Complex Numbers and Quadratic Equations Test 29

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Complex Numbers and Quadratic Equations Test 29

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Weekly Quiz Competition

Question 1
1 / -0

If $$|Z|=2,|z_{2}|=3,|z_{3}=4|$$ and $$|z_{1}+z_{2}+z_{3}|=5$$ then $$|4z_{2}z_{3}+9z_{3}z_{1}+16z_{1}z_{2}|=$$

A
$$20$$

B
$$24$$

C
$$48$$

D
$$120$$

Solution

Solution:- (D) 120
$$\left| {z}_{1} \right| = 2, \left| {z}_{2} \right| = 3, \left| {z}_{3} \right| = 4, \left| {z}_{1} + {z}_{2} + {z}_{3} \right| = 5$$
$$\left| 4 {z}_{2} {z}_{3} + 9 {z}_{3} {z}_{1} + 16 {z}_{1} {z}_{2} \right| = \cfrac{\left| {z}_{1} \right| \left| {z}_{2} \right| \left| {z}_{3} \right|}{\left| {z}_{1} \right| \left| {z}_{2} \right| \left| {z}_{3} \right|} \left| 4 {z}_{2} {z}_{3} + 9 {z}_{3} {z}_{1} + 16 {z}_{1} {z}_{2} \right|$$
$$= \left| {z}_{1} \right| \left| {z}_{2} \right| \left| {z}_{3} \right| \left| \cfrac{4}{{z}_{1}} + \cfrac{9}{{z}_{2}} + \cfrac{16}{{z}_{3}} \right|$$
$$= \left( 2 \times 3 \times 4 \right) \left| \cfrac{4 \bar{{z}_{1}}}{{\left| {z}_{1} \right|}^{2}} + \cfrac{9 \bar{{z}_{2}}}{{\left| {z}_{2} \right|}^{2}} + \cfrac{16 \bar{{z}_{3}}}{{\left| {z}_{3} \right|}^{2}}\right|$$
$$= 24 \left| \cfrac{4 \bar{{z}_{1}}}{{2}^{2}} + \cfrac{9 \bar{{z}_{2}}}{{3}^{2}} + \cfrac{16 \bar{{z}_{3}}}{{4}^{2}} \right|$$
$$= 24 \left| \cfrac{4 \bar{{z}_{1}}}{4} + \cfrac{9 \bar{{z}_{2}}}{9} + \cfrac{16 \bar{{z}_{3}}}{16} \right|$$
$$= 24 \left| \bar{{z}_{1}} + \bar{{z}_{2}} + \bar{{z}_{3}} \right|$$
$$= 24 \left| \overline{{z}_{1} + {z}_{2} + {z}_{3}} \right|$$
$$= 24 \times 5$$
$$= 120$$
Hence, $$\left| 4 {z}_{2} {z}_{3} + 9 {z}_{3} {z}_{1} + 16 {z}_{1} {z}_{2} \right| = 120$$.
Question 2
1 / -0

If $$\dfrac{x - 3}{3 + i} + \dfrac{y - 3}{3 - i} = i $$ where $$x , y \in R$$ then

A
$$x = 2$$ & $$y = -8$$

B
$$x = -2$$ & $$y = 8$$

C
$$x = -2$$ & $$y = -6$$

D
$$x = 2$$ & $$y = 8$$

Solution

$$\dfrac{x-3}{3+i}+\dfrac{y-3}{3-i}=i$$
$$\dfrac{(x-3)(3-i)}{10}+\dfrac{(y-3)(3+i)}{10}=i$$
$$3(x-3)+3(y-3)-i((x-3+(3-y))=10i$$
$$3(x+y-6)-i(x-y)=10i$$
compare real and imaginary point
$$x+y=6$$
$$y-x=10$$
$$\Rightarrow y=8, x=-2$$
Question 3
1 / -0

Consider two quadratic expressions $$f(x)=ax^2+bx+c$$ and $$g(x)=ax^2+px+q$$, $$(b\neq q)$$ such that their discriminants are equal. If $$f(x)=g(x)$$ has a root $$x=\alpha$$, then?

A
$$\alpha$$ will be A.M. of the roots of $$f(x)=0$$ and $$g(x)=0$$

B
$$\alpha$$ will be A.M. of the roots of $$f(x)=0$$

C
$$\alpha$$ will be A.M. of the roots of $$g(x)=0$$

D
None of these

Solution

Discriminants equal means
$$b^2 - 4ac = p^2-4aq$$
Also given $$x=\alpha$$ for $$f(x) = g(x)$$
$$b\alpha + c = p\alpha + q$$
$$\alpha (b-p) = q-c$$
$$\alpha = \cfrac{q-c}{b-p}$$
As we have,
$$b^2 - p^2 = 4a(c-q)$$
$$b+p = -4a\alpha$$
$$\alpha = -\cfrac{b+p}{4a}$$
Let the roots of $$f(x)$$ be $$x_1,x_2$$ and the roots of $$g(x)$$ be $$x_3,x_4$$
$$x_1 + x_2 = -\cfrac{b}{a}$$
$$x_3 + x_4 = -\cfrac{p}{a}$$
AM of $$x_1,x_2,x_3,x_4$$ is $$\cfrac{x_1+x_2+x_3+x_4}{4}$$
$$\cfrac{x_1+x_2+x_3+x_4}{4} = -\cfrac{b+p}{4a} = \alpha$$
Question 4
1 / -0

The complex number $$\dfrac{1 + 2i}{1 - i}$$ lies in which quadrant of the complex plane.

A
First

B
Second

C
Third

D
Fourth

Solution

$$\dfrac{1+2i}{1-i}$$
$$\Rightarrow \dfrac{1+2i}{1-i}\times \dfrac{1+i}{1+i}$$
$$=\dfrac{1+i+2i-2i^2}{1-i^2}=\dfrac{1+3i-2}{2}$$
$$=\dfrac{-1+3i}{2}$$
$$\therefore$$ It lies in $$2^{nd}$$ Quadrant.
Question 5
1 / -0

Given $$\left| z \right| =4$$ and $$Argz=\dfrac{5z}{6}$$, then $$z$$ is

A
$$2\sqrt{3}+2i$$

B
$$2\sqrt{3}-2i$$

C
$$-2\sqrt{3}+2i$$

D
$$-\sqrt{3}+i$$

Solution

Given that

$$\Rightarrow |z|=4$$ and $$Arg \space z=\dfrac{5\pi}{6}$$

$$\Rightarrow $$Let $$z=x+iy$$

$$\Rightarrow |z|=4$$

$$\Rightarrow \sqrt{x^2+y^2}=4$$

$$\Rightarrow {x^2+y^2}=16$$ ...$$(1)$$

$$\Rightarrow Arg \space z=\dfrac{5\pi}{6}$$

$$\Rightarrow \tan ^{-1}(\dfrac{y}{x})=\dfrac{5\pi}{6}$$

$$\Rightarrow \dfrac{y}{x}=\tan (\dfrac{5\pi}{6})$$

$$\Rightarrow \dfrac{y}{x}=-\dfrac{1}{\sqrt 3}$$

$$\Rightarrow x^2=3y^2$$

Substituting this in $$(1)$$ we get,

$$\Rightarrow x^2+\dfrac{x^2}{3}=16$$

$$\Rightarrow \dfrac{4x^2}{3}=16$$

$$\Rightarrow x^2=12$$

$$\Rightarrow x=\pm 2\sqrt 3$$

$$\Rightarrow y=\mp 2$$

$$\Rightarrow \theta$$ lies in $$2^{nd} Quadrant$$

Hence, $$\Rightarrow z=-2\sqrt 3+2i$$
Question 6
1 / -0

The locus of $$z$$ such that $$\left| {\dfrac{{z + i}}{{z - 1}}} \right| = 2$$

A
straight line

B
circle with radius $$2$$

C
circle with radius $$\frac{{2\sqrt 2 }}{3}$$

D
none of these

Solution

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

Put, $$z = x + iy$$

$$\therefore\,\left|\dfrac{x + iy+i}{x + iy-1}\right|=2$$

$$\Rightarrow\,\left|\dfrac{x + i\left(y+1\right)}{\left(x-1\right)+iy}\right|=2$$

$$\Rightarrow\,\left|x + i\left(y+1\right)\right|=2\left|\left(x-1\right)+iy\right|$$

$$\Rightarrow\,\sqrt{{x}^{2}+{\left(y+1\right)}^{2}}=2\sqrt{{\left(x-1\right)}^{2}+{y}^{2}}$$

Squaring both sides, we get
$$\Rightarrow\,{x}^{2}+{\left(y+1\right)}^{2}=4{\left(x-1\right)}^{2}+4{y}^{2}$$

$$\Rightarrow\,{x}^{2}-{\left(2x-2\right)}^{2}={\left(2y\right)}^{2}-{\left(y+1\right)}^{2}$$

$$\Rightarrow\,\left(x-2x+2\right)\left(x+2x-2\right)=\left(2y-y-1\right)\left(2y+y+1\right)$$

$$\Rightarrow\,\left(-x+2\right)\left(3x-2\right)=\left(y-1\right)\left(3y+1\right)$$
$$\Rightarrow\,-3{x}^{2}+2x+6x-4=3{y}^{2}+y-3y-1$$

$$\Rightarrow\,-3{x}^{2}+8x-4=3{y}^{2}-2y-1$$

$$\Rightarrow\,-3{x}^{2}+8x-4-3{y}^{2}+2y+1=0$$

$$\Rightarrow\,3{x}^{2}+3{y}^{2}-8x-2y+3=0$$

$$\Rightarrow\,{x}^{2}+{y}^{2}-\dfrac{8}{3}x-\dfrac{2}{3}y+1=0$$ is of the form $${x}^{2}+{y}^{2}+2gx+2fy+c=0$$

where $$g=\dfrac{4}{3},\,f=\dfrac{1}{3}$$ and $$c=1$$

Radius$$=r=\sqrt{{\left(\dfrac{4}{3}\right)}^{2}+{\left(\dfrac{1}{3}\right)}^{2}+1}=\sqrt{\dfrac{16}{9}+\dfrac{1}{9}-1}=\sqrt{\dfrac{17-9}{9}}=\dfrac{\sqrt{8}}{3}=\dfrac{2\sqrt{2}}{3}$$

The locus of $$z$$ is a circle with radius $$\dfrac{2\sqrt{2}}{3}$$
Question 7
1 / -0

$$\left(\dfrac{1 + i}{1 - i}\right)^4 + \left(\dfrac{1 - i}{1 + i}\right)^4 = $$

A
$$0$$

B
$$1$$

C
$$2$$

D
$$4$$

Solution

$$\left( \dfrac{1+i}{1-i} \right)^{4}+ \left( \dfrac{1-i}{1+i} \right)^{4}$$
$$= \left( \dfrac{(1+i)^{2}}{1-i^{2}} \right)^{4}+ \left( \dfrac{(1-i)^{2}}{1-i^{2}} \right)^{4}$$
$$=\dfrac{1}{2^{4}} (1+i^{2} +2 i)^{4}+ \dfrac{1}{2^{4}} (1+i^{2}-2i)^{4}$$
$$=\dfrac{1}{2^{4}} (1-1+ 2i)^{4}+ \dfrac{1}{2^{4}} (1-1-2i)^{4}$$
$$=\dfrac{2^{4}}{2^{4}} (i^{4})+ \dfrac{2^{4}}{2^{4}} (i)^{4}$$
$$= 1+1 =2$$
Question 8
1 / -0

In the graph of the parametric equations $$\begin{cases} x={ t }^{ 2 }+t \\ y={ t }^{ 2 }-t \end{cases}$$ for integral values of t.

A
$$x\ge 0$$

B
$$x\ge -\cfrac{1}{4}$$

C
$$x$$ is any real number

D
$$x\ge -1$$

E
$$x\le 1$$

Solution

$$x=t^2+t = t(t+1)$$
$$x<0\, \forall \, t\in (-1,0)$$

$$\therefore x<0\Rightarrow t=$$ not an integer.

$$\therefore x\geqslant 0$$ for all integral values of $$t$$
A option is cvorrect.
Question 9
1 / -0

If $$\left| {z - 1} \right| = 2$$, then the value of $$z\overline z - z - \overline z $$ is equal to:

A
$$-3$$

B
$$4$$

C
$$3$$

D
$$-4$$

Solution

Given $$|z-1|=2$$
we know that $$|z|^{2}=z.\bar{z}$$
so taking square root on both side.
$$|z-1|^{2}=4$$
$$(z-1)(\bar{z-1})=4$$
$$(z-1)(\bar{z}-1)=4$$
$$z.\bar{z}-\bar{z}-z+1=4$$
$$z.\bar{z}-z-\bar{z}=3$$
so answer is 3.
Question 10
1 / -0

If $$a < b < c < d$$, then the roots of the equation $$(x-a)(x-c)+2(x-b)(x-d)=0$$ are?

A
Real and distinct

B
Real and equal

C
Imaginary

D
None of these

Solution

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Complex Numbers and Quadratic Equations Test 29

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Question 1

$$20$$

$$24$$

$$48$$

$$120$$

Solution

Question 2

$$x = 2$$ & $$y = -8$$

$$x = -2$$ & $$y = 8$$

$$x = -2$$ & $$y = -6$$

$$x = 2$$ & $$y = 8$$

Solution

Question 3

$$\alpha$$ will be A.M. of the roots of $$f(x)=0$$ and $$g(x)=0$$

$$\alpha$$ will be A.M. of the roots of $$f(x)=0$$

$$\alpha$$ will be A.M. of the roots of $$g(x)=0$$

None of these

Solution

Question 4

First

Second

Third

Fourth

Solution

Question 5

$$2\sqrt{3}+2i$$

$$2\sqrt{3}-2i$$

$$-2\sqrt{3}+2i$$

$$-\sqrt{3}+i$$

Solution

Question 6

straight line

circle with radius $$2$$

circle with radius $$\frac{{2\sqrt 2 }}{3}$$

none of these

Solution

Question 7

$$0$$

$$1$$

$$2$$

$$4$$

Solution

Question 8

$$x\ge 0$$

$$x\ge -\cfrac{1}{4}$$

$$x$$ is any real number

$$x\ge -1$$

$$x\le 1$$

Solution

Question 9

$$-3$$

$$4$$

$$3$$

$$-4$$

Solution

Question 10

Real and distinct

Real and equal

Imaginary

None of these

Solution

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