QBManager

Complex Numbers and Quadratic Equation

118232 A root of the equation $17 x^{2} + 17 x \tan (2 \tan^{- 1} \frac{1}{5} - \frac{π}{4}) - 10 = 0$ is

1

\frac{10}{17}

2 -1

3

- \frac{7}{17}

4 1

Explanation:

D Given,
$17 x^{2} + 17 x \tan (2 \tan^{- 1} \frac{1}{5} - \frac{π}{4}) - 10 = 0$
We know that,
$2 \tan^{- 1} x = \tan^{- 1} \frac{2 x}{1 - x^{2}}$
$2 \tan^{- 1} \frac{1}{5} = \tan^{- 1} \frac{2 \times \frac{1}{5}}{1 - {(\frac{1}{5})}^{2}} = \tan^{- 1} \frac{5}{12}$
So, $\tan (2 \tan^{- 1} \frac{1}{5} - \frac{π}{4})$
$= \tan (\tan^{- 1} \frac{5}{12} - \tan^{- 1} (1))$
$= \tan [\tan^{- 1} \frac{(\frac{5}{12} - 1)}{1 + \frac{5}{12} \times 1}]$
$= \tan [\tan^{- 1} (\frac{- 7}{17})]$
$= \frac{- 7}{17}$
Now,
$17 x^{2} + 17 x \times (\frac{- 7}{17}) - 10 = 0$
$17 x^{2} - 7 x - 10 = 0$
$17 x^{2} - 17 x + 10 x - 10 = 0$
$17 x (x - 1) + 10 (x - 1) = 0$
$(x - 1) (17 x + 10) = 0$
$x = 1 & \frac{- 10}{17}$

BCECE-2015

Complex Numbers and Quadratic Equation

118233 If $α$ and $β$ be the roots of $x^{2} + b x + 1 = 0$ . then, the equation whose roots are $- (α + \frac{1}{β})$ and $- (β + \frac{1}{α})$ is

1

x^{2} = 0

2

x^{2} + 2 b x + 4 = 0

3

x^{2} - 2 b x + 4 = 0

4

x^{2} - 2 b x + 1 = 0

Explanation:

C Given,
$x^{2} + b x + 1 = 0$ and $α, β$ are roots
Then,
$α + β = - b$
$α \cdot β = 1$
Now, the equation whose roots are
$- (α + \frac{1}{β}) and - (β + \frac{1}{α})$
So, sum of roots
$= - (α + \frac{1}{β}) - (β + \frac{1}{α})$
$= - [(α + β) + \frac{1}{β} + \frac{1}{α}]$
$= - [α + β + \frac{α + β}{α β}]$
$= - [- b + \frac{(- b)}{1}]$
$= 2 b$
Product of roots
$= - (α + \frac{1}{β}) (- (β + \frac{1}{α}))$
$= α \cdot β + 1 + 1 + \frac{1}{α β}$
$= 1 + 1 + 1 + 1 = 4$
Therefore equation is $x^{2} - ($ sum of roots) $x +$ (product of roots) $= 0$
$x^{2} - (2 b) x + 4 = 0$
$x^{2} - 2 b x + 4 = 0$

BCECE-2015

Complex Numbers and Quadratic Equation

118255 Let $a, b, c$ be three real numbers such that $a +$ $2 b + 4 c = 0$ . Then the equation $a x^{2} + b x + c = 0$

1 has both the roots complex

2 has its roots lying within

- 1 < x < 0

3 has one of the roots equal to

\frac{1}{2}

4 has its roots lying within

2 < x < 6

Explanation:

C Given
$a^{2} + b x + c = 0$
$a + 2 b + 4 c = 0$
$(\frac{1}{4}) a + (\frac{1}{2}) b + c = 0$
${(\frac{1}{2})}^{2} a + (\frac{1}{2}) b + c = 0$
$on comparing with {ax}^{2} + bx + c = 0$
$x = \frac{1}{2}$ on comparing with ${ax}^{2} + bx + c = 0$

WB JEE-2011

Complex Numbers and Quadratic Equation

118234 If $α$ and $β$ are the roots of $x^{2} + 5 x + 4 = 0$ , then equation whose roots are $\frac{α + 2}{3}$ and $\frac{β + 2}{3}$ , is

1

9 x^{2} + 3 x + 2 = 0

2

9 x^{2} - 3 x + 2 = 0

3

9 x^{2} + 3 x - 2 = 0

4

9 x^{2} - 3 x - 2 = 0

Explanation:

C Given,
$x^{2} + 5 x + 4 = 0 and roots are α, β$
Then, sum of roots $(α + β)$
$α + β = - 5$
Product of roots $= α \cdot β$
$α \cdot β = 4$
Now, the equation whose roots are
$\frac{α + 2}{3}, \frac{β + 2}{3}$
So, sum of roots
$= \frac{α + 2}{3} + \frac{β + 2}{3}$
$= \frac{α + β + 4}{3}$
$= \frac{- 5 + 4}{3} = \frac{- 1}{3}$
Product of roots
$= \frac{(α + 2)}{3} \times \frac{(β + 2)}{3}$
$= \frac{α β + 2 (α + β) + 4}{9}$
$= \frac{4 + 2 (- 5) + 4}{9} = \frac{- 2}{9}$
Therefore, equation is $x^{2} - ($ sum of roots) $x +$ product of roots $= 0$
$x^{2} - (\frac{- 1}{3}) x + (\frac{- 2}{9}) = 0$
$9 x^{2} + 3 x - 2 = 0$

BCECE-2014

Complex Numbers and Quadratic Equation

118232 A root of the equation $17 x^{2} + 17 x \tan (2 \tan^{- 1} \frac{1}{5} - \frac{π}{4}) - 10 = 0$ is

1

\frac{10}{17}

2 -1

3

- \frac{7}{17}

4 1

Explanation:

D Given,
$17 x^{2} + 17 x \tan (2 \tan^{- 1} \frac{1}{5} - \frac{π}{4}) - 10 = 0$
We know that,
$2 \tan^{- 1} x = \tan^{- 1} \frac{2 x}{1 - x^{2}}$
$2 \tan^{- 1} \frac{1}{5} = \tan^{- 1} \frac{2 \times \frac{1}{5}}{1 - {(\frac{1}{5})}^{2}} = \tan^{- 1} \frac{5}{12}$
So, $\tan (2 \tan^{- 1} \frac{1}{5} - \frac{π}{4})$
$= \tan (\tan^{- 1} \frac{5}{12} - \tan^{- 1} (1))$
$= \tan [\tan^{- 1} \frac{(\frac{5}{12} - 1)}{1 + \frac{5}{12} \times 1}]$
$= \tan [\tan^{- 1} (\frac{- 7}{17})]$
$= \frac{- 7}{17}$
Now,
$17 x^{2} + 17 x \times (\frac{- 7}{17}) - 10 = 0$
$17 x^{2} - 7 x - 10 = 0$
$17 x^{2} - 17 x + 10 x - 10 = 0$
$17 x (x - 1) + 10 (x - 1) = 0$
$(x - 1) (17 x + 10) = 0$
$x = 1 & \frac{- 10}{17}$

BCECE-2015

Complex Numbers and Quadratic Equation

118233 If $α$ and $β$ be the roots of $x^{2} + b x + 1 = 0$ . then, the equation whose roots are $- (α + \frac{1}{β})$ and $- (β + \frac{1}{α})$ is

1

x^{2} = 0

2

x^{2} + 2 b x + 4 = 0

3

x^{2} - 2 b x + 4 = 0

4

x^{2} - 2 b x + 1 = 0

Explanation:

C Given,
$x^{2} + b x + 1 = 0$ and $α, β$ are roots
Then,
$α + β = - b$
$α \cdot β = 1$
Now, the equation whose roots are
$- (α + \frac{1}{β}) and - (β + \frac{1}{α})$
So, sum of roots
$= - (α + \frac{1}{β}) - (β + \frac{1}{α})$
$= - [(α + β) + \frac{1}{β} + \frac{1}{α}]$
$= - [α + β + \frac{α + β}{α β}]$
$= - [- b + \frac{(- b)}{1}]$
$= 2 b$
Product of roots
$= - (α + \frac{1}{β}) (- (β + \frac{1}{α}))$
$= α \cdot β + 1 + 1 + \frac{1}{α β}$
$= 1 + 1 + 1 + 1 = 4$
Therefore equation is $x^{2} - ($ sum of roots) $x +$ (product of roots) $= 0$
$x^{2} - (2 b) x + 4 = 0$
$x^{2} - 2 b x + 4 = 0$

BCECE-2015

Complex Numbers and Quadratic Equation

118255 Let $a, b, c$ be three real numbers such that $a +$ $2 b + 4 c = 0$ . Then the equation $a x^{2} + b x + c = 0$

1 has both the roots complex

2 has its roots lying within

- 1 < x < 0

3 has one of the roots equal to

\frac{1}{2}

4 has its roots lying within

2 < x < 6

Explanation:

C Given
$a^{2} + b x + c = 0$
$a + 2 b + 4 c = 0$
$(\frac{1}{4}) a + (\frac{1}{2}) b + c = 0$
${(\frac{1}{2})}^{2} a + (\frac{1}{2}) b + c = 0$
$on comparing with {ax}^{2} + bx + c = 0$
$x = \frac{1}{2}$ on comparing with ${ax}^{2} + bx + c = 0$

WB JEE-2011

Complex Numbers and Quadratic Equation

118234 If $α$ and $β$ are the roots of $x^{2} + 5 x + 4 = 0$ , then equation whose roots are $\frac{α + 2}{3}$ and $\frac{β + 2}{3}$ , is

1

9 x^{2} + 3 x + 2 = 0

2

9 x^{2} - 3 x + 2 = 0

3

9 x^{2} + 3 x - 2 = 0

4

9 x^{2} - 3 x - 2 = 0

Explanation:

C Given,
$x^{2} + 5 x + 4 = 0 and roots are α, β$
Then, sum of roots $(α + β)$
$α + β = - 5$
Product of roots $= α \cdot β$
$α \cdot β = 4$
Now, the equation whose roots are
$\frac{α + 2}{3}, \frac{β + 2}{3}$
So, sum of roots
$= \frac{α + 2}{3} + \frac{β + 2}{3}$
$= \frac{α + β + 4}{3}$
$= \frac{- 5 + 4}{3} = \frac{- 1}{3}$
Product of roots
$= \frac{(α + 2)}{3} \times \frac{(β + 2)}{3}$
$= \frac{α β + 2 (α + β) + 4}{9}$
$= \frac{4 + 2 (- 5) + 4}{9} = \frac{- 2}{9}$
Therefore, equation is $x^{2} - ($ sum of roots) $x +$ product of roots $= 0$
$x^{2} - (\frac{- 1}{3}) x + (\frac{- 2}{9}) = 0$
$9 x^{2} + 3 x - 2 = 0$

BCECE-2014

Complex Numbers and Quadratic Equation

118232 A root of the equation $17 x^{2} + 17 x \tan (2 \tan^{- 1} \frac{1}{5} - \frac{π}{4}) - 10 = 0$ is

1

\frac{10}{17}

2 -1

3

- \frac{7}{17}

4 1

Explanation:

D Given,
$17 x^{2} + 17 x \tan (2 \tan^{- 1} \frac{1}{5} - \frac{π}{4}) - 10 = 0$
We know that,
$2 \tan^{- 1} x = \tan^{- 1} \frac{2 x}{1 - x^{2}}$
$2 \tan^{- 1} \frac{1}{5} = \tan^{- 1} \frac{2 \times \frac{1}{5}}{1 - {(\frac{1}{5})}^{2}} = \tan^{- 1} \frac{5}{12}$
So, $\tan (2 \tan^{- 1} \frac{1}{5} - \frac{π}{4})$
$= \tan (\tan^{- 1} \frac{5}{12} - \tan^{- 1} (1))$
$= \tan [\tan^{- 1} \frac{(\frac{5}{12} - 1)}{1 + \frac{5}{12} \times 1}]$
$= \tan [\tan^{- 1} (\frac{- 7}{17})]$
$= \frac{- 7}{17}$
Now,
$17 x^{2} + 17 x \times (\frac{- 7}{17}) - 10 = 0$
$17 x^{2} - 7 x - 10 = 0$
$17 x^{2} - 17 x + 10 x - 10 = 0$
$17 x (x - 1) + 10 (x - 1) = 0$
$(x - 1) (17 x + 10) = 0$
$x = 1 & \frac{- 10}{17}$

BCECE-2015

Complex Numbers and Quadratic Equation

118233 If $α$ and $β$ be the roots of $x^{2} + b x + 1 = 0$ . then, the equation whose roots are $- (α + \frac{1}{β})$ and $- (β + \frac{1}{α})$ is

1

x^{2} = 0

2

x^{2} + 2 b x + 4 = 0

3

x^{2} - 2 b x + 4 = 0

4

x^{2} - 2 b x + 1 = 0

Explanation:

C Given,
$x^{2} + b x + 1 = 0$ and $α, β$ are roots
Then,
$α + β = - b$
$α \cdot β = 1$
Now, the equation whose roots are
$- (α + \frac{1}{β}) and - (β + \frac{1}{α})$
So, sum of roots
$= - (α + \frac{1}{β}) - (β + \frac{1}{α})$
$= - [(α + β) + \frac{1}{β} + \frac{1}{α}]$
$= - [α + β + \frac{α + β}{α β}]$
$= - [- b + \frac{(- b)}{1}]$
$= 2 b$
Product of roots
$= - (α + \frac{1}{β}) (- (β + \frac{1}{α}))$
$= α \cdot β + 1 + 1 + \frac{1}{α β}$
$= 1 + 1 + 1 + 1 = 4$
Therefore equation is $x^{2} - ($ sum of roots) $x +$ (product of roots) $= 0$
$x^{2} - (2 b) x + 4 = 0$
$x^{2} - 2 b x + 4 = 0$

BCECE-2015

Complex Numbers and Quadratic Equation

118255 Let $a, b, c$ be three real numbers such that $a +$ $2 b + 4 c = 0$ . Then the equation $a x^{2} + b x + c = 0$

1 has both the roots complex

2 has its roots lying within

- 1 < x < 0

3 has one of the roots equal to

\frac{1}{2}

4 has its roots lying within

2 < x < 6

Explanation:

C Given
$a^{2} + b x + c = 0$
$a + 2 b + 4 c = 0$
$(\frac{1}{4}) a + (\frac{1}{2}) b + c = 0$
${(\frac{1}{2})}^{2} a + (\frac{1}{2}) b + c = 0$
$on comparing with {ax}^{2} + bx + c = 0$
$x = \frac{1}{2}$ on comparing with ${ax}^{2} + bx + c = 0$

WB JEE-2011

Complex Numbers and Quadratic Equation

118234 If $α$ and $β$ are the roots of $x^{2} + 5 x + 4 = 0$ , then equation whose roots are $\frac{α + 2}{3}$ and $\frac{β + 2}{3}$ , is

1

9 x^{2} + 3 x + 2 = 0

2

9 x^{2} - 3 x + 2 = 0

3

9 x^{2} + 3 x - 2 = 0

4

9 x^{2} - 3 x - 2 = 0

Explanation:

C Given,
$x^{2} + 5 x + 4 = 0 and roots are α, β$
Then, sum of roots $(α + β)$
$α + β = - 5$
Product of roots $= α \cdot β$
$α \cdot β = 4$
Now, the equation whose roots are
$\frac{α + 2}{3}, \frac{β + 2}{3}$
So, sum of roots
$= \frac{α + 2}{3} + \frac{β + 2}{3}$
$= \frac{α + β + 4}{3}$
$= \frac{- 5 + 4}{3} = \frac{- 1}{3}$
Product of roots
$= \frac{(α + 2)}{3} \times \frac{(β + 2)}{3}$
$= \frac{α β + 2 (α + β) + 4}{9}$
$= \frac{4 + 2 (- 5) + 4}{9} = \frac{- 2}{9}$
Therefore, equation is $x^{2} - ($ sum of roots) $x +$ product of roots $= 0$
$x^{2} - (\frac{- 1}{3}) x + (\frac{- 2}{9}) = 0$
$9 x^{2} + 3 x - 2 = 0$

BCECE-2014

Complex Numbers and Quadratic Equation

118232 A root of the equation $17 x^{2} + 17 x \tan (2 \tan^{- 1} \frac{1}{5} - \frac{π}{4}) - 10 = 0$ is

1

\frac{10}{17}

2 -1

3

- \frac{7}{17}

4 1

Explanation:

D Given,
$17 x^{2} + 17 x \tan (2 \tan^{- 1} \frac{1}{5} - \frac{π}{4}) - 10 = 0$
We know that,
$2 \tan^{- 1} x = \tan^{- 1} \frac{2 x}{1 - x^{2}}$
$2 \tan^{- 1} \frac{1}{5} = \tan^{- 1} \frac{2 \times \frac{1}{5}}{1 - {(\frac{1}{5})}^{2}} = \tan^{- 1} \frac{5}{12}$
So, $\tan (2 \tan^{- 1} \frac{1}{5} - \frac{π}{4})$
$= \tan (\tan^{- 1} \frac{5}{12} - \tan^{- 1} (1))$
$= \tan [\tan^{- 1} \frac{(\frac{5}{12} - 1)}{1 + \frac{5}{12} \times 1}]$
$= \tan [\tan^{- 1} (\frac{- 7}{17})]$
$= \frac{- 7}{17}$
Now,
$17 x^{2} + 17 x \times (\frac{- 7}{17}) - 10 = 0$
$17 x^{2} - 7 x - 10 = 0$
$17 x^{2} - 17 x + 10 x - 10 = 0$
$17 x (x - 1) + 10 (x - 1) = 0$
$(x - 1) (17 x + 10) = 0$
$x = 1 & \frac{- 10}{17}$

BCECE-2015

Complex Numbers and Quadratic Equation

118233 If $α$ and $β$ be the roots of $x^{2} + b x + 1 = 0$ . then, the equation whose roots are $- (α + \frac{1}{β})$ and $- (β + \frac{1}{α})$ is

1

x^{2} = 0

2

x^{2} + 2 b x + 4 = 0

3

x^{2} - 2 b x + 4 = 0

4

x^{2} - 2 b x + 1 = 0

Explanation:

C Given,
$x^{2} + b x + 1 = 0$ and $α, β$ are roots
Then,
$α + β = - b$
$α \cdot β = 1$
Now, the equation whose roots are
$- (α + \frac{1}{β}) and - (β + \frac{1}{α})$
So, sum of roots
$= - (α + \frac{1}{β}) - (β + \frac{1}{α})$
$= - [(α + β) + \frac{1}{β} + \frac{1}{α}]$
$= - [α + β + \frac{α + β}{α β}]$
$= - [- b + \frac{(- b)}{1}]$
$= 2 b$
Product of roots
$= - (α + \frac{1}{β}) (- (β + \frac{1}{α}))$
$= α \cdot β + 1 + 1 + \frac{1}{α β}$
$= 1 + 1 + 1 + 1 = 4$
Therefore equation is $x^{2} - ($ sum of roots) $x +$ (product of roots) $= 0$
$x^{2} - (2 b) x + 4 = 0$
$x^{2} - 2 b x + 4 = 0$

BCECE-2015

Complex Numbers and Quadratic Equation

118255 Let $a, b, c$ be three real numbers such that $a +$ $2 b + 4 c = 0$ . Then the equation $a x^{2} + b x + c = 0$

1 has both the roots complex

2 has its roots lying within

- 1 < x < 0

3 has one of the roots equal to

\frac{1}{2}

4 has its roots lying within

2 < x < 6

Explanation:

C Given
$a^{2} + b x + c = 0$
$a + 2 b + 4 c = 0$
$(\frac{1}{4}) a + (\frac{1}{2}) b + c = 0$
${(\frac{1}{2})}^{2} a + (\frac{1}{2}) b + c = 0$
$on comparing with {ax}^{2} + bx + c = 0$
$x = \frac{1}{2}$ on comparing with ${ax}^{2} + bx + c = 0$

WB JEE-2011

Complex Numbers and Quadratic Equation

118234 If $α$ and $β$ are the roots of $x^{2} + 5 x + 4 = 0$ , then equation whose roots are $\frac{α + 2}{3}$ and $\frac{β + 2}{3}$ , is

1

9 x^{2} + 3 x + 2 = 0

2

9 x^{2} - 3 x + 2 = 0

3

9 x^{2} + 3 x - 2 = 0

4

9 x^{2} - 3 x - 2 = 0

Explanation:

C Given,
$x^{2} + 5 x + 4 = 0 and roots are α, β$
Then, sum of roots $(α + β)$
$α + β = - 5$
Product of roots $= α \cdot β$
$α \cdot β = 4$
Now, the equation whose roots are
$\frac{α + 2}{3}, \frac{β + 2}{3}$
So, sum of roots
$= \frac{α + 2}{3} + \frac{β + 2}{3}$
$= \frac{α + β + 4}{3}$
$= \frac{- 5 + 4}{3} = \frac{- 1}{3}$
Product of roots
$= \frac{(α + 2)}{3} \times \frac{(β + 2)}{3}$
$= \frac{α β + 2 (α + β) + 4}{9}$
$= \frac{4 + 2 (- 5) + 4}{9} = \frac{- 2}{9}$
Therefore, equation is $x^{2} - ($ sum of roots) $x +$ product of roots $= 0$
$x^{2} - (\frac{- 1}{3}) x + (\frac{- 2}{9}) = 0$
$9 x^{2} + 3 x - 2 = 0$

BCECE-2014

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