Complex Numbers and Quadratic Equations Test 34

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Complex Numbers and Quadratic Equations Test 34

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Weekly Quiz Competition

Question 1
1 / -0

The modulus of $$\dfrac { \left( 3+2i \right) ^{ 2 } }{ \left( 4-3i \right) } $$ is:

A
$$\frac { 13 }{ 5 } $$

B
$$\frac { 11 }{ 5 } $$

C
$$\frac { 9 }{ 5 } $$

D
$$\frac { 7 }{ 5 } $$

Solution
Question 2
1 / -0

Number of complex numbers $$z$$ such that $$|z|=1$$ and $$\left|\dfrac {z}{z}+\dfrac {\bar {z}}{z}\right|=1$$ is

A
$$4$$

B
$$1$$

C
$$8$$

D
$$more\ then\ 8$$

Solution

Since $$z$$ is equidistant from $$1,-1,i$$, we can say that $$z$$is center of circle with $$1,-1,i$$ lying on it's circumference and so there is one one circle possible passing through $$3$$ points. So, the answer is $$1$$.
Question 3
1 / -0

If arg $$\left( z \right) < 0,$$ then arg $$\left( { - z} \right)-arg(z)$$

A
$$\pi $$

B
$$ - \pi $$

C
$$ - \dfrac{\pi }{2}$$

D
$$\dfrac{\pi }{2}$$

Solution
Question 4
1 / -0

Purely imaginary then find the sum of statement i $$a,b$$

A
$$\dfrac {5\pi}{6}$$

B
$$\pi$$

C
$$\dfrac {3\pi}{4}$$

D
$$\dfrac {2\pi}{3}$$
Question 5
1 / -0

If $$\alpha$$ and $$\beta$$ are the roots of $${ 4x }^{ 2 }-16x+c=0,$$ c>0 such that $$1<\alpha <2<\beta <3$$, then the no.of integer values of c is

A
$$17$$

B
$$14$$

C
$$18$$

D
$$15$$

Solution

Given, $$\alpha$$ and $$\beta$$ are roots of $$4x^2-16x+x=0$$
where $$a=4$$
$$b=-16$$
$$x > 0$$
now, $$\alpha\beta =\dfrac{a}{a}$$
$$\Rightarrow \alpha\beta =\dfrac{c}{4}$$
$$\Rightarrow \beta =\dfrac{c}{4\alpha}$$ ………(i)
we have $$2 < \beta < 3$$
$$\Rightarrow 2 < \dfrac{c}{4\alpha} < 3$$ [from $$(1)$$]
$$\Rightarrow 8 < \dfrac{c}{\alpha} < 12$$
$$\Rightarrow 8\alpha < c < 12\alpha$$ ………(ii)
and $$1 < \alpha < 2$$
$$\Rightarrow \alpha > 1\Rightarrow 8\alpha > 8$$ ……….(iii)
and $$\alpha < 2$$
$$\Rightarrow 12\alpha < 24$$ ………..(iv)
Substituting (iii) and (iv) in (ii) we get
$$8 < 8\alpha < c < 12\alpha < 24$$ $$[\alpha =9, 10, 11, 12, 13, 14, 15, 16, 13, 18, 19, 20, 21, 22, 23]$$
$$\Rightarrow 8 < c < 24$$
Therefore number of integer values of c is $$15$$.
Question 6
1 / -0

$$\sqrt { \left( \log _ { 3 } \tan x \right) }$$ is real for:

A
$$n \pi + \pi / 4 \leq x < n \pi + \pi / 2$$

B
$$n \pi < x < n \pi + \pi / 2$$

C
$$n \pi \pm \pi / 4 \leq x < n \pi \pm \pi / 2$$

D
None of these.
Question 7
1 / -0

Arg $$\left\{ {\sin \frac{{8\pi }}{5} + i\left( {1 + \cos \frac{{8\pi }}{5}} \right)} \right\}$$ is equal to

A
$$\frac{{3\pi }}{{10}}$$

B
$$\frac{{7\pi }}{{10}}$$

C
$$\frac{{4\pi }}{5}$$

D
$$\frac{{3\pi }}{5}$$

Solution
Question 8
1 / -0

Let 'z' be a complex number satisfying $$|z-2-i|\le 5,$$ Then |z-14-6i| lies in

A
{8,18}

B
{2,8}

C
{0,2}

D
{3,7}

Solution

Here, $$|z-2-i|\leq 5$$ ………..$$(1)$$
Now, $$|z-14-6i|=|z-2-i-12-5i|=|(z-2-i)-(12+5i)|$$
$$\leq |z-2-i|+|12+5i|$$
$$\leq 5+|12+5i|$$
$$=5+\sqrt{(12)^2+(5)^2}$$
$$=5+\sqrt{144+25}$$
$$=5+\sqrt{169}$$
$$=5+13$$
$$=18$$.
$$\therefore |z-14-6i|\leq 18$$
and $$|z-14-6i|=|z-2-i-12-5i|$$
$$=|-(12+5i)+(z-2-i)|$$
$$\geq ||-(12+5i)|-|z-2-i||$$
$$\geq |12+5i|-5$$ $$[\because |z-2-i|\leq 5$$ $$\Rightarrow -|z-2-i|\geq -5]$$
$$=13-5$$
$$=8$$
$$\therefore |z-14-6i|\geq 8$$
$$\therefore |z-14-6i|$$ lies in $$\{8, 18\}$$.
Question 9
1 / -0

IF $$z_1=1+i,z_2=1-i$$ find $$z_1z_2$$

A
$$z_1+z_2$$

B
$$z_1-z_2$$

C
$$z_1/z_2$$

D
None.

Solution

$$z_1=1+i$$

$$z_2=1-i$$

$$z_1z_2=1-(-1)=2$$

$$=1+1$$

$$=1-i+1+i$$

$$z_1z_2=z_1+z_2$$
Question 10
1 / -0

The value of the sum $$\sum _{ n=1 }^{ 13 }{ ({ i }^{ n }+{ i }^{ n+1 }) } $$ , where $$i=\sqrt { -1 } $$ , equals

A
$$i$$

B
$$i-1$$

C
$$-i$$

D
$$0$$

Solution

As we know that $$i+i^2+i^3+i^4=0$$
$$\displaystyle\sum ^{13}_{n=1} (i^n+i^{n+1})=\sum ^{13}_{n=1}(1+i)(i^n)=(1+i)\sum ^{13}_{n=1} i^{n}=(1+i)(0+0+i^{13})=(1+i)i=i+i^2=i-1$$