QBManager

Complex Numbers and Quadratic Equation

118007 If $\alpha, \beta$ are the roots of $x^2-2 x+4=0$, for $n \in$ $\mathrm{N}$, what is the value of $\boldsymbol α^{n} + \boldsymbol β^{n} =$

1

2^{n + 2} \cos (\frac{n π}{3})

2

2^{n + 1} \cos (\frac{n π}{3})

3

2^{n + 1} \cos (\frac{n π}{6})

4

2^{n + 2} \cos (\frac{n π}{6})

Explanation:

B Given that, $α$ and $β$ are the roots of the equation $x^{2} - 2 x + 4 = 0$
$∴ α + β = 2$
$α \cdot β = 4$
So, $(α - β) = \sqrt{(α + β)^{2} - 4 α β}$
$(α - β) = \sqrt{(2)^{2} - 4 \times 4}$
$α - β = \sqrt{- 12}$
$(α - β) = 2 \sqrt{3} i$
Adding equation (i) and (ii), we get -
$2 α = 2 + 2 \sqrt{3 i}$
$α = 2 [\frac{1}{2} + \frac{\sqrt{3}}{2} i]$
$α = 2 [\cos (\frac{π}{3}) + i \sin (\frac{π}{3})]$
On putting the value of $α$ in equation (i) we get -
$β = \frac{1}{2} [2 - 2 \sqrt{3} i]$
$β = 2 [\frac{1}{2} - \frac{\sqrt{3}}{2} i]$
$= 2 [\cos (\frac{π}{3}) - i \sin (\frac{π}{3})]$
$α^{1} + β^{1} = 2^{1} [2 \cos (\frac{1 π}{3})]$
$α^{n} + β^{n} = 2^{n} [2 \cos (\frac{n π}{3})]$
$α^{n} + β^{n} = 2^{n + 1} \cos (\frac{n π}{3})$

AP EAMCET - 18.09.2020 Shift - II

Complex Numbers and Quadratic Equation

118008 The value of
$1 + \sum_{k = 0}^{14} {\cos \frac{(2 k + 1)}{15} π + i \sin \frac{(2 k + 1)}{15} π} i$

1 0

2 -1

3 1

4 i

Explanation:

C $1 + \sum_{k = 0}^{14} {\cos \frac{(2 k + 1)}{15} π + i \sin \frac{(2 k + 1)}{15} π}$
We know that,
$\cos θ + i \sin θ = e^{i θ}$
Then, $1 + \sum_{k = 0}^{14} e^{\frac{i (2 k + 1) π}{15}}$
$= 1 + (α + α^{3} + α^{5} + α^{7} + \dots . . α^{29})$
$α = e^{i π / 15} \Rightarrow α^{30} = e^{2 i π} = 1$
It is G.P series,
$a = α, r = α^{2}$
$= 1 + α (\frac{1 - {(α^{2})}^{15}}{1 - α^{2}})$
$= 1 + α (\frac{1 - α^{30}}{1 - α^{2}})$
$= 1 + α (\frac{1 - 1}{1 - α^{2}})$
$= 1 + 0$
$= 1$
$(∵ α^{30} = 1)$

AMU-2005

Complex Numbers and Quadratic Equation

118009 The value of
$1 + \sum_{k = 0}^{14} {\cos \frac{(2 k + 1)}{15} π + i \sin \frac{(2 k + 1)}{15} π}$ is

1 0

2 -1

3 1

4

i

Explanation:

C $1 + \sum_{k = 0}^{14} {\cos \frac{(2 k + 1)}{15} π + i \sin \frac{(2 k + 1)}{15} π}$
We know that, $\cos θ + i \sin θ = e^{i θ}$
Then, $1 + \sum_{k = 0}^{14} e^{\frac{i (2 k + 1) π}{15}}$
$= 1 + (α + α^{3} + α^{5} + α^{7} + \dots . . α^{29})$
$α = e^{i π / 15} \Rightarrow α^{30} = e^{2 i π} = 1$
It is G.P series,
$a = α, r = α^{2}$
$= 1 + α (\frac{1 - {(α^{2})}^{15}}{1 - α^{2}})$
$= 1 + α (\frac{1 - α^{30}}{1 - α^{2}})$
$= 1 + α (\frac{1 - 1}{1 - α^{2}})$
$= 1 + 0$
$= 1$
$(∵ α^{30} = 1)$

AMU-2005

Complex Numbers and Quadratic Equation

118017 If $x + \frac{1}{x} = 2 \cos θ$ , then for any integer $n$ , $x^{n} + \frac{1}{x^{n}} =$

1

2 \cos n θ

2

2 \sin n θ

3

2 i \cos n θ

4

2 i \sin n θ

Explanation:

A Given,
$x + \frac{1}{x} = 2 \cos θ$
$x^{2} + 1 = 2 x \cos θ$
$x^{2} - 2 x \cos θ + 1 = 0$
$∴ x = \frac{2 \cos θ \pm \sqrt{(2 \cos θ)^{2} - 4}}{2}]$
$= \cos θ \pm i \sin θ$
Let us take $x = \cos θ + i \sin θ$
$∴ \frac{1}{(x)} = \cos θ - i \sin θ$
$∴ x^{n} + \frac{1}{x^{n}} = (\cos θ + i \sin θ)^{n} + (\cos θ - i \sin θ)^{n}$
$= \cos n θ + i \sin n θ + \cos n θ - i \sin n θ$
$2 \cos n θ$

WB JEE-2011

Complex Numbers and Quadratic Equation

118007 If $α, β$ are the roots of $x^{2} - 2 x + 4 = 0$ , for $n \in$ $N$ , what is the value of $\boldsymbol α^{n} + \boldsymbol β^{n} =$

1

2^{n + 2} \cos (\frac{n π}{3})

2

2^{n + 1} \cos (\frac{n π}{3})

3

2^{n + 1} \cos (\frac{n π}{6})

4

2^{n + 2} \cos (\frac{n π}{6})

Explanation:

B Given that, $α$ and $β$ are the roots of the equation $x^{2} - 2 x + 4 = 0$
$∴ α + β = 2$
$α \cdot β = 4$
So, $(α - β) = \sqrt{(α + β)^{2} - 4 α β}$
$(α - β) = \sqrt{(2)^{2} - 4 \times 4}$
$α - β = \sqrt{- 12}$
$(α - β) = 2 \sqrt{3} i$
Adding equation (i) and (ii), we get -
$2 α = 2 + 2 \sqrt{3 i}$
$α = 2 [\frac{1}{2} + \frac{\sqrt{3}}{2} i]$
$α = 2 [\cos (\frac{π}{3}) + i \sin (\frac{π}{3})]$
On putting the value of $α$ in equation (i) we get -
$β = \frac{1}{2} [2 - 2 \sqrt{3} i]$
$β = 2 [\frac{1}{2} - \frac{\sqrt{3}}{2} i]$
$= 2 [\cos (\frac{π}{3}) - i \sin (\frac{π}{3})]$
$α^{1} + β^{1} = 2^{1} [2 \cos (\frac{1 π}{3})]$
$α^{n} + β^{n} = 2^{n} [2 \cos (\frac{n π}{3})]$
$α^{n} + β^{n} = 2^{n + 1} \cos (\frac{n π}{3})$

AP EAMCET - 18.09.2020 Shift - II

Complex Numbers and Quadratic Equation

118008 The value of
$1 + \sum_{k = 0}^{14} {\cos \frac{(2 k + 1)}{15} π + i \sin \frac{(2 k + 1)}{15} π} i$

1 0

2 -1

3 1

4 i

Explanation:

C $1 + \sum_{k = 0}^{14} {\cos \frac{(2 k + 1)}{15} π + i \sin \frac{(2 k + 1)}{15} π}$
We know that,
$\cos θ + i \sin θ = e^{i θ}$
Then, $1 + \sum_{k = 0}^{14} e^{\frac{i (2 k + 1) π}{15}}$
$= 1 + (α + α^{3} + α^{5} + α^{7} + \dots . . α^{29})$
$α = e^{i π / 15} \Rightarrow α^{30} = e^{2 i π} = 1$
It is G.P series,
$a = α, r = α^{2}$
$= 1 + α (\frac{1 - {(α^{2})}^{15}}{1 - α^{2}})$
$= 1 + α (\frac{1 - α^{30}}{1 - α^{2}})$
$= 1 + α (\frac{1 - 1}{1 - α^{2}})$
$= 1 + 0$
$= 1$
$(∵ α^{30} = 1)$

AMU-2005

Complex Numbers and Quadratic Equation

118009 The value of
$1 + \sum_{k = 0}^{14} {\cos \frac{(2 k + 1)}{15} π + i \sin \frac{(2 k + 1)}{15} π}$ is

1 0

2 -1

3 1

4

i

Explanation:

C $1 + \sum_{k = 0}^{14} {\cos \frac{(2 k + 1)}{15} π + i \sin \frac{(2 k + 1)}{15} π}$
We know that, $\cos θ + i \sin θ = e^{i θ}$
Then, $1 + \sum_{k = 0}^{14} e^{\frac{i (2 k + 1) π}{15}}$
$= 1 + (α + α^{3} + α^{5} + α^{7} + \dots . . α^{29})$
$α = e^{i π / 15} \Rightarrow α^{30} = e^{2 i π} = 1$
It is G.P series,
$a = α, r = α^{2}$
$= 1 + α (\frac{1 - {(α^{2})}^{15}}{1 - α^{2}})$
$= 1 + α (\frac{1 - α^{30}}{1 - α^{2}})$
$= 1 + α (\frac{1 - 1}{1 - α^{2}})$
$= 1 + 0$
$= 1$
$(∵ α^{30} = 1)$

AMU-2005

Complex Numbers and Quadratic Equation

118017 If $x + \frac{1}{x} = 2 \cos θ$ , then for any integer $n$ , $x^{n} + \frac{1}{x^{n}} =$

1

2 \cos n θ

2

2 \sin n θ

3

2 i \cos n θ

4

2 i \sin n θ

Explanation:

A Given,
$x + \frac{1}{x} = 2 \cos θ$
$x^{2} + 1 = 2 x \cos θ$
$x^{2} - 2 x \cos θ + 1 = 0$
$∴ x = \frac{2 \cos θ \pm \sqrt{(2 \cos θ)^{2} - 4}}{2}]$
$= \cos θ \pm i \sin θ$
Let us take $x = \cos θ + i \sin θ$
$∴ \frac{1}{(x)} = \cos θ - i \sin θ$
$∴ x^{n} + \frac{1}{x^{n}} = (\cos θ + i \sin θ)^{n} + (\cos θ - i \sin θ)^{n}$
$= \cos n θ + i \sin n θ + \cos n θ - i \sin n θ$
$2 \cos n θ$

WB JEE-2011

Complex Numbers and Quadratic Equation

118007 If $α, β$ are the roots of $x^{2} - 2 x + 4 = 0$ , for $n \in$ $N$ , what is the value of $\boldsymbol α^{n} + \boldsymbol β^{n} =$

1

2^{n + 2} \cos (\frac{n π}{3})

2

2^{n + 1} \cos (\frac{n π}{3})

3

2^{n + 1} \cos (\frac{n π}{6})

4

2^{n + 2} \cos (\frac{n π}{6})

Explanation:

B Given that, $α$ and $β$ are the roots of the equation $x^{2} - 2 x + 4 = 0$
$∴ α + β = 2$
$α \cdot β = 4$
So, $(α - β) = \sqrt{(α + β)^{2} - 4 α β}$
$(α - β) = \sqrt{(2)^{2} - 4 \times 4}$
$α - β = \sqrt{- 12}$
$(α - β) = 2 \sqrt{3} i$
Adding equation (i) and (ii), we get -
$2 α = 2 + 2 \sqrt{3 i}$
$α = 2 [\frac{1}{2} + \frac{\sqrt{3}}{2} i]$
$α = 2 [\cos (\frac{π}{3}) + i \sin (\frac{π}{3})]$
On putting the value of $α$ in equation (i) we get -
$β = \frac{1}{2} [2 - 2 \sqrt{3} i]$
$β = 2 [\frac{1}{2} - \frac{\sqrt{3}}{2} i]$
$= 2 [\cos (\frac{π}{3}) - i \sin (\frac{π}{3})]$
$α^{1} + β^{1} = 2^{1} [2 \cos (\frac{1 π}{3})]$
$α^{n} + β^{n} = 2^{n} [2 \cos (\frac{n π}{3})]$
$α^{n} + β^{n} = 2^{n + 1} \cos (\frac{n π}{3})$

AP EAMCET - 18.09.2020 Shift - II

Complex Numbers and Quadratic Equation

118008 The value of
$1 + \sum_{k = 0}^{14} {\cos \frac{(2 k + 1)}{15} π + i \sin \frac{(2 k + 1)}{15} π} i$

1 0

2 -1

3 1

4 i

Explanation:

C $1 + \sum_{k = 0}^{14} {\cos \frac{(2 k + 1)}{15} π + i \sin \frac{(2 k + 1)}{15} π}$
We know that,
$\cos θ + i \sin θ = e^{i θ}$
Then, $1 + \sum_{k = 0}^{14} e^{\frac{i (2 k + 1) π}{15}}$
$= 1 + (α + α^{3} + α^{5} + α^{7} + \dots . . α^{29})$
$α = e^{i π / 15} \Rightarrow α^{30} = e^{2 i π} = 1$
It is G.P series,
$a = α, r = α^{2}$
$= 1 + α (\frac{1 - {(α^{2})}^{15}}{1 - α^{2}})$
$= 1 + α (\frac{1 - α^{30}}{1 - α^{2}})$
$= 1 + α (\frac{1 - 1}{1 - α^{2}})$
$= 1 + 0$
$= 1$
$(∵ α^{30} = 1)$

AMU-2005

Complex Numbers and Quadratic Equation

118009 The value of
$1 + \sum_{k = 0}^{14} {\cos \frac{(2 k + 1)}{15} π + i \sin \frac{(2 k + 1)}{15} π}$ is

1 0

2 -1

3 1

4

i

Explanation:

C $1 + \sum_{k = 0}^{14} {\cos \frac{(2 k + 1)}{15} π + i \sin \frac{(2 k + 1)}{15} π}$
We know that, $\cos θ + i \sin θ = e^{i θ}$
Then, $1 + \sum_{k = 0}^{14} e^{\frac{i (2 k + 1) π}{15}}$
$= 1 + (α + α^{3} + α^{5} + α^{7} + \dots . . α^{29})$
$α = e^{i π / 15} \Rightarrow α^{30} = e^{2 i π} = 1$
It is G.P series,
$a = α, r = α^{2}$
$= 1 + α (\frac{1 - {(α^{2})}^{15}}{1 - α^{2}})$
$= 1 + α (\frac{1 - α^{30}}{1 - α^{2}})$
$= 1 + α (\frac{1 - 1}{1 - α^{2}})$
$= 1 + 0$
$= 1$
$(∵ α^{30} = 1)$

AMU-2005

Complex Numbers and Quadratic Equation

118017 If $x + \frac{1}{x} = 2 \cos θ$ , then for any integer $n$ , $x^{n} + \frac{1}{x^{n}} =$

1

2 \cos n θ

2

2 \sin n θ

3

2 i \cos n θ

4

2 i \sin n θ

Explanation:

A Given,
$x + \frac{1}{x} = 2 \cos θ$
$x^{2} + 1 = 2 x \cos θ$
$x^{2} - 2 x \cos θ + 1 = 0$
$∴ x = \frac{2 \cos θ \pm \sqrt{(2 \cos θ)^{2} - 4}}{2}]$
$= \cos θ \pm i \sin θ$
Let us take $x = \cos θ + i \sin θ$
$∴ \frac{1}{(x)} = \cos θ - i \sin θ$
$∴ x^{n} + \frac{1}{x^{n}} = (\cos θ + i \sin θ)^{n} + (\cos θ - i \sin θ)^{n}$
$= \cos n θ + i \sin n θ + \cos n θ - i \sin n θ$
$2 \cos n θ$

WB JEE-2011

Complex Numbers and Quadratic Equation

118007 If $α, β$ are the roots of $x^{2} - 2 x + 4 = 0$ , for $n \in$ $N$ , what is the value of $\boldsymbol α^{n} + \boldsymbol β^{n} =$

1

2^{n + 2} \cos (\frac{n π}{3})

2

2^{n + 1} \cos (\frac{n π}{3})

3

2^{n + 1} \cos (\frac{n π}{6})

4

2^{n + 2} \cos (\frac{n π}{6})

Explanation:

B Given that, $α$ and $β$ are the roots of the equation $x^{2} - 2 x + 4 = 0$
$∴ α + β = 2$
$α \cdot β = 4$
So, $(α - β) = \sqrt{(α + β)^{2} - 4 α β}$
$(α - β) = \sqrt{(2)^{2} - 4 \times 4}$
$α - β = \sqrt{- 12}$
$(α - β) = 2 \sqrt{3} i$
Adding equation (i) and (ii), we get -
$2 α = 2 + 2 \sqrt{3 i}$
$α = 2 [\frac{1}{2} + \frac{\sqrt{3}}{2} i]$
$α = 2 [\cos (\frac{π}{3}) + i \sin (\frac{π}{3})]$
On putting the value of $α$ in equation (i) we get -
$β = \frac{1}{2} [2 - 2 \sqrt{3} i]$
$β = 2 [\frac{1}{2} - \frac{\sqrt{3}}{2} i]$
$= 2 [\cos (\frac{π}{3}) - i \sin (\frac{π}{3})]$
$α^{1} + β^{1} = 2^{1} [2 \cos (\frac{1 π}{3})]$
$α^{n} + β^{n} = 2^{n} [2 \cos (\frac{n π}{3})]$
$α^{n} + β^{n} = 2^{n + 1} \cos (\frac{n π}{3})$

AP EAMCET - 18.09.2020 Shift - II

Complex Numbers and Quadratic Equation

118008 The value of
$1 + \sum_{k = 0}^{14} {\cos \frac{(2 k + 1)}{15} π + i \sin \frac{(2 k + 1)}{15} π} i$

1 0

2 -1

3 1

4 i

Explanation:

C $1 + \sum_{k = 0}^{14} {\cos \frac{(2 k + 1)}{15} π + i \sin \frac{(2 k + 1)}{15} π}$
We know that,
$\cos θ + i \sin θ = e^{i θ}$
Then, $1 + \sum_{k = 0}^{14} e^{\frac{i (2 k + 1) π}{15}}$
$= 1 + (α + α^{3} + α^{5} + α^{7} + \dots . . α^{29})$
$α = e^{i π / 15} \Rightarrow α^{30} = e^{2 i π} = 1$
It is G.P series,
$a = α, r = α^{2}$
$= 1 + α (\frac{1 - {(α^{2})}^{15}}{1 - α^{2}})$
$= 1 + α (\frac{1 - α^{30}}{1 - α^{2}})$
$= 1 + α (\frac{1 - 1}{1 - α^{2}})$
$= 1 + 0$
$= 1$
$(∵ α^{30} = 1)$

AMU-2005

Complex Numbers and Quadratic Equation

118009 The value of
$1 + \sum_{k = 0}^{14} {\cos \frac{(2 k + 1)}{15} π + i \sin \frac{(2 k + 1)}{15} π}$ is

1 0

2 -1

3 1

4

i

Explanation:

C $1 + \sum_{k = 0}^{14} {\cos \frac{(2 k + 1)}{15} π + i \sin \frac{(2 k + 1)}{15} π}$
We know that, $\cos θ + i \sin θ = e^{i θ}$
Then, $1 + \sum_{k = 0}^{14} e^{\frac{i (2 k + 1) π}{15}}$
$= 1 + (α + α^{3} + α^{5} + α^{7} + \dots . . α^{29})$
$α = e^{i π / 15} \Rightarrow α^{30} = e^{2 i π} = 1$
It is G.P series,
$a = α, r = α^{2}$
$= 1 + α (\frac{1 - {(α^{2})}^{15}}{1 - α^{2}})$
$= 1 + α (\frac{1 - α^{30}}{1 - α^{2}})$
$= 1 + α (\frac{1 - 1}{1 - α^{2}})$
$= 1 + 0$
$= 1$
$(∵ α^{30} = 1)$

AMU-2005

Complex Numbers and Quadratic Equation

118017 If $x + \frac{1}{x} = 2 \cos θ$ , then for any integer $n$ , $x^{n} + \frac{1}{x^{n}} =$

1

2 \cos n θ

2

2 \sin n θ

3

2 i \cos n θ

4

2 i \sin n θ

Explanation:

A Given,
$x + \frac{1}{x} = 2 \cos θ$
$x^{2} + 1 = 2 x \cos θ$
$x^{2} - 2 x \cos θ + 1 = 0$
$∴ x = \frac{2 \cos θ \pm \sqrt{(2 \cos θ)^{2} - 4}}{2}]$
$= \cos θ \pm i \sin θ$
Let us take $x = \cos θ + i \sin θ$
$∴ \frac{1}{(x)} = \cos θ - i \sin θ$
$∴ x^{n} + \frac{1}{x^{n}} = (\cos θ + i \sin θ)^{n} + (\cos θ - i \sin θ)^{n}$
$= \cos n θ + i \sin n θ + \cos n θ - i \sin n θ$
$2 \cos n θ$

WB JEE-2011

Question Bank Series

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