Complex Numbers and Quadratic Equations Test 26

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Complex Numbers and Quadratic Equations Test 26

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Weekly Quiz Competition

Question 1
1 / -0

If $${ a }^{ 2 }+{ b }^{ 2 }=1$$, then $$\dfrac {\left( 1+b+ia \right) }{\left( 1+b-ia \right)} $$ is

A
$$1$$

B
$$2$$

C
$$b+ia$$

D
$$a+ib$$

Solution

$$a^2+b^2=1$$

$$\implies 1-a^2=b^2$$

$$\dfrac{1+b+ia}{1+b-ia}$$

$$=\dfrac{1+b+ia}{1+b-ia}\times \dfrac{1+b+ia}{1+b+ia}$$

$$=\dfrac{(1+b)^2-a^2+2ia(1+b)}{(1+b)^2+a^2}$$

$$=\dfrac{b^2+2b+1-a^2+2ia(1+b)}{1+2b+b^2+a^2}$$

$$=\dfrac{2b^2+2b+2ia(1+b)}{2+2b}$$

$$=\dfrac{(2b+2ia)(1+b)}{2(1+b)}$$

$$=b+ia$$

Answer-(C)
Question 2
1 / -0

The real part of $$(1-\cos\theta +2i \sin\theta)^{-1}$$ is?

A
$$\displaystyle\frac{1}{3+5\cos\theta}$$

B
$$\displaystyle\frac{1}{5-3\cos\theta}$$

C
$$\displaystyle\frac{1}{3-5\cos\theta}$$

D
$$\displaystyle\frac{1}{5+3\cos\theta}$$

Solution

$$(1-\cos\theta +2i \sin\theta)^{-1}$$
$$=\dfrac{1}{(1-\cos\theta +2i \sin\theta)}$$
$$=\dfrac{1}{(2\sin^2\dfrac{\theta}{2} +4i \sin\dfrac{\theta}{2}.\cos \dfrac{\theta}{2})}$$
$$=\dfrac{1}{2\sin\dfrac{\theta}{2}}\dfrac{1}{(\sin\dfrac{\theta}{2} +2i \cos\dfrac{\theta}{2})}$$
$$=\dfrac{1}{2\sin\dfrac{\theta}{2}}\dfrac{\left(\sin\dfrac{\theta}{2}-2i\cos \dfrac{\theta}{2}\right)}{(\sin^2\dfrac{\theta}{2} +4 \cos^2\dfrac{\theta}{2})}$$
$$=\dfrac{1}{\sin\dfrac{\theta}{2}}\dfrac{\left(\sin\dfrac{\theta}{2}-2i\cos \dfrac{\theta}{2}\right)}{(2\sin^2\dfrac{\theta}{2} +8 \cos^2\dfrac{\theta}{2})}$$
$$=\dfrac{\left(1-2i\cot \dfrac{\theta}{2}\right)}{\{1-\cos \theta+4(1+\cos \theta)\}}$$
$$=\dfrac{\left(1-2i\cot \dfrac{\theta}{2}\right)}{5+3\cos \theta}$$
So the real part is $$\dfrac{1}{5+3\cos \theta}$$..
Question 3
1 / -0

If in applying the quardratic formula to a quadratic equation
$$f(x) = ax^2 + bx + c = 0$$, it happens that $$c = b^2/4a$$, then the graph of $$y = f(x)$$ will certainly:

A
have a maximum

B
have a minimum

C
tangent to the $$x$$-axis

D
be tangent to the $$y$$-axis

E
lie in one quadrant only

Solution

The condition $$c = b^2/4a$$ implies equal real coefficients, i.e., the curve touches the $$x$$-axis at exactly one point and, therefore, the graph is tangent to the $$x$$-axis.
Question 4
1 / -0

If $$\mid{z_1}\mid=2$$, $$\mid{z_2}\mid=3$$, $$\mid{z_3}\mid=4$$ and $$\mid{z_1+z_2+z_3}\mid=2$$, then the value of $$\mid{4z_2z_3+9z_3z_1+16z_1z_2}\mid$$.

A
24

B
48

C
96

D
120

Solution

$$z_1\bar{z_1}=|z_1|^2=1$$$$,z_2\bar{z_2}=|z_2|^2=9$$$$,z_3\bar{z_3}=|z_3|^2=6$$
$$|4z_2z_3+9z_3z_1+16z_1z_2|$$
$$\implies |\bar{z_1}z_1z_2z_3$$$$+\bar{z_2}z_1z_2z_3$$$$+\bar{z_3}z_1z_2z_3|$$
$$\implies |z_1||z_2||z_3||\bar{z_1}+\bar{z_2}+\bar{z_3}|$$
$$\implies 24\times$$ $$|\bar{z_1}+\bar{z_2}+\bar{z_3}|$$
$$\implies 48$$
Question 5
1 / -0

Evaluate:
$${ \left( \dfrac { cos\dfrac { \pi }{ 8 } -isin\dfrac { \pi }{ 8 } }{ cos\dfrac { \pi }{ 8 } +isin\dfrac { \pi }{ 8 } } \right) }^{ 4 }$$

A
$$1$$

B
$$-1$$

C
$$2$$

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

Solution

We know that $$e^{i\theta}=cos\theta+isin\theta$$ and $$e^{-1\theta}=cos\theta-isin\theta$$

$${ \left( \dfrac { cos\dfrac { \pi }{ 8 } -isin\dfrac { \pi }{ 8 } }{ cos\dfrac { \pi }{ 8 } +isin\dfrac { \pi }{ 8 } } \right) }^{ 4 }$$

$$=\left(\dfrac{e^{-i(\pi/8)}}{e^{i(\pi/8)}}\right)^4=(e^{-i(\pi/4)})^4=e^{-i\pi}=cos\pi-isin\pi=-1$$
Question 6
1 / -0

Let $$z_1$$ and $$z_2$$ are two complex numbers such that $$(1-i)z_1=2z_2$$ and $$arg(z_1z_2)=\dfrac{\pi}{2}$$ then $$arg(z_2)$$ is equals to:

A
$$\dfrac{3 \pi}{8}$$

B
$$\dfrac{\pi}{8}$$

C
$$\dfrac{5 \pi}{8}$$

D
$$\dfrac{-7 \pi}{8}$$

Solution

$$(1-i)z_1=2z_2$$
$$\cfrac{z_2}{z_1}=\cfrac{1}{2}-\cfrac{i}{2}$$
Let $$z_1=r_1e^{i\theta_1}\ and\ z_=r_1r^{\theta_2}$$
$$arg(\cfrac{z_2}{z_1})=\tan^{-1}\cfrac{-1/2}{1/2}=-\pi/4$$
$$\theta_2-\theta_1=-\pi/4$$ and $$arg(z_1z_2)=\pi/2$$ (given)
$$\implies \theta_2-\theta_1=-\pi/4$$ and $$\theta_2
+\theta_1=\pi/2$$
$$\implies 2\theta_2=\pi/4,\theta_2=\pi/8$$
$$\implies arg(z_2)=\pi/8$$
Question 7
1 / -0

Let $$z$$ be a complex number such that $$\left| z+\dfrac { 1 }{ z } \right| =2$$.
If $$\left| z \right| ={ r }_{ 1 }$$ and $$\left| \dfrac { 1 }{ z } \right| =$$ $${r}_{2}$$ for $$\arg z=\dfrac { \pi }{ 4 }$$ then
$$\left| { r }_{ 1 }-{ r }_{ 2 } \right| =$$

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

B
$$1$$

C
$$\sqrt { 2 }$$

D
$$2$$

Solution

$$ \left| z+\dfrac { 1 }{ z } \right| ^{ 2 }={ r }^{ 2 }+\dfrac { 1 }{ { r }^{ 2 } } +2r.\dfrac { 1 }{ r } \cos { \left( \theta +\theta \right) } =4$$
or $${ r }^{ 2 }+\dfrac { 1 }{ { r }^{ 2 } } =4-2\cos { 2\theta }$$
$$\therefore\quad \left( r-\dfrac { 1 }{ r } \right) ^{ 2 }=2\left( 1-\cos { 2\theta } \right) =4\sin { ^{ 2 }\theta }$$
$$\left| { r }_{ 1 }-{ r }_{ 2 } \right| =\left| r-\dfrac { 1 }{ r } \right| =2\sin { \theta } =2\sin { \dfrac { \pi }{ 4 } } =\sqrt { 2 }$$
Question 8
1 / -0

If $${z_1}$$ and $${z_2}$$ are two non-zero complex number such that $$\left| {{{{z_1}} \over {{z_2}}}} \right|$$ = 2 and $$\arg \left( {{z_1}{z_2}} \right) = {{3\pi } \over 2}$$ , then $${{\overline {{z_1}} } \over {{z_2}}}$$ is equal to

A
2i

B
-2

C
-2i

D
2

Solution

$$Let\quad { z }_{ 1 }=2r{ e }^{ i{ \theta }_{ 1 } },{ z }_{ 2 }=r{ e }^{ i{ \theta }_{ 2 } }\\ arg\left( { z }_{ 1 }{ z }_{ 2 } \right) ={ \theta }_{ 1 }+{ \theta }_{ 2 }=\cfrac { 3\pi }{ 2 } \\ \bar { { z }_{ 1 } } =2r{ e }^{ -i{ \theta }_{ 1 } }\\ \cfrac { \bar { { z }_{ 1 } } }{ { z }_{ 2 } } =\cfrac { 2r{ e }^{ -i{ \theta }_{ 1 } } }{ r{ e }^{ i{ \theta }_{ 2 } } } =2r{ e }^{ -i({ \theta }_{ 1 }+{ \theta }_{ 2 }) }=2r{ e }^{ -i\cfrac { 3\pi }{ 2 } }=2i$$
Question 9
1 / -0

When will the quadratic equation $$ax^2+bx+c=0$$ have Real Roots?

A
$$b^2- 4ac \ge 0$$

B
$$b^2-4ac \le 0$$

C
$$b^2-4ac < 0$$

D
None of the above.

Solution

We learnt that the quadratic equation $$ax^2+bx+c=0$$ will have Real Roots when $$b^2-4ac0$$ Option $$a$$ is correct.
Question 10
1 / -0

If the value of '$$b^2-4ac$$'is equal to zero, the quadratic equation $$ax^2+bx+c=0$$ will have

A
Two real roots which are equal

B
Two Distinct Real Roots.

C
No Real Roots.

D
No Roots or Solutions.

Solution

$$ax^{2}+bx+c=0$$
For roots, $$D \equiv b^{2}-4ac$$
and given that $$b^{2}-4ac=0$$
$$\Rightarrow D=0$$
SO, equation will have only one root as,
$$x=\cfrac{-b\pm\sqrt{b^{2}-4ac}}{2(a)}$$
$$x=\cfrac{-b\pm 0}{2a}$$
$$x=\cfrac{-b}{2a}$$ Only one real root.

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Complex Numbers and Quadratic Equations Test 26

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Question 1

$$1$$

$$2$$

$$b+ia$$

$$a+ib$$

Solution

Question 2

$$\displaystyle\frac{1}{3+5\cos\theta}$$

$$\displaystyle\frac{1}{5-3\cos\theta}$$

$$\displaystyle\frac{1}{3-5\cos\theta}$$

$$\displaystyle\frac{1}{5+3\cos\theta}$$

Solution

Question 3

have a maximum

have a minimum

tangent to the $$x$$-axis

be tangent to the $$y$$-axis

lie in one quadrant only

Solution

Question 4

24

48

96

120

Solution

Question 5

$$1$$

$$-1$$

$$2$$

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

Solution

Question 6

$$\dfrac{3 \pi}{8}$$

$$\dfrac{\pi}{8}$$

$$\dfrac{5 \pi}{8}$$

$$\dfrac{-7 \pi}{8}$$

Solution

Question 7

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

$$1$$

$$\sqrt { 2 }$$

$$2$$

Solution

Question 8

2i

-2

-2i

2

Solution

Question 9

$$b^2- 4ac \ge 0$$

$$b^2-4ac \le 0$$

$$b^2-4ac < 0$$

None of the above.

Solution

Question 10

Two real roots which are equal

Two Distinct Real Roots.

No Real Roots.

No Roots or Solutions.

Solution

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