QBManager

Complex Numbers and Quadratic Equation

118165 If both the roots of the quadratic equation $x^{2} -$ $2 k x + k^{2} + k - 5 = 0$ are less than 5 , then $k$ lies in the interval

1

[4, 5]

2

(- \infty, 4)

3

(6, \infty)

4

(5, 6]

Explanation:

B Given both the root of the equation $x^{2} - 2 kx$ $+ k^{2} + k - 5 = 0$
$∴$ Roots are $x = \frac{2 k \pm \sqrt{4 k^{2} - 4 (k^{2} + k - 5)}}{2}$
$= \frac{2 k \pm \sqrt{4 k^{2} - 4 k^{2} - 4 k + 20}}{2}$
$= \frac{2 k \pm 2 \sqrt{5 - k}}{2} = k \pm \sqrt{5 - k}$
Given, that both the roots are less than 5 .
$∴ k + \sqrt{5 - k} < 5 \Rightarrow \sqrt{5 - k} < 5 - k$
$o (5 - k) < (5 - k)^{2}$
$r (5 - k)^{2} - (5 - k) > 0$
$\Rightarrow (5 - k) (5 - k - 1) > 0$
$\Rightarrow (5 - k) (4 - k) > 0$
$\Rightarrow 5 - k > 0 and 4 - k > 0$
$\Rightarrow k < 5 and k < 4$
$∴ k < 4, so k lies in the interval (- \infty, 4)$
We can also check the other option,
$k - \sqrt{5 - k} < 5$
$\Rightarrow (k - 5) < \sqrt{5 - k}$
$\Rightarrow (k - 5)^{2} < 5 - k$
$\Rightarrow (k - 5)^{2} < - (k - 5)$
$\Rightarrow (k - 5)^{2} + k - 5 < 0$
$(k - 5) (k - 4) < 0$
$∴ k \in (4, 5)$ So, we get $k \in (- \infty, 4)$

AIEEE-2005

Complex Numbers and Quadratic Equation

118166 Sachin and Rahul attempted to solve a quadratic equation. Sachin made a mistake in writing down the constant term and ended up in roots $(4, 3)$ . Rahul made a mistake in writing down coefficient of $x$ to get roots $(3, 2)$ . The correct roots of equation are

1

- 4, - 3

2 6,1

3 4,3

4

- 6, - 1

Explanation:

B Sachin made mistake in interpreting constant term and found roots 4,3
$∴ 4 + 3 = - b / a$ when the equation is
$a x^{2} + b x + c = 0$
$x^{2} - \frac{b}{a} x + \frac{c}{a} = 0$
or
Rahul made num take in interpreting coefficient of $x$ and found roots $(3, 2)$
$∴ 3 \times 2 = c / a \Rightarrow c / a = 6$
$We know α + β = - b / c = 7$
$α β = c / a = 6$
$∴ (α - β)^{2} = (α + β)^{2} - 4 α β$
$= (7)^{2} - 4 \times 6$
$= 49 - 24 = 25$
$∴ α - β = \pm 5$
$α + β = 7$
$h l i n e 2 α = 12$
$α = 6$
$and β = 7 - 6 = 1$
$∴ The correct root are 6, 1$
$h l i n e r i g h t .$ $∴$ The correct root are 6,1 .

AIEEE-2011

Complex Numbers and Quadratic Equation

118167 The equation $e^{\sin x} - e^{- \sin x} - 4 = 0$ has

1 infinite number of real roots

2 no real root

3 exactly one real root

4 exactly four real roots

Explanation:

B Given the equation
$e^{\sin x} - e^{- \sin x} = 4$
$Let e^{\sin x} = t$
$∴ t - \frac{1}{t} = 4$
$\Rightarrow t^{2} - 1 = 4 t$
$t^{2} - 4 t - 1 = 0$
$∴ t = \frac{4 \pm \sqrt{16 + 4}}{2} \frac{4 \pm \sqrt{20}}{2}$
$= 2 \pm \sqrt{5}$
$∴ e^{\sin x} = (2 \pm \sqrt{5})$
$∴ \sin x = (2 \pm \sqrt{5}) > 1$
$It has no real solutions.$

AIEEE-2012

Complex Numbers and Quadratic Equation

118168 If $a \in R$ and the equation $- 3 (x - [x])^{2} + 2 (x -$ $[x]) + a^{2} = 0$ (where, $[x]$ denotes the greatest integer $\leq x$ ) has no integral solution, then all possible values of a lie in the interval

1

(- 1, 0) \cup (0, 1)

2

(1, 2)

3

(- 2, - 1)

4

(- \infty, - 2) \cup (2, \infty)

Explanation:

A Given, $a \in R$ ,
$- 3 (x - [x])^{2} + 2 (x - [x]) + a^{2} = 0$ ,
$[x]$ denotes the greater integer $\leq x$ .
$x - [x] = t$ , so we get-
$- 3 t^{2} + 2 t + a^{2} = 0$
$3 t^{2} - 2 t - a^{2} = 0$
$t = \frac{2 \pm \sqrt{4 + 12 a^{2}}}{6} = x - [x]$
$0 < \frac{2 \pm \sqrt{4 + 12 a^{2}}}{6} < 1$
$0 < 2 \pm \sqrt{4 + 12 a^{2}} < 6$
$0 < \pm \sqrt{4 + 12 a^{2}} < 4$
$0 < 4 + 12 a^{2} < 16$
$0 < 12 a^{2} < 12$
$0 < a^{2} < 1$ or $0 < (a - 1) (a + 1) < 1$
But, the given equation has no integral solution.
$∴ a \in (- 1, 0) \cup (0, 1) (∵ a \neq 0)$

JEE Main-2014

Complex Numbers and Quadratic Equation

118165 If both the roots of the quadratic equation $x^{2} -$ $2 k x + k^{2} + k - 5 = 0$ are less than 5 , then $k$ lies in the interval

1

[4, 5]

2

(- \infty, 4)

3

(6, \infty)

4

(5, 6]

Explanation:

B Given both the root of the equation $x^{2} - 2 kx$ $+ k^{2} + k - 5 = 0$
$∴$ Roots are $x = \frac{2 k \pm \sqrt{4 k^{2} - 4 (k^{2} + k - 5)}}{2}$
$= \frac{2 k \pm \sqrt{4 k^{2} - 4 k^{2} - 4 k + 20}}{2}$
$= \frac{2 k \pm 2 \sqrt{5 - k}}{2} = k \pm \sqrt{5 - k}$
Given, that both the roots are less than 5 .
$∴ k + \sqrt{5 - k} < 5 \Rightarrow \sqrt{5 - k} < 5 - k$
$o (5 - k) < (5 - k)^{2}$
$r (5 - k)^{2} - (5 - k) > 0$
$\Rightarrow (5 - k) (5 - k - 1) > 0$
$\Rightarrow (5 - k) (4 - k) > 0$
$\Rightarrow 5 - k > 0 and 4 - k > 0$
$\Rightarrow k < 5 and k < 4$
$∴ k < 4, so k lies in the interval (- \infty, 4)$
We can also check the other option,
$k - \sqrt{5 - k} < 5$
$\Rightarrow (k - 5) < \sqrt{5 - k}$
$\Rightarrow (k - 5)^{2} < 5 - k$
$\Rightarrow (k - 5)^{2} < - (k - 5)$
$\Rightarrow (k - 5)^{2} + k - 5 < 0$
$(k - 5) (k - 4) < 0$
$∴ k \in (4, 5)$ So, we get $k \in (- \infty, 4)$

AIEEE-2005

Complex Numbers and Quadratic Equation

118166 Sachin and Rahul attempted to solve a quadratic equation. Sachin made a mistake in writing down the constant term and ended up in roots $(4, 3)$ . Rahul made a mistake in writing down coefficient of $x$ to get roots $(3, 2)$ . The correct roots of equation are

1

- 4, - 3

2 6,1

3 4,3

4

- 6, - 1

Explanation:

B Sachin made mistake in interpreting constant term and found roots 4,3
$∴ 4 + 3 = - b / a$ when the equation is
$a x^{2} + b x + c = 0$
$x^{2} - \frac{b}{a} x + \frac{c}{a} = 0$
or
Rahul made num take in interpreting coefficient of $x$ and found roots $(3, 2)$
$∴ 3 \times 2 = c / a \Rightarrow c / a = 6$
$We know α + β = - b / c = 7$
$α β = c / a = 6$
$∴ (α - β)^{2} = (α + β)^{2} - 4 α β$
$= (7)^{2} - 4 \times 6$
$= 49 - 24 = 25$
$∴ α - β = \pm 5$
$α + β = 7$
$h l i n e 2 α = 12$
$α = 6$
$and β = 7 - 6 = 1$
$∴ The correct root are 6, 1$
$h l i n e r i g h t .$ $∴$ The correct root are 6,1 .

AIEEE-2011

Complex Numbers and Quadratic Equation

118167 The equation $e^{\sin x} - e^{- \sin x} - 4 = 0$ has

1 infinite number of real roots

2 no real root

3 exactly one real root

4 exactly four real roots

Explanation:

B Given the equation
$e^{\sin x} - e^{- \sin x} = 4$
$Let e^{\sin x} = t$
$∴ t - \frac{1}{t} = 4$
$\Rightarrow t^{2} - 1 = 4 t$
$t^{2} - 4 t - 1 = 0$
$∴ t = \frac{4 \pm \sqrt{16 + 4}}{2} \frac{4 \pm \sqrt{20}}{2}$
$= 2 \pm \sqrt{5}$
$∴ e^{\sin x} = (2 \pm \sqrt{5})$
$∴ \sin x = (2 \pm \sqrt{5}) > 1$
$It has no real solutions.$

AIEEE-2012

Complex Numbers and Quadratic Equation

118168 If $a \in R$ and the equation $- 3 (x - [x])^{2} + 2 (x -$ $[x]) + a^{2} = 0$ (where, $[x]$ denotes the greatest integer $\leq x$ ) has no integral solution, then all possible values of a lie in the interval

1

(- 1, 0) \cup (0, 1)

2

(1, 2)

3

(- 2, - 1)

4

(- \infty, - 2) \cup (2, \infty)

Explanation:

A Given, $a \in R$ ,
$- 3 (x - [x])^{2} + 2 (x - [x]) + a^{2} = 0$ ,
$[x]$ denotes the greater integer $\leq x$ .
$x - [x] = t$ , so we get-
$- 3 t^{2} + 2 t + a^{2} = 0$
$3 t^{2} - 2 t - a^{2} = 0$
$t = \frac{2 \pm \sqrt{4 + 12 a^{2}}}{6} = x - [x]$
$0 < \frac{2 \pm \sqrt{4 + 12 a^{2}}}{6} < 1$
$0 < 2 \pm \sqrt{4 + 12 a^{2}} < 6$
$0 < \pm \sqrt{4 + 12 a^{2}} < 4$
$0 < 4 + 12 a^{2} < 16$
$0 < 12 a^{2} < 12$
$0 < a^{2} < 1$ or $0 < (a - 1) (a + 1) < 1$
But, the given equation has no integral solution.
$∴ a \in (- 1, 0) \cup (0, 1) (∵ a \neq 0)$

JEE Main-2014

Complex Numbers and Quadratic Equation

118165 If both the roots of the quadratic equation $x^{2} -$ $2 k x + k^{2} + k - 5 = 0$ are less than 5 , then $k$ lies in the interval

1

[4, 5]

2

(- \infty, 4)

3

(6, \infty)

4

(5, 6]

Explanation:

B Given both the root of the equation $x^{2} - 2 kx$ $+ k^{2} + k - 5 = 0$
$∴$ Roots are $x = \frac{2 k \pm \sqrt{4 k^{2} - 4 (k^{2} + k - 5)}}{2}$
$= \frac{2 k \pm \sqrt{4 k^{2} - 4 k^{2} - 4 k + 20}}{2}$
$= \frac{2 k \pm 2 \sqrt{5 - k}}{2} = k \pm \sqrt{5 - k}$
Given, that both the roots are less than 5 .
$∴ k + \sqrt{5 - k} < 5 \Rightarrow \sqrt{5 - k} < 5 - k$
$o (5 - k) < (5 - k)^{2}$
$r (5 - k)^{2} - (5 - k) > 0$
$\Rightarrow (5 - k) (5 - k - 1) > 0$
$\Rightarrow (5 - k) (4 - k) > 0$
$\Rightarrow 5 - k > 0 and 4 - k > 0$
$\Rightarrow k < 5 and k < 4$
$∴ k < 4, so k lies in the interval (- \infty, 4)$
We can also check the other option,
$k - \sqrt{5 - k} < 5$
$\Rightarrow (k - 5) < \sqrt{5 - k}$
$\Rightarrow (k - 5)^{2} < 5 - k$
$\Rightarrow (k - 5)^{2} < - (k - 5)$
$\Rightarrow (k - 5)^{2} + k - 5 < 0$
$(k - 5) (k - 4) < 0$
$∴ k \in (4, 5)$ So, we get $k \in (- \infty, 4)$

AIEEE-2005

Complex Numbers and Quadratic Equation

118166 Sachin and Rahul attempted to solve a quadratic equation. Sachin made a mistake in writing down the constant term and ended up in roots $(4, 3)$ . Rahul made a mistake in writing down coefficient of $x$ to get roots $(3, 2)$ . The correct roots of equation are

1

- 4, - 3

2 6,1

3 4,3

4

- 6, - 1

Explanation:

B Sachin made mistake in interpreting constant term and found roots 4,3
$∴ 4 + 3 = - b / a$ when the equation is
$a x^{2} + b x + c = 0$
$x^{2} - \frac{b}{a} x + \frac{c}{a} = 0$
or
Rahul made num take in interpreting coefficient of $x$ and found roots $(3, 2)$
$∴ 3 \times 2 = c / a \Rightarrow c / a = 6$
$We know α + β = - b / c = 7$
$α β = c / a = 6$
$∴ (α - β)^{2} = (α + β)^{2} - 4 α β$
$= (7)^{2} - 4 \times 6$
$= 49 - 24 = 25$
$∴ α - β = \pm 5$
$α + β = 7$
$h l i n e 2 α = 12$
$α = 6$
$and β = 7 - 6 = 1$
$∴ The correct root are 6, 1$
$h l i n e r i g h t .$ $∴$ The correct root are 6,1 .

AIEEE-2011

Complex Numbers and Quadratic Equation

118167 The equation $e^{\sin x} - e^{- \sin x} - 4 = 0$ has

1 infinite number of real roots

2 no real root

3 exactly one real root

4 exactly four real roots

Explanation:

B Given the equation
$e^{\sin x} - e^{- \sin x} = 4$
$Let e^{\sin x} = t$
$∴ t - \frac{1}{t} = 4$
$\Rightarrow t^{2} - 1 = 4 t$
$t^{2} - 4 t - 1 = 0$
$∴ t = \frac{4 \pm \sqrt{16 + 4}}{2} \frac{4 \pm \sqrt{20}}{2}$
$= 2 \pm \sqrt{5}$
$∴ e^{\sin x} = (2 \pm \sqrt{5})$
$∴ \sin x = (2 \pm \sqrt{5}) > 1$
$It has no real solutions.$

AIEEE-2012

Complex Numbers and Quadratic Equation

118168 If $a \in R$ and the equation $- 3 (x - [x])^{2} + 2 (x -$ $[x]) + a^{2} = 0$ (where, $[x]$ denotes the greatest integer $\leq x$ ) has no integral solution, then all possible values of a lie in the interval

1

(- 1, 0) \cup (0, 1)

2

(1, 2)

3

(- 2, - 1)

4

(- \infty, - 2) \cup (2, \infty)

Explanation:

A Given, $a \in R$ ,
$- 3 (x - [x])^{2} + 2 (x - [x]) + a^{2} = 0$ ,
$[x]$ denotes the greater integer $\leq x$ .
$x - [x] = t$ , so we get-
$- 3 t^{2} + 2 t + a^{2} = 0$
$3 t^{2} - 2 t - a^{2} = 0$
$t = \frac{2 \pm \sqrt{4 + 12 a^{2}}}{6} = x - [x]$
$0 < \frac{2 \pm \sqrt{4 + 12 a^{2}}}{6} < 1$
$0 < 2 \pm \sqrt{4 + 12 a^{2}} < 6$
$0 < \pm \sqrt{4 + 12 a^{2}} < 4$
$0 < 4 + 12 a^{2} < 16$
$0 < 12 a^{2} < 12$
$0 < a^{2} < 1$ or $0 < (a - 1) (a + 1) < 1$
But, the given equation has no integral solution.
$∴ a \in (- 1, 0) \cup (0, 1) (∵ a \neq 0)$

JEE Main-2014

Complex Numbers and Quadratic Equation

118165 If both the roots of the quadratic equation $x^{2} -$ $2 k x + k^{2} + k - 5 = 0$ are less than 5 , then $k$ lies in the interval

1

[4, 5]

2

(- \infty, 4)

3

(6, \infty)

4

(5, 6]

Explanation:

B Given both the root of the equation $x^{2} - 2 kx$ $+ k^{2} + k - 5 = 0$
$∴$ Roots are $x = \frac{2 k \pm \sqrt{4 k^{2} - 4 (k^{2} + k - 5)}}{2}$
$= \frac{2 k \pm \sqrt{4 k^{2} - 4 k^{2} - 4 k + 20}}{2}$
$= \frac{2 k \pm 2 \sqrt{5 - k}}{2} = k \pm \sqrt{5 - k}$
Given, that both the roots are less than 5 .
$∴ k + \sqrt{5 - k} < 5 \Rightarrow \sqrt{5 - k} < 5 - k$
$o (5 - k) < (5 - k)^{2}$
$r (5 - k)^{2} - (5 - k) > 0$
$\Rightarrow (5 - k) (5 - k - 1) > 0$
$\Rightarrow (5 - k) (4 - k) > 0$
$\Rightarrow 5 - k > 0 and 4 - k > 0$
$\Rightarrow k < 5 and k < 4$
$∴ k < 4, so k lies in the interval (- \infty, 4)$
We can also check the other option,
$k - \sqrt{5 - k} < 5$
$\Rightarrow (k - 5) < \sqrt{5 - k}$
$\Rightarrow (k - 5)^{2} < 5 - k$
$\Rightarrow (k - 5)^{2} < - (k - 5)$
$\Rightarrow (k - 5)^{2} + k - 5 < 0$
$(k - 5) (k - 4) < 0$
$∴ k \in (4, 5)$ So, we get $k \in (- \infty, 4)$

AIEEE-2005

Complex Numbers and Quadratic Equation

118166 Sachin and Rahul attempted to solve a quadratic equation. Sachin made a mistake in writing down the constant term and ended up in roots $(4, 3)$ . Rahul made a mistake in writing down coefficient of $x$ to get roots $(3, 2)$ . The correct roots of equation are

1

- 4, - 3

2 6,1

3 4,3

4

- 6, - 1

Explanation:

B Sachin made mistake in interpreting constant term and found roots 4,3
$∴ 4 + 3 = - b / a$ when the equation is
$a x^{2} + b x + c = 0$
$x^{2} - \frac{b}{a} x + \frac{c}{a} = 0$
or
Rahul made num take in interpreting coefficient of $x$ and found roots $(3, 2)$
$∴ 3 \times 2 = c / a \Rightarrow c / a = 6$
$We know α + β = - b / c = 7$
$α β = c / a = 6$
$∴ (α - β)^{2} = (α + β)^{2} - 4 α β$
$= (7)^{2} - 4 \times 6$
$= 49 - 24 = 25$
$∴ α - β = \pm 5$
$α + β = 7$
$h l i n e 2 α = 12$
$α = 6$
$and β = 7 - 6 = 1$
$∴ The correct root are 6, 1$
$h l i n e r i g h t .$ $∴$ The correct root are 6,1 .

AIEEE-2011

Complex Numbers and Quadratic Equation

118167 The equation $e^{\sin x} - e^{- \sin x} - 4 = 0$ has

1 infinite number of real roots

2 no real root

3 exactly one real root

4 exactly four real roots

Explanation:

B Given the equation
$e^{\sin x} - e^{- \sin x} = 4$
$Let e^{\sin x} = t$
$∴ t - \frac{1}{t} = 4$
$\Rightarrow t^{2} - 1 = 4 t$
$t^{2} - 4 t - 1 = 0$
$∴ t = \frac{4 \pm \sqrt{16 + 4}}{2} \frac{4 \pm \sqrt{20}}{2}$
$= 2 \pm \sqrt{5}$
$∴ e^{\sin x} = (2 \pm \sqrt{5})$
$∴ \sin x = (2 \pm \sqrt{5}) > 1$
$It has no real solutions.$

AIEEE-2012

Complex Numbers and Quadratic Equation

118168 If $a \in R$ and the equation $- 3 (x - [x])^{2} + 2 (x -$ $[x]) + a^{2} = 0$ (where, $[x]$ denotes the greatest integer $\leq x$ ) has no integral solution, then all possible values of a lie in the interval

1

(- 1, 0) \cup (0, 1)

2

(1, 2)

3

(- 2, - 1)

4

(- \infty, - 2) \cup (2, \infty)

Explanation:

A Given, $a \in R$ ,
$- 3 (x - [x])^{2} + 2 (x - [x]) + a^{2} = 0$ ,
$[x]$ denotes the greater integer $\leq x$ .
$x - [x] = t$ , so we get-
$- 3 t^{2} + 2 t + a^{2} = 0$
$3 t^{2} - 2 t - a^{2} = 0$
$t = \frac{2 \pm \sqrt{4 + 12 a^{2}}}{6} = x - [x]$
$0 < \frac{2 \pm \sqrt{4 + 12 a^{2}}}{6} < 1$
$0 < 2 \pm \sqrt{4 + 12 a^{2}} < 6$
$0 < \pm \sqrt{4 + 12 a^{2}} < 4$
$0 < 4 + 12 a^{2} < 16$
$0 < 12 a^{2} < 12$
$0 < a^{2} < 1$ or $0 < (a - 1) (a + 1) < 1$
But, the given equation has no integral solution.
$∴ a \in (- 1, 0) \cup (0, 1) (∵ a \neq 0)$

JEE Main-2014

Question Bank Series

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Question Bank Series

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