Complex Numbers and Quadratic Equations Test 27

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Complex Numbers and Quadratic Equations Test 27

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Weekly Quiz Competition

Question 1
1 / -0

If z satisfies $$\left| {z - 1} \right| < \left| {z + 3} \right|$$ then $$w = 2z + 3 - i$$ , ( where $$w = 2z + 3 - i$$ ) satisfies:

A
$$\left| {w - 5 - i} \right| < \left| {w + 3i} \right|$$

B
$$\left| {w - 5} \right| < \left| {w + 3} \right|$$

C
$$\left( {iw} \right) > 1$$

D
$$\left| {\arg \left( {w - 1} \right)} \right| < {\pi \over 2}$$

Solution
Question 2
1 / -0

Real part of $$\dfrac{(1 + i)^2}{3 - i} =$$

A
$$-1/5$$

B
$$1/5$$

C
$$1/10$$

D
$$-1/10$$

Solution

Given $$\dfrac{(1 + i)^2}{3 - i}$$

$$=\dfrac{2i}{3 - i}$$ [Since $$(i+1)^2=2i$$]

$$=\dfrac{2i(3+i)}{3^2 - i^2}$$ [After rationalizing the denominator]

$$=\dfrac{2i(3+i)}{10}$$

$$=\dfrac{(3i-1)}{5}$$.

Now real part of the given complex number is $$-\dfrac{1}{5}$$.
Question 3
1 / -0

If $$\dfrac{2z_1}{3z_2}$$ is a purely imaginary number,then $$\left|\dfrac{z_1-z_2}{z_1+z_2}\right|=$$

A
$$3/2$$

B
$$1$$

C
$$2/3$$

D
$$4/9$$

Solution

$$\dfrac{2z_1}{3z_2}=k'i$$
$$\dfrac{z_1}{z_2}=\dfrac{3k'i}{2}$$
$$\dfrac{z_1}{z_2}=ki$$ where $$\dfrac{3k'}{2}=k$$
$$z_1=kiz_2$$
Thus $$\left|\dfrac{z_1-z_2}{z_1+z_2}\right|=\left|\dfrac{kiz_2-z_2}{kiz_2+z_2}\right|$$

$$=\left|\dfrac{z_2(ki-1)}{z_2(ki+1)}\right|$$

$$=\dfrac{|ki-1|}{|ki+1|}$$

$$=\dfrac{\sqrt{k^2+1^2}}{\sqrt{k^2+1^2}}=1$$
Question 4
1 / -0

Find the real number $$x$$ if $$(x-2i)(1+i)$$ is purely imaginary.

A
$$2$$

B
$$-2$$

C
$$4$$

D
$$-4$$

Solution

$$(x-2i)(1+i)$$
$$ = x+ix -2i-2i^2$$
$$= (x+2) -i(x-2)$$ is purely imaginary if (x+2)=0
$$\therefore x=-2$$
Question 5
1 / -0

If the equation $${ x }^{ 2 }+nx+n=0,n\epsilon I$$, has integral roots then $${ n }^{ 2 }-4n$$ can assume

A
no integral value

B
one integral value

C
two integral value

D
three integral value

Solution

$${ x }^{ 2 }+nx+n=0,n\epsilon 1$$ Discriminant $$={ n }^{ 2 }-4n$$ For integral roots, $${ n }^{ 2 }-4n$$ must be a perfect square.This happens when $${ n }^{ 2 }-4n={ p }^{ 2 }$$ for some $$p\epsilon I$$ here if $$p=0$$ then $$(n-4)n=0\Rightarrow n=4$$ or $$n=0$$ ($$2$$ integral values)
$${ n }^{ 2 }-4n-p^{ 2 }=0$$ to have integral solution$$\Rightarrow $$Discriminant $$={ (-4) }^{ 2 }+4{ p }^{ 2 }$$ must be a perfect square$$=4({ p }^{ 2 }+1)$$ must be a perfect square, happens only when $$p=0$$
Question 6
1 / -0

If, for quadratic equation $$ax^2+bx +c=0, b^2-4ac \nleq 0$$ and a, b, c are all positive, then

A
Both the roots of the equation are positive

B
One root is positive; another is negative

C
Both the roots are negative

D
One root is negetive; another can be positive or negative

Solution

$$ax^{2}+bx+c=0$$
and $$ b^{2}-4ac\nleq 0$$
$$\Rightarrow b^{2}-4ac>0$$
Roots of the Equation, $$x=\cfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$$
as $$a, b, c$$ are all positive and we know $$\sqrt{b^{2}-4ac}$$ is positive
$$\Rightarrow $$
one root is negative. and another can be positive or negative.
Question 7
1 / -0

The value of $$\dfrac{1}{i} + \dfrac{1}{{{i^2}}} + \dfrac{1}{{{i^3}}} + ... + \dfrac{1}{{i^{102}}}$$ is equal to

A
$$ - 1 - i$$

B
$$ - 1 + i$$

C
$$ 1 - i$$

D
$$1 + i$$

Solution

Let $$t=\cfrac { 1 }{ i } +\cfrac { 1 }{ { i }^{ 2 } } +.....+\cfrac { 1 }{ { i }^{ 102 } } $$
$${ i }^{ 102 }t={ i }^{ 101 }+{ i }^{ 100 }+....+{ i }^{ 0 }$$
Using sum of GP, $$a={ i }^{ 0 }=1$$
$$r=i$$
$$n={ 102 }$$
$${ i }^{ 102 }t=\cfrac { 1\left( { i }^{ 102 }-1 \right) }{ i-1 } =\cfrac { { i }^{ 2 }-1 }{ i-1 } =\cfrac { -2\left( i+1 \right) }{ \left( i-1 \right) \left( i+1 \right) } $$
$${ i }^{ 102 }t=i+1\Rightarrow { i }^{ 2 }t=i+1$$
$$t=-i-1$$
Question 8
1 / -0

If $$i^2= -1$$, then $$1+ i^2+ i^4 +i^6+i^8 +.............to ( 2n +1)$$ terms is equal to

A
$$0$$

B
$$1$$

C
$$3i$$

D
$$4i$$

Solution

$$ 1+ i^2 + i^4 + ......(2n+1)terms$$
First term $$ a = i^2$$
Common ratio $$ r = i^2$$
$$ 1 + \cfrac{a(r^n-1)}{r-1}$$
$$ = 1 - \cfrac{i^2(i^{2(2n)} - 1)}{i-1}$$
$$ = 1 - \cfrac{i^2(i^4-1)}{i-1}$$ Since $$ i^4 = 1$$
$$ =1$$
Question 9
1 / -0

$$i \, \log \left(\dfrac{x - i}{x + i}\right)$$ is equal to

A
$$2i\log (x-i)-i\log (x^2+1)$$

B
$$2i\log (x-i)+i\log (x^2+1)$$

C
$$2i\log (x+i)-3i\log (x^2+1)$$

D
$$2i\log (x-i)-i\log (x^2+i)$$

Solution

$$i\log \left ( \cfrac{x-i}{x+i} \right )$$

$$=i\log \left ( \cfrac{x-i}{x+i}\times \cfrac{x-i}{x-i}\right )$$

$$=i\log \left ( \cfrac{(x-i)^{2}}{x^{2}-i^{2}} \right )$$

$$=i\log \left ( \cfrac{(x-i)^{2}}{x^{2}+1} \right )$$

$$=i\log (x-i)^{2}-i\log \left ( x^{2}+1 \right )$$

$$=2i\log (x-i)-i\log \left ( x^{2}+1 \right )$$
Question 10
1 / -0

The probability of choosing randomly a number c from the set $$\{1, 2, 3, ..........9\} $$ such that the quadratic equation $$x^2+ 4x +c=0$$ has real roots is:

A
$$\dfrac{1}{9}$$

B
$$\dfrac{2}{9}$$

C
$$\dfrac{3}{9}$$

D
$$\dfrac{4}{9}$$

Solution

$$x^2+4x+c=0$$
For roots to be real
$$0\ge 0$$
$$16-4c\ge c$$
$$4\ge c$$
$$\therefore \{1,2,3,4\}$$
$$\therefore$$ Probability $$=\cfrac{4}{9}$$

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Complex Numbers and Quadratic Equations Test 27

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Complex Numbers and Quadratic Equations Test 27

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Question 1

$$\left| {w - 5 - i} \right| < \left| {w + 3i} \right|$$

$$\left| {w - 5} \right| < \left| {w + 3} \right|$$

$$\left( {iw} \right) > 1$$

$$\left| {\arg \left( {w - 1} \right)} \right| < {\pi \over 2}$$

Solution

Question 2

$$-1/5$$

$$1/5$$

$$1/10$$

$$-1/10$$

Solution

Question 3

$$3/2$$

$$1$$

$$2/3$$

$$4/9$$

Solution

Question 4

$$2$$

$$-2$$

$$4$$

$$-4$$

Solution

Question 5

no integral value

one integral value

two integral value

three integral value

Solution

Question 6

Both the roots of the equation are positive

One root is positive; another is negative

Both the roots are negative

One root is negetive; another can be positive or negative

Solution

Question 7

$$ - 1 - i$$

$$ - 1 + i$$

$$ 1 - i$$

$$1 + i$$

Solution

Question 8

$$0$$

$$1$$

$$3i$$

$$4i$$

Solution

Question 9

$$2i\log (x-i)-i\log (x^2+1)$$

$$2i\log (x-i)+i\log (x^2+1)$$

$$2i\log (x+i)-3i\log (x^2+1)$$

$$2i\log (x-i)-i\log (x^2+i)$$

Solution

Question 10

$$\dfrac{1}{9}$$

$$\dfrac{2}{9}$$

$$\dfrac{3}{9}$$

$$\dfrac{4}{9}$$

Solution

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