Complex Numbers and Quadratic Equations Test -8

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Complex Numbers and Quadratic Equations Test -8

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Weekly Quiz Competition

Question 1
1 / -0

The equation $$\sqrt {3x^2+x+5}=x-3$$, where $$x$$ is real, has

A
no solution.

B
exactly one solution.

C
exactly two solutions.

D
exactly four solutions.

Solution

Now, $$3x^2+x+5 \geqslant 0$$

This is because $$b^2-4ac=1-4\times 5\times 3$$

$$=-59$$ (roots are imaginary)

$$3x^2+x+5 =(x-3)^2=x^2+9-6x$$ and $$x-3 \geqslant 0$$

$$2x^2+7x-4=0$$

$$2x^2+8x-x-4=0$$

$$2x(x+4)-1(x+4)=0$$

$$(2x-1)(x+4)=0$$

$$x=\dfrac12 \ and -4$$ but $$x\geq 3$$

$$\therefore$$ No solution.
Question 2
1 / -0

Let $$\alpha,\ \beta$$ be real and $$\mathrm{z}$$ be a complex number. If $$\mathrm{z}^{2}+\alpha \mathrm{z}+\beta=0$$ has two distinct roots on the line $$Re(z) =1$$, then it is necessary that:

A
$$\beta\in(0,1)$$

B
$$\beta\in(-1,0)$$

C
$$|\beta|=1$$

D
$$\beta\in(1, \infty)$$
Question 3
1 / -0

Let z be a complex number such that $$\left|\dfrac{z-i}{z+2i}\right|=1$$ and $$|z|=\dfrac{5}{2}$$. Then the value of $$|z+3i|$$ is?

A
$$\dfrac{7}{2}$$

B
$$\dfrac{15}{4}$$

C
$$2\sqrt{3}$$

D
$$\sqrt{10}$$

Solution

Given $$\left|\dfrac{z - i}{z + 2 i } \right| = 1$$

$$\Rightarrow |z - i| = |z + 2i|$$

Let $$z = x + iy$$

So, $$x^2 + (y- 1)^2 = x^2 + (y + 2)^2$$

$$-2y + 1 = 4y + 4$$

$$6y = -3 \Rightarrow y = \dfrac{-1}{2}$$

$$x^2 + y^2 = \dfrac{25}{4} \Rightarrow x^2 = \dfrac{24}{4} = 6$$

$$z = \pm \sqrt{6} - \dfrac{i}{2}$$

$$|z + 3i| = \sqrt{6 + \dfrac{25}{4}} = \sqrt{\dfrac{49}{4}} = \dfrac{7}{2}$$

$$\therefore$$ $$\boxed{|2 + 3i| = \dfrac{7}{2}}.....Answer$$

Hence option $$'A'$$ is the answer.
Question 4
1 / -0

If $$z = x + iy$$ and $$\omega = \dfrac{(1 -iz)}{(z-i)}$$, then $$\left|\omega\right| = 1$$ implies that in the complex plane

A
z lies on the imaginary axis

B
z lies on the real axis

C
z lies on the unit circle

D
none of these

Solution

Given $$w=\dfrac { 1-iz }{ z-i } $$ and

$$\left| w \right| =1$$

$$\Rightarrow \left| \dfrac { 1-iz }{ z-i } \right| =1$$ ...(1)

Substitute $$z=x+iy$$ in equation (1)

$$\Rightarrow \left| \dfrac { 1-i\left( x+iy \right) }{ \left( x+iy \right) -i } \right| =1$$

$$\Rightarrow \left| 1+y-ix \right| =\left| x+i\left( y-1 \right) \right| \\ \Rightarrow { \left( 1+y \right) }^{ 2 }+{ x }^{ 2 }={ x }^{ 2 }+{ \left( y-1 \right) }^{ 2 }\\ \Rightarrow y=0$$

Therefore z lies on the real axis.

Ans: B
Question 5
1 / -0

Let z be a complex number such that the imaginary part of z is nonzero and a = $$z^2 + z + 1$$ is real. Then a cannot take the value

A
$$-1$$

B
$$\dfrac{1}{3}$$

C
$$\dfrac{1}{2}$$

D
$$\dfrac{3}{4}$$

Solution

Let $$z=\alpha+i\beta$$. Then

$$Im(z^2+z+1)=2\alpha\beta+\beta=0 $$

$$\Rightarrow \alpha=-\dfrac{1}{2}$$ since $$\beta\neq0$$. Then

$$a=\alpha^2-\beta^2+\alpha+1$$

$$a=\dfrac{3}{4}+\beta$$

$$a\neq\dfrac{3}{4}$$. $$[\because \beta\neq0]$$
Question 6
1 / -0

Express $$\dfrac{1}{(1 - cos \theta + 2 i sin \theta)}$$ in the form $$x + iy$$

A
$$\left(\displaystyle \frac{1}{5 + 3 cos \theta}\right) + \left(\displaystyle \frac{2 cot \theta/2}{5 + 3 cos \theta}\right)i$$

B
$$\left(\displaystyle \frac{1}{5 - 3 cos \theta}\right) + \left(\displaystyle \frac{-2 cot \theta/2}{5 - 3 cos \theta}\right)i$$

C
$$\left(\displaystyle \frac{1}{5 + 3 cos \theta}\right) + \left(\displaystyle \frac{-2 cot \theta/2}{5 + 3 cos \theta}\right)i$$

D
$$\left(\displaystyle \frac{1}{5 - 3 cos \theta}\right) + \left(\displaystyle \frac{2 cot \theta/2}{5 - 3 cos \theta}\right)i$$

Solution

To Express $$ \dfrac{1}{(1-cos\theta +2isin\theta )}$$ in the form $$x+iy$$
We have , $$ \dfrac{1}{(1-cos\theta +2isin\theta )}$$
$$ = \dfrac{1}{(1-cos\theta) +2isin\theta } \times \dfrac{(1-cos\theta) - 2isin\theta }{(1-cos\theta ) - 2isin\theta }$$

$$ = \dfrac{(1-cos\theta -2isin\theta )}{(1-cos\theta )^{2}+(2sin\theta )^{2}}$$
$$ = \dfrac{(1-cos\theta -2isin\theta )}{1 + cos^{2}\theta -2cos\theta +4sin^{2}\theta }$$
$$ = \dfrac{(1-cos\theta )}{(1-cos\theta )( 3cos\theta + 5)} - \dfrac{2isin\theta }{(1-cos\theta )( 3cos\theta + 5)} $$

$$ = \dfrac{1}{( 3cos\theta + 5)} - \dfrac{2isin\dfrac{\theta }{2}cos\dfrac{\theta }{2} }{2sin^{2}\dfrac{\theta }{2}( 3cos\theta + 5)}$$
$$ = \dfrac{1}{( 3cos\theta + 5)} - \dfrac{2cot\dfrac{\theta }{2} }{( 3cos\theta + 5)}i$$

$$ = \dfrac{1}{( 5 + 3cos\theta )} + \dfrac{(-2cot{\theta }/{2} )}{( 5 + 3cos\theta )}i$$

Hence , Option C
Question 7
1 / -0

$$\displaystyle \left|\dfrac{\sqrt{3}+i}{(1+i)(1+\sqrt{3}i)}\right|=$$

A
$$1$$

B
$$\sqrt{2}$$

C
$$\dfrac{1}{2}$$

D
$$\dfrac{1}{\sqrt{2}}$$

Solution

The value of $$\displaystyle \left |\frac{\sqrt{3} + i}{(1 + i)(1 + \sqrt{3}i)} \right |$$
$$ = \dfrac{|\sqrt{3} + i|}{|1 + i||1 + \sqrt{3}i|} = \dfrac{\sqrt{3 + 1}}{\sqrt{1 + 1} \times \sqrt{1 + 3}} $$
$$= \dfrac{2}{2\sqrt{2}} $$
$$= \dfrac{1}{\sqrt{2}}$$
Question 8
1 / -0

If $$a, b$$ and $$c$$ are real numbers then the roots of the equation $$(x - a)(x - b) + (x - b)(x - c) + (x - c)(x - a) = 0$$ are always

A
Real

B
Imaginary

C
Positive

D
Negative

Solution

Given equation is $$(x - a)(x - b) + (x - b)(x - c) + (x - c)(x - a) = 0$$

$$\Rightarrow 3x^{2} - 2(b + a + c) x + ab + bc + ca = 0$$

Now, here $$A = 3, B = -2(a + b + c)$$

$$C = ab + bc + ca$$

Therefore, $$ D = \sqrt {B^{2} - 4AC}$$

$$= \sqrt {(-2(a + b + c))^{2} - 4(3)(ab + bc + ca)}$$

$$= \sqrt {4(a + b +c)^{2} - 12(ab + bc + ca)}$$

$$= 2\sqrt {a^{2} + b^{2} + c^{2} - ab - bc - ca}$$

$$= 2\sqrt {\dfrac {1}{2}\left \{(a - b)^{2} - (b - c)^{2} + (c - a)^{2} \right \}} \geq 0$$

This is always $$\geq 0$$, we have real roots for the equation $$(x - a)(x - b) + (x - b)(x - c) + (x - c)(x - a) = 0$$
Question 9
1 / -0

$$ax^2+bx+c=0$$, where a, b, c are real, has real roots if.

A
a, b, c are integers

B
$$b^2 > 4ac$$

C
$$ac < 0$$ and b is zero

D
$$c=0$$

Solution

Given equation is $$ax^2+bx+c=0$$ where $$a.b.c$$ are real

Roots of this equation is given by $$x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$$
$$x$$ is real, if $$\sqrt{b^2-4ac}$$ is real.

if $$b^{2}-4ac>0$$

if $$b^{2}>4ac$$

Thus, $$x$$ is real iff $$b^2>4ac$$.

Hence, option B is correct.
Question 10
1 / -0

If $$z =3+5i$$, then $$z^3+z+198=$$

A
$$3 - 15i$$

B
$$-3 - 15i$$

C
$$-3 + 15i$$

D
$$3 + 15i$$

Solution

$$z=3+5i$$

$$z^{3}=(3+5i)^{3}$$

$$=3^{3}+3.3^{2}(5i)+3.3(5i)^{2}+(5i)^{3}$$

$$=27-125i+135i-225$$

$$=-225+27+(135-125)i$$

$$=-198+10i$$

$$\therefore z^{3}+z+198$$

$$=-198+10i+3+5i+198$$

$$=3+15i$$

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Complex Numbers and Quadratic Equations Test -8

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Question 1

no solution.

exactly one solution.

exactly two solutions.

exactly four solutions.

Solution

Question 2

$$\beta\in(0,1)$$

$$\beta\in(-1,0)$$

$$|\beta|=1$$

$$\beta\in(1, \infty)$$

Question 3

$$\dfrac{7}{2}$$

$$\dfrac{15}{4}$$

$$2\sqrt{3}$$

$$\sqrt{10}$$

Solution

Question 4

z lies on the imaginary axis

z lies on the real axis

z lies on the unit circle

none of these

Solution

Question 5

$$-1$$

$$\dfrac{1}{3}$$

$$\dfrac{1}{2}$$

$$\dfrac{3}{4}$$

Solution

Question 6

$$\left(\displaystyle \frac{1}{5 + 3 cos \theta}\right) + \left(\displaystyle \frac{2 cot \theta/2}{5 + 3 cos \theta}\right)i$$

$$\left(\displaystyle \frac{1}{5 - 3 cos \theta}\right) + \left(\displaystyle \frac{-2 cot \theta/2}{5 - 3 cos \theta}\right)i$$

$$\left(\displaystyle \frac{1}{5 + 3 cos \theta}\right) + \left(\displaystyle \frac{-2 cot \theta/2}{5 + 3 cos \theta}\right)i$$

$$\left(\displaystyle \frac{1}{5 - 3 cos \theta}\right) + \left(\displaystyle \frac{2 cot \theta/2}{5 - 3 cos \theta}\right)i$$

Solution

Question 7

$$1$$

$$\sqrt{2}$$

$$\dfrac{1}{2}$$

$$\dfrac{1}{\sqrt{2}}$$

Solution

Question 8

Real

Imaginary

Positive

Negative

Solution

Question 9

a, b, c are integers

$$b^2 > 4ac$$

$$ac < 0$$ and b is zero

$$c=0$$

Solution

Question 10

$$3 - 15i$$

$$-3 - 15i$$

$$-3 + 15i$$

$$3 + 15i$$

Solution

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