Complex Numbers and Quadratic Equations Test 22

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Complex Numbers and Quadratic Equations Test 22

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Weekly Quiz Competition

Question 1
1 / -0

If $$x + i y = \dfrac{3}{2 + cos \theta + i sin \theta}$$, then $$x^2 + y^2$$ is equal to

A
$$3x - 4$$

B
$$4x - 3$$

C
$$4x +3$$

D
None of these

Solution

$$x + i y = \dfrac{3}{2 + cos \theta + i sin \theta} = \dfrac{3 (2 + cos \theta - i sin \theta)}{(2 + cos \theta)^2 - i^2 sin^2 \theta}$$
$$= \dfrac{3(2+cos \theta-i sin \theta)}{(2+cos \theta)^2+sin \theta}$$

$$= \dfrac{6 + 3 cos \theta - 3 i sin \theta}{4 + cos^2 \theta + 4 cos \theta + sin^2 \theta}$$

$$= \dfrac{6 + 3 cos \theta - 3 i sin \theta}{5 + 4 cos \theta}$$

$$= \left( \dfrac{6 + 3 cos \theta}{5 + 4 cos \theta} \right ) + \left( \dfrac{-3 sin \theta}{5 + 4 cos \theta} \right )$$
On equating real and imaginary parts, we get
$$x = \dfrac{3 (2 + cos \theta)}{5 + 4 cos \theta}$$
and $$y = \dfrac{-3 sin \theta}{5 + 4 cos \theta}$$

$$\therefore x^2 + y^2 = \dfrac{9 [4 + cos^2 \theta + 4 cos \theta + sin^2 \theta]}{(5 + 4 cos \theta)^2}$$

$$= \dfrac{9}{5 + 4 cos \theta} = 4 \left( \dfrac{6 + 3 cos \theta}{5 + 4 cos^2 \theta} \right ) - 3$$
$$= 4x - 3$$
Hence, option B is correct.
Question 2
1 / -0

Let $$z = x + iy$$, where $$x$$ and $$y$$ are real. The points $$(x, y)$$ in the $$X-Y$$ plane for which $$\dfrac {z + i}{z - i}$$ is purely imaginary lie on

A
A straight line

B
An ellipse

C
A hyperbola

D
A circle

Solution

Let $$z = x + iy$$
$$\therefore \dfrac {z + i}{(z - i)}=\dfrac{x+iy+i}{x+iy-i}$$
$$= \dfrac {(x + i(y + 1))}{(x - i(1 - y))}\times \dfrac {(x + i(1 - y))}{(x + i(1 - y))}$$
$$=\dfrac {x^{2} + (y^{2} - 1)+2ix}{x^{2} + (1 - y)^{2}} $$
Since, $$\dfrac{z+i}{z-i}$$ should be purely imaginary
$$\therefore Re\left (\dfrac {z + i}{z - i}\right ) = 0$$
$$\implies \dfrac {x^{2} + (y^{2} - 1)}{x^{2} + (1 - y)^{2}} = 0$$
$$\implies x^{2}+y^{2}-1=0$$
$$\implies x^{2} + y^{2} = 1$$ which is the equation of a circle
Hence, $$(x,y)$$ lie on a circle.
Question 3
1 / -0

The roots of the equation $${ x }^{ 2 }-8x+16=0$$

A
Are imaginary

B
Are distinct and real

C
Are equal and real

D
Cannot be ascertained

Solution

Given equation is $$x^2-8x+16=0$$
Discriminant is $${(-8)}^2-4\times 1\times 16=64-64=0$$
Since the discriminant is zero, the quadratic equation has equal and real roots.
Question 4
1 / -0

The expression $$\dfrac {(1 + i)^{n}}{(1 - i)^{n - 2}}$$ equals.

A
$$-i^{n + 1}$$

B
$$i^{n + 1}$$

C
$$-2i^{n + 1}$$

D
$$1$$

Solution

$$(1 - i) = (1-i)\times \dfrac{(1+i)}{(1+i)} = \dfrac{1-i^{2}}{(1+i)} = \dfrac {2}{(1 + i)}$$
$$\therefore \dfrac {(1 + i)^{n}}{(1 - i)^{n - 2}} = \dfrac {(1 + i)^{n}}{2^{n - 2}} . (1 + i)^{n - 2}$$
$$= \dfrac {(1 + i)^{2(n - 1)}}{2^{n - 2}}$$

$$= \dfrac {(1+i^{2}+2i)^{n - 1}}{2^{n - 2}}$$

$$= \dfrac {(2i)^{n - 1}}{2^{n - 2}}$$
$$= 2i^{n - 1} = -2i^{n + 1}$$
Hence, $$\therefore \dfrac {(1 + i)^{n}}{(1 - i)^{n - 2}} = - 2 i^{n+1}$$
Question 5
1 / -0

The value of $$a$$ for which the quadratic equation $$3{x}^{2} + 2{a}^{2}x + 2x + {a}^{2} - 3a + 2 = 0$$ possesses roots of opposite signs lies in

A
$$(-\infty, 1)$$

B
$$\left(-\infty, 0\right)$$

C
$$\left(1, 2\right)$$

D
$$\left(\dfrac{3}{2}, 2\right)$$

Solution

$$3{ x }^{ 2 }+2{ a }^{ 2 }x+2x+{ a }^{ 2 }-3a+2=0$$

$$3{ x }^{ 2 }+2(a^{ 2 }+1)x+({ a }^{ 2 }-3a+2)=0$$

Condition for real distinct roots $$\Rightarrow \quad D>0$$
$${ b }^{ 2 }-4ac>0$$

$$4{ \left( { a }^{ 2 }+1 \right) }^{ 2 }-4\times 3\times ({ a }^{ 2 }-3a+2)>0$$

$$4({ a }^{ 4 }+1+2{ a }^{ 2 })-(12{ a }^{ 2 }-36a+24)>0$$

$$4{ a }^{ 4 }+4+8{ a }^{ 2 }-12{ a }^{ 2 }+36a-24>0$$

$$4{ a }^{ 4 }-4{ a }^{ 2 }+36a-20>0$$

$${ a }^{ 4 }-{ a }^{ 2 }+9a-5>0\longrightarrow (i)$$

If roots are of opposite signs then product of roots $$<0$$

$$\cfrac { constant }{ coefficient\quad of\quad { x }^{ 2 } } <0\Rightarrow \cfrac { { a }^{ 2 }-3a+2 }{ 3 } <0$$

$${ a }^{ 2 }-3a+2<0$$

$${ a }^{ 2 }-2a-a+2<0$$

$$a\left( a-2 \right) -1\left( a-2 \right) <0\Rightarrow \left( a-2 \right) \left( a-1 \right) <0$$

$$\left( a-2 \right) >0$$ and $$\left( a-1 \right) <0$$ or $$\left( a-2 \right) <0$$ and $$\left( a-1 \right) >0$$

$$a>2$$ and $$a<1$$ or $$a<2$$ and $$a>1$$

$$a<1$$ and $$a>2\Rightarrow 2<a<1$$ which is not possible

$$\therefore \quad a>2$$ & $$a<1$$ is rejected

$$\therefore a\in \left( 1,2 \right) $$

Hence, C is correct.
Question 6
1 / -0

If $$z_1$$ and $$z_2$$ are complex numbers with $$|z_1|=|z_2|$$, then which of the following is/are correct?
1. $$z_1=z_2$$
2. Real part of $$z_1 =$$ Real part of $$z_2$$
3. Imaginary part of $$z_1 =$$ Imaginary part of $$z_2$$
Select the correct answer using the statements given below :

A
1 only

B
2 only

C
3 only

D
None

Solution

Solution:
We have, $$|z_1|=|z_2|$$
Let,
$$z_1=x_1+iy_1$$ and $$z_2=x_2+iy_2$$
$$\because|z_1|=|z_2|$$
$$\therefore x_1^2+y_1^2=x_2^2+y_2^2$$
or, $$(x_1^2-x_2^2)+(y_1^2-y_2^2)=0$$
or, $$x_1^2-x_2^2=0$$ and $$y_1^2-y_2^2=0$$
or, $$x_1=\pm x_2$$ and $$y_1=\pm y_2$$
Let, $$z_1=1+i$$ then $$z_2=-1-i$$
$$|z_1|=|z_2|=\sqrt2$$
But, $$Re(z_1)\neq Re(z_2)$$ and $$Im(z_1)\neq Im(z_2)$$ and $$z_1\neq z_2$$
Hence, D is the correct option.
Question 7
1 / -0

Let a, b be non-zero real numbers. Which of the following statements about the quadratic equation $$ax^2 + (a+b) x + b = 0$$ is necessarily true?
(I) It has at least one negative root.
(II) It has at least one positive root.
(III) Both its roots are real.

A
(I) and (II) only

B
(I) and (III) only

C
(II) and (III) only

D
All of them

Solution

Roots of the given equation are,

$$=\dfrac { -(a+b)\pm \sqrt { (a+b)^{ 2 }-4ab } }{ 2a } $$

$$ =\dfrac { -(a+b)\pm (a-b) }{ 2a }$$

We get the roots as,
$$ -\dfrac { b }{ a } , -1$$
So, both the roots are real and atleast one negative root.
Hence, B is correct.
Question 8
1 / -0

The quadratic equation $${ x }^{ 2 }+bx+4=0$$ will have real roots if

A
$$b\le -4$$ only

B
$$b\ge 4$$ only

C
$$-4 < b < 4$$

D
$$b\le -4,b\ge 4$$

Solution

In any quadratic equation, real roots exist if discriminant is greater than or equal to zero.
So for $$x^2+bx+4=0$$
Discriminant is $$={ b }^{ 2 }-4\times 1\times 4$$
So, $$b^{2}-16\geq 0$$
$$\Rightarrow b\ge 4,b\le -4$$
Question 9
1 / -0

The roots of the equation $$2a^2x^2 - 2abx +b^2=0$$ when a < 0 and b > 0 are :

A
Sometimes complex

B
Always irrational

C
Always complex

D
Always real

Solution

For an equation $$ ax^2 \ + \ bx \ +c =0 $$, where $$ a \neq \ 0 $$ to have real roots, D must be greater than or equal to 0, where
$$D = b^2-4ac$$

In the given equation, $$2a^2 x^2 \ - \ 2abx \ + \ b^2 =0$$
$$D=(-2ab)^2-4\times(2a^2)\times(b^2)$$
$$\Rightarrow D= -6 a^2 b^2$$
$$\because a,b \neq 0 \ \Rightarrow -6a^2 b^2$$ is always negative
We see that $$D$$ is strictly negative
Hence, the equation always has complex roots.
Question 10
1 / -0

If ................. then the roots of the quadratic equation are equal.

A
$$D=0$$

B
$$D\ne 0$$

C
$$D< 0$$

D
$$D> 0$$

Solution

There are two values of $$x$$ which satisfy a quadratic equation.

If the equation reads $$ax^2 + bx + c = 0$$, the determinant is $$D = \sqrt{b^2 - 4ac}$$

The two values of $$x$$ are given by $$x = \cfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \cfrac{-b \pm \sqrt{D}}{2a}$$

As can be concluded from the equation, we will have equal roots if $$D = 0$$

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Complex Numbers and Quadratic Equations Test 22

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SHARING IS CARING

Question 1

$$3x - 4$$

$$4x - 3$$

$$4x +3$$

None of these

Solution

Question 2

A straight line

An ellipse

A hyperbola

A circle

Solution

Question 3

Are imaginary

Are distinct and real

Are equal and real

Cannot be ascertained

Solution

Question 4

$$-i^{n + 1}$$

$$i^{n + 1}$$

$$-2i^{n + 1}$$

$$1$$

Solution

Question 5

$$(-\infty, 1)$$

$$\left(-\infty, 0\right)$$

$$\left(1, 2\right)$$

$$\left(\dfrac{3}{2}, 2\right)$$

Solution

Question 6

1 only

2 only

3 only

None

Solution

Question 7

(I) and (II) only

(I) and (III) only

(II) and (III) only

All of them

Solution

Question 8

$$b\le -4$$ only

$$b\ge 4$$ only

$$-4 < b < 4$$

$$b\le -4,b\ge 4$$

Solution

Question 9

Sometimes complex

Always irrational

Always complex

Always real

Solution

Question 10

$$D=0$$

$$D\ne 0$$

$$D< 0$$

$$D> 0$$

Solution

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