C Given,
$x^{2} + (3 - a) x + 1 = 2 a$
$x^{2} + (3 - a) x + 1 - 2 a = 0$
Let, $α, β$ are the roots of the above,
Then, $α + β = (a - 3)$
$α β = 1 - 2 a$
$α^{2} + β^{2} = (α + β)^{2} - 2 α β$
$= (a - 3)^{2} - 2 (1 - 2 a)$
$= a^{2} + 9 - 6 a - 2 + 4 a$
$= a^{2} - 2 a + 7$
$= (a - 1)^{2} + 7 - 1 = (a - 1)^{2} + 6$
$∴ Minimum value of α^{2} + β^{2} = 6 [∵ (a - 1)^{2}] = 0$

JEE Main-26.07.2022

Complex Numbers and Quadratic Equation

118108 For $x^{2} - (a + 3) | x | + 4 = 0$ to have real solutions, the range of a is

1

(- \infty, - 7] \cup [1, \infty)

2

(- 3, \infty)

3

(- \infty, - 7]

4

[1, \infty)

Explanation:

A Given, $x^{2} - (a + 3) | x | + 4 = 0$
Case I:- $x > 0$ , then $| x | = x$
$∴ x^{2} - (a + 3) x + 4 = 0$
But $x$ is real i.e. real solution.
$D = (a + 3)^{2} - 16 \geq 0$
$D = (a + 3 + 4) (a + 3 - 4) \geq 0$
$= (a + 7) (a - 1) \geq 0$
$= a \geq 1 or a \leq - 7 But x > 0$
$Case II : - x < 0 \Rightarrow | x | = - x$
$∴ x^{2} - (a + 3) (- x) + 4 = 0$
$\Rightarrow x^{2} + (a + 3) x + 4 = 0$
$Discriminant (D) = (a + 3)^{2} - 16 \geq 0$
$∴ We get, a \geq 1 or a \leq - 7$
$∴ a \in (- \infty, - 7) \cup (1, \infty)$

AMU-2019

Complex Numbers and Quadratic Equation

118109 The number of real solutions of equation $3 (x^{2} + \frac{1}{x^{2}}) - 2 (x + \frac{1}{x}) + 5 = 0$ is

1 4

2 3

3 2

4 0

Explanation:

D Given,
$3 (x^{2} + \frac{1}{x^{2}}) - 2 (x + \frac{1}{x}) + 5 = 0$
$3 \cdot [{(x + \frac{1}{x})}^{2} - 2] - 2 (x + \frac{1}{x}) + 5 = 0$
Let, $(x + \frac{1}{x}) = t$ we get-
or
$3 [t^{2} - 2] - 2 t + 5 = 0$
$3 t^{2} - 6 - 2 t + 5 = 0$
$3 t^{2} - 2 t - 1 = 0$
$3 t^{2} - 3 t + t - 1 = 0$
$3 t (t - 1) + 1 (t - 1) = 0$
$(3 t + 1) (t - 1) = 0$
So, we get,
$3 t + 1 = 0, t = - 1 / 3$
$x + \frac{1}{x} = 1$
$x^{2} + 1 = x$
or
$x^{2} - x + 1 = 0$
$p = (- 1)^{2} - 4 = - 3 < 0$
$\Rightarrow$ No real solution.
and $3 (x + \frac{1}{x}) + 1 = 0$
$\Rightarrow 3 (x^{2} + 1) + x = 0$
$\Rightarrow 3 x^{2} + 3 + x = 0$
$∴ D = (1)^{2} - 4 \times 3 \times 3 = - 35$
$\Rightarrow$ No real solution
$∴$ The number of real solutions is zero.

JEE Main-24.01.2023

Complex Numbers and Quadratic Equation

118110 Let $α$ be a root of the equation $1 + x^{2} + x^{4} = 0$ . Then the value of $α^{1011 +} α^{2022]} - α^{3033}$ is equal to :

1 1

2

α

3

1 + α

4

1 + 2 α

Explanation:

A Given,
$1 + x^{2} + x^{4} = 0 .$
Let, $ω$ is root ( $ω$ is cube root of unity)
Then, $ω^{1011} + ω^{2022} - ω^{3033}$
$= {(ω^{3})}^{337} + {(ω^{3})}^{674} - {(ω^{3})}^{1011}$
$= 1 + 1 - 1 = 1$

JEE Main-29.06.2022

Complex Numbers and Quadratic Equation

118107 The minimum value of the sum of the squares of the roots of $x^{2} + (3 - a) x + 1 = 2 a$ is :

1 4

2 5

3 6

4 8

Explanation:

C Given,
$x^{2} + (3 - a) x + 1 = 2 a$
$x^{2} + (3 - a) x + 1 - 2 a = 0$
Let, $α, β$ are the roots of the above,
Then, $α + β = (a - 3)$
$α β = 1 - 2 a$
$α^{2} + β^{2} = (α + β)^{2} - 2 α β$
$= (a - 3)^{2} - 2 (1 - 2 a)$
$= a^{2} + 9 - 6 a - 2 + 4 a$
$= a^{2} - 2 a + 7$
$= (a - 1)^{2} + 7 - 1 = (a - 1)^{2} + 6$
$∴ Minimum value of α^{2} + β^{2} = 6 [∵ (a - 1)^{2}] = 0$

JEE Main-26.07.2022

Complex Numbers and Quadratic Equation

118108 For $x^{2} - (a + 3) | x | + 4 = 0$ to have real solutions, the range of a is

1

(- \infty, - 7] \cup [1, \infty)

2

(- 3, \infty)

3

(- \infty, - 7]

4

[1, \infty)

Explanation:

A Given, $x^{2} - (a + 3) | x | + 4 = 0$
Case I:- $x > 0$ , then $| x | = x$
$∴ x^{2} - (a + 3) x + 4 = 0$
But $x$ is real i.e. real solution.
$D = (a + 3)^{2} - 16 \geq 0$
$D = (a + 3 + 4) (a + 3 - 4) \geq 0$
$= (a + 7) (a - 1) \geq 0$
$= a \geq 1 or a \leq - 7 But x > 0$
$Case II : - x < 0 \Rightarrow | x | = - x$
$∴ x^{2} - (a + 3) (- x) + 4 = 0$
$\Rightarrow x^{2} + (a + 3) x + 4 = 0$
$Discriminant (D) = (a + 3)^{2} - 16 \geq 0$
$∴ We get, a \geq 1 or a \leq - 7$
$∴ a \in (- \infty, - 7) \cup (1, \infty)$

AMU-2019

Complex Numbers and Quadratic Equation

118109 The number of real solutions of equation $3 (x^{2} + \frac{1}{x^{2}}) - 2 (x + \frac{1}{x}) + 5 = 0$ is

1 4

2 3

3 2

4 0

Explanation:

D Given,
$3 (x^{2} + \frac{1}{x^{2}}) - 2 (x + \frac{1}{x}) + 5 = 0$
$3 \cdot [{(x + \frac{1}{x})}^{2} - 2] - 2 (x + \frac{1}{x}) + 5 = 0$
Let, $(x + \frac{1}{x}) = t$ we get-
or
$3 [t^{2} - 2] - 2 t + 5 = 0$
$3 t^{2} - 6 - 2 t + 5 = 0$
$3 t^{2} - 2 t - 1 = 0$
$3 t^{2} - 3 t + t - 1 = 0$
$3 t (t - 1) + 1 (t - 1) = 0$
$(3 t + 1) (t - 1) = 0$
So, we get,
$3 t + 1 = 0, t = - 1 / 3$
$x + \frac{1}{x} = 1$
$x^{2} + 1 = x$
or
$x^{2} - x + 1 = 0$
$p = (- 1)^{2} - 4 = - 3 < 0$
$\Rightarrow$ No real solution.
and $3 (x + \frac{1}{x}) + 1 = 0$
$\Rightarrow 3 (x^{2} + 1) + x = 0$
$\Rightarrow 3 x^{2} + 3 + x = 0$
$∴ D = (1)^{2} - 4 \times 3 \times 3 = - 35$
$\Rightarrow$ No real solution
$∴$ The number of real solutions is zero.

JEE Main-24.01.2023

Complex Numbers and Quadratic Equation

118110 Let $α$ be a root of the equation $1 + x^{2} + x^{4} = 0$ . Then the value of $α^{1011 +} α^{2022]} - α^{3033}$ is equal to :

1 1

2

α

3

1 + α

4

1 + 2 α

Explanation:

A Given,
$1 + x^{2} + x^{4} = 0 .$
Let, $ω$ is root ( $ω$ is cube root of unity)
Then, $ω^{1011} + ω^{2022} - ω^{3033}$
$= {(ω^{3})}^{337} + {(ω^{3})}^{674} - {(ω^{3})}^{1011}$
$= 1 + 1 - 1 = 1$

JEE Main-29.06.2022

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