QBManager

Complex Numbers and Quadratic Equation

118051 The number of real solution of the equation
$| x^{2} | - 3 | x | + 2 = 0$ is

1 3

2 2

3 1

4 4

Explanation:

D We have :-
$| x^{2} | - 3 | x | + 2 = 0$
Let $y = | x |$
$x = \pm y$
So, eq $^{n}$ (i) becomes :-
$y^{2} - 3 y + 2 = 0$
$(y - 2) (y - 1) = 0$
$y = 2, 1$
From (ii), we have
$x = \pm 2, \pm 1$ $∴ 4$ solutions exist.

SRM JEEE-2013

Complex Numbers and Quadratic Equation

118052 The solution of the equation $9^{x} + 78 = 3^{2 x + 3}$ is

1 2

2 3

3

1 / 3

4

1 / 2

Explanation:

D Given :-
$9^{x} + 78 = 3^{2 x + 3}$
$3^{2 x} + 78 = 3^{2 x} \cdot 27$
Put $3^{2 x} = y$
$∴$ $y + 78 = 27 y$
$y = 3$
$3^{2 x} = 3$
$S o, 2 x = 1$
$x = \frac{1}{2}$

SRM JEEE-2012

Complex Numbers and Quadratic Equation

118055 If the roots of the equation $x^{3} + a x^{2} + b x + c = 0$ are in A.P., then $2 a^{3} - 9 a b =$

1

9 c

2

18 c

3

27 c

4

- 27 c

Explanation:

D Given, equation is : $x^{3} + {ax}^{2} + bx + c = 0$
Then, sum of roots,
$α + β + γ = - a$
sum of product of roots taking two at a time,
$(α β + β γ + γ α) = b$
product of roots,
$α β γ = - c$
According to the question, roots are in A.P.
$∴ β - α = γ - β$
$2 β = α + γ$
$∵ α + β + γ = - a$
$∴ (α + γ) + β = - a$
$2 β + β = - a$
$3 β = - a \Rightarrow β = \frac{- a}{3}$
Since, $β$ is root of ${eq}^{n} x^{3} + {ax}^{2} + bx + c = 0$
$∴ β^{3} + a β^{2} + b β + c = 0$
put $β = \frac{- a}{3}$ in above eq $^{n}$
${(\frac{- a}{3})}^{3} + a {(\frac{- a}{3})}^{2} + b (\frac{- a}{3}) + c = 0$
$\frac{- a^{3}}{27} + \frac{a^{3}}{9} - \frac{a b}{3} + c = 0$
$\frac{- a^{3} + 3 a^{3} - 9 a b + 27 c}{27} = 0$
$2 a^{3} - 9 a b = - 27 c$

Karnataka CET-2013

Complex Numbers and Quadratic Equation

118057 If $α, β$ and $γ$ are roots of $x^{3} - 2 x + 1 = 0$ , then the value of $Σ (\frac{1}{α + β - γ})$ is

1

\frac{- 1}{2}

2 -1

3 0

4

\frac{1}{2}

Explanation:

B Given
$x^{3} - 2 x + 1 = 0$
$α + β + γ = 0$
$α β + β γ + γ α = - 2$
$α β γ = - 1$
$\sum (\frac{1}{α + β - γ}) = \frac{1}{α + β - γ} + \frac{1}{β + γ - α} + \frac{1}{α + γ - β}$
$= \frac{- 1}{2} (\frac{1}{α} + \frac{1}{β} + \frac{1}{γ})$
$\frac{- 1}{2} (\frac{α β + β γ + γ α}{α β γ}) = \frac{- 1}{2} \times \frac{- 2}{- 1} = - 1$

Karnataka CET-2011

Complex Numbers and Quadratic Equation

118051 The number of real solution of the equation
$| x^{2} | - 3 | x | + 2 = 0$ is

1 3

2 2

3 1

4 4

Explanation:

D We have :-
$| x^{2} | - 3 | x | + 2 = 0$
Let $y = | x |$
$x = \pm y$
So, eq $^{n}$ (i) becomes :-
$y^{2} - 3 y + 2 = 0$
$(y - 2) (y - 1) = 0$
$y = 2, 1$
From (ii), we have
$x = \pm 2, \pm 1$ $∴ 4$ solutions exist.

SRM JEEE-2013

Complex Numbers and Quadratic Equation

118052 The solution of the equation $9^{x} + 78 = 3^{2 x + 3}$ is

1 2

2 3

3

1 / 3

4

1 / 2

Explanation:

D Given :-
$9^{x} + 78 = 3^{2 x + 3}$
$3^{2 x} + 78 = 3^{2 x} \cdot 27$
Put $3^{2 x} = y$
$∴$ $y + 78 = 27 y$
$y = 3$
$3^{2 x} = 3$
$S o, 2 x = 1$
$x = \frac{1}{2}$

SRM JEEE-2012

Complex Numbers and Quadratic Equation

118055 If the roots of the equation $x^{3} + a x^{2} + b x + c = 0$ are in A.P., then $2 a^{3} - 9 a b =$

1

9 c

2

18 c

3

27 c

4

- 27 c

Explanation:

D Given, equation is : $x^{3} + {ax}^{2} + bx + c = 0$
Then, sum of roots,
$α + β + γ = - a$
sum of product of roots taking two at a time,
$(α β + β γ + γ α) = b$
product of roots,
$α β γ = - c$
According to the question, roots are in A.P.
$∴ β - α = γ - β$
$2 β = α + γ$
$∵ α + β + γ = - a$
$∴ (α + γ) + β = - a$
$2 β + β = - a$
$3 β = - a \Rightarrow β = \frac{- a}{3}$
Since, $β$ is root of ${eq}^{n} x^{3} + {ax}^{2} + bx + c = 0$
$∴ β^{3} + a β^{2} + b β + c = 0$
put $β = \frac{- a}{3}$ in above eq $^{n}$
${(\frac{- a}{3})}^{3} + a {(\frac{- a}{3})}^{2} + b (\frac{- a}{3}) + c = 0$
$\frac{- a^{3}}{27} + \frac{a^{3}}{9} - \frac{a b}{3} + c = 0$
$\frac{- a^{3} + 3 a^{3} - 9 a b + 27 c}{27} = 0$
$2 a^{3} - 9 a b = - 27 c$

Karnataka CET-2013

Complex Numbers and Quadratic Equation

118057 If $α, β$ and $γ$ are roots of $x^{3} - 2 x + 1 = 0$ , then the value of $Σ (\frac{1}{α + β - γ})$ is

1

\frac{- 1}{2}

2 -1

3 0

4

\frac{1}{2}

Explanation:

B Given
$x^{3} - 2 x + 1 = 0$
$α + β + γ = 0$
$α β + β γ + γ α = - 2$
$α β γ = - 1$
$\sum (\frac{1}{α + β - γ}) = \frac{1}{α + β - γ} + \frac{1}{β + γ - α} + \frac{1}{α + γ - β}$
$= \frac{- 1}{2} (\frac{1}{α} + \frac{1}{β} + \frac{1}{γ})$
$\frac{- 1}{2} (\frac{α β + β γ + γ α}{α β γ}) = \frac{- 1}{2} \times \frac{- 2}{- 1} = - 1$

Karnataka CET-2011

Complex Numbers and Quadratic Equation

118051 The number of real solution of the equation
$| x^{2} | - 3 | x | + 2 = 0$ is

1 3

2 2

3 1

4 4

Explanation:

D We have :-
$| x^{2} | - 3 | x | + 2 = 0$
Let $y = | x |$
$x = \pm y$
So, eq $^{n}$ (i) becomes :-
$y^{2} - 3 y + 2 = 0$
$(y - 2) (y - 1) = 0$
$y = 2, 1$
From (ii), we have
$x = \pm 2, \pm 1$ $∴ 4$ solutions exist.

SRM JEEE-2013

Complex Numbers and Quadratic Equation

118052 The solution of the equation $9^{x} + 78 = 3^{2 x + 3}$ is

1 2

2 3

3

1 / 3

4

1 / 2

Explanation:

D Given :-
$9^{x} + 78 = 3^{2 x + 3}$
$3^{2 x} + 78 = 3^{2 x} \cdot 27$
Put $3^{2 x} = y$
$∴$ $y + 78 = 27 y$
$y = 3$
$3^{2 x} = 3$
$S o, 2 x = 1$
$x = \frac{1}{2}$

SRM JEEE-2012

Complex Numbers and Quadratic Equation

118055 If the roots of the equation $x^{3} + a x^{2} + b x + c = 0$ are in A.P., then $2 a^{3} - 9 a b =$

1

9 c

2

18 c

3

27 c

4

- 27 c

Explanation:

D Given, equation is : $x^{3} + {ax}^{2} + bx + c = 0$
Then, sum of roots,
$α + β + γ = - a$
sum of product of roots taking two at a time,
$(α β + β γ + γ α) = b$
product of roots,
$α β γ = - c$
According to the question, roots are in A.P.
$∴ β - α = γ - β$
$2 β = α + γ$
$∵ α + β + γ = - a$
$∴ (α + γ) + β = - a$
$2 β + β = - a$
$3 β = - a \Rightarrow β = \frac{- a}{3}$
Since, $β$ is root of ${eq}^{n} x^{3} + {ax}^{2} + bx + c = 0$
$∴ β^{3} + a β^{2} + b β + c = 0$
put $β = \frac{- a}{3}$ in above eq $^{n}$
${(\frac{- a}{3})}^{3} + a {(\frac{- a}{3})}^{2} + b (\frac{- a}{3}) + c = 0$
$\frac{- a^{3}}{27} + \frac{a^{3}}{9} - \frac{a b}{3} + c = 0$
$\frac{- a^{3} + 3 a^{3} - 9 a b + 27 c}{27} = 0$
$2 a^{3} - 9 a b = - 27 c$

Karnataka CET-2013

Complex Numbers and Quadratic Equation

118057 If $α, β$ and $γ$ are roots of $x^{3} - 2 x + 1 = 0$ , then the value of $Σ (\frac{1}{α + β - γ})$ is

1

\frac{- 1}{2}

2 -1

3 0

4

\frac{1}{2}

Explanation:

B Given
$x^{3} - 2 x + 1 = 0$
$α + β + γ = 0$
$α β + β γ + γ α = - 2$
$α β γ = - 1$
$\sum (\frac{1}{α + β - γ}) = \frac{1}{α + β - γ} + \frac{1}{β + γ - α} + \frac{1}{α + γ - β}$
$= \frac{- 1}{2} (\frac{1}{α} + \frac{1}{β} + \frac{1}{γ})$
$\frac{- 1}{2} (\frac{α β + β γ + γ α}{α β γ}) = \frac{- 1}{2} \times \frac{- 2}{- 1} = - 1$

Karnataka CET-2011

Complex Numbers and Quadratic Equation

118051 The number of real solution of the equation
$| x^{2} | - 3 | x | + 2 = 0$ is

1 3

2 2

3 1

4 4

Explanation:

D We have :-
$| x^{2} | - 3 | x | + 2 = 0$
Let $y = | x |$
$x = \pm y$
So, eq $^{n}$ (i) becomes :-
$y^{2} - 3 y + 2 = 0$
$(y - 2) (y - 1) = 0$
$y = 2, 1$
From (ii), we have
$x = \pm 2, \pm 1$ $∴ 4$ solutions exist.

SRM JEEE-2013

Complex Numbers and Quadratic Equation

118052 The solution of the equation $9^{x} + 78 = 3^{2 x + 3}$ is

1 2

2 3

3

1 / 3

4

1 / 2

Explanation:

D Given :-
$9^{x} + 78 = 3^{2 x + 3}$
$3^{2 x} + 78 = 3^{2 x} \cdot 27$
Put $3^{2 x} = y$
$∴$ $y + 78 = 27 y$
$y = 3$
$3^{2 x} = 3$
$S o, 2 x = 1$
$x = \frac{1}{2}$

SRM JEEE-2012

Complex Numbers and Quadratic Equation

118055 If the roots of the equation $x^{3} + a x^{2} + b x + c = 0$ are in A.P., then $2 a^{3} - 9 a b =$

1

9 c

2

18 c

3

27 c

4

- 27 c

Explanation:

D Given, equation is : $x^{3} + {ax}^{2} + bx + c = 0$
Then, sum of roots,
$α + β + γ = - a$
sum of product of roots taking two at a time,
$(α β + β γ + γ α) = b$
product of roots,
$α β γ = - c$
According to the question, roots are in A.P.
$∴ β - α = γ - β$
$2 β = α + γ$
$∵ α + β + γ = - a$
$∴ (α + γ) + β = - a$
$2 β + β = - a$
$3 β = - a \Rightarrow β = \frac{- a}{3}$
Since, $β$ is root of ${eq}^{n} x^{3} + {ax}^{2} + bx + c = 0$
$∴ β^{3} + a β^{2} + b β + c = 0$
put $β = \frac{- a}{3}$ in above eq $^{n}$
${(\frac{- a}{3})}^{3} + a {(\frac{- a}{3})}^{2} + b (\frac{- a}{3}) + c = 0$
$\frac{- a^{3}}{27} + \frac{a^{3}}{9} - \frac{a b}{3} + c = 0$
$\frac{- a^{3} + 3 a^{3} - 9 a b + 27 c}{27} = 0$
$2 a^{3} - 9 a b = - 27 c$

Karnataka CET-2013

Complex Numbers and Quadratic Equation

118057 If $α, β$ and $γ$ are roots of $x^{3} - 2 x + 1 = 0$ , then the value of $Σ (\frac{1}{α + β - γ})$ is

1

\frac{- 1}{2}

2 -1

3 0

4

\frac{1}{2}

Explanation:

B Given
$x^{3} - 2 x + 1 = 0$
$α + β + γ = 0$
$α β + β γ + γ α = - 2$
$α β γ = - 1$
$\sum (\frac{1}{α + β - γ}) = \frac{1}{α + β - γ} + \frac{1}{β + γ - α} + \frac{1}{α + γ - β}$
$= \frac{- 1}{2} (\frac{1}{α} + \frac{1}{β} + \frac{1}{γ})$
$\frac{- 1}{2} (\frac{α β + β γ + γ α}{α β γ}) = \frac{- 1}{2} \times \frac{- 2}{- 1} = - 1$

Karnataka CET-2011