Complex Numbers and Quadratic Equations Test 28

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Complex Numbers and Quadratic Equations Test 28

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Weekly Quiz Competition

Question 1
1 / -0

If $$a+ ib= \sum_{k=1}^{101} i^k $$, then $$(a, b)$$ equals

A
$$(0, 1)$$

B
$$(1, 0)$$

C
$$(0,- 1)$$

D
$$(1, 1)$$

Solution

Given:-
$$a + ib = \Sigma^{101}_{k = 1}{{i}^{k}} \longrightarrow \left( 1 \right)$$

Solution:-

$$\Sigma^{101}_{k = 1}{{i}^{k}} = i + {i}^{2} + {i}^{3} + ............... + {i}^{101}$$

As te above series is in the form of G.P. with common ration $$(r)=i$$, first term $$\left( a \right) = i$$ and no. of terms $$\left( n \right) = 101$$

$$\therefore \; {S}_{101} = \cfrac{ i \left( 1 - {i}^{101} \right)}{\left( 1 - i \right)} \; \left\{ \because {S}_{n} = \cfrac{a \left( 1 - {r}^{n} \right)}{\left( 1 - r \right)}\; \text{ for } r < 1 \right\}$$

$$\Rightarrow \; {S}_{n} = \cfrac{i \left( 1 - {i}^{4 \times 25}.i \right)}{\left( 1 - i \right)}$$

$$\Rightarrow \; {S}_{n} = \cfrac{ i \left( 1 - i \right)}{\left( 1 - i \right)} \; \left\{ \because {i}^{4n} = 1 \right\}$$

$$\Rightarrow \; {S}_{n} = i$$

$$\therefore \; \Sigma^{101}_{k = 1}{{i}^{k}} = i \longrightarrow \left( 2 \right)$$

Now, from $${eq}^{n} \left( 1 \right) \& \left( 2 \right)$$, we have,

$$a + ib = i$$

$$\Rightarrow \; a + ib = 0 + i.1$$

On comparing real and imaginary parts, we have

$$a = 0 \; \& \; b = 1$$

Hence, $$\left( a, b \right) = \left( 0, 1 \right)$$.
Question 2
1 / -0

If $$z = -3- i,$$ find $$|z|$$.

A
$$\sqrt{10}$$

B
$$\sqrt{9}$$

C
$$\sqrt{8}$$

D
$$\sqrt{7}$$

Solution

$$Re(z)=-3\\Im(z)=-1\\ \bar{z}=-3+i\\|z|=3^2+1^2=10$$
Question 3
1 / -0

A particle starts from a point $$z_0= I + i$$, where $$i
=\sqrt{-1}$$ It moves horizontally away from origin by $$2$$ units and then
vertically away from origin by $$3$$ units to reach a point$$ z_1$$. From $$z_1$$
particle moves $$\sqrt{5}$$ units in the direction of $$2\hat i + \hat j$$ and
then it moves through an angle of $$\cos e{c^{ - 1}}\sqrt 2 $$ in anticlockwise
direction of a circle with centre at origin to reach a point $$z_2$$ . The arg $$z_2$$ is given by

A
$${\sec ^{ - 1}}2$$

B
$${\cot ^{ - 1}}0$$

C
$${\sin ^{ - 1}}\left( {\dfrac{{\sqrt 3 - 1}}{{2\sqrt 2 }}} \right)$$

D
$${\cos ^{ - 1}}\left( {\dfrac{{ - 1}}{2}} \right)$$

Solution
Question 4
1 / -0

Find the nature of the roots for the equation $$n^{2}+2\sqrt{2} n+1=0$$

A
Rational & District

B
Irrational & District

C
Real & District

D
Imaginary

Solution

Solution :- Given equation is
$$ n^{2}+2\sqrt{2n}+1 = 0 $$
By quadratic root formula
$$ \dfrac{-b\pm \sqrt{b^{2}-4ac}}{2a} $$
here $$ a = 1, b = 2\sqrt{2}, c = 1 $$
$$ x = \dfrac{-2\sqrt{2}\pm \sqrt{(2\sqrt{2})^{2}4.1.1}}{2.1} $$
$$ \Rightarrow x = \dfrac{-2\sqrt{2}\pm \sqrt{8-4}}{2} = \dfrac{-2\sqrt{2}\pm \sqrt{4}}{2} $$
$$ \Rightarrow x = \dfrac{-2\sqrt{2}\pm 2}{2} = -\sqrt{2}\pm 1 $$
So roots are $$ (-\sqrt{2}+1), (-\sqrt{2}-1) $$
$$ \therefore $$ So roots are irrational and district
So correct answer is B.
Question 5
1 / -0

For the equation $$|x|^2+|x|-6=0$$, the roots are

A
real and equal

B
real and sum $$0$$

C
real with sum $$1$$

D
real with product $$0$$

Solution

$$\Rightarrow$$ The given equation is $$|x|^2+|x|-6=0$$ ---- ( 1 )
$$\Rightarrow$$ When $$x\ge 0$$, then $$|x|=x$$, so equation ( 1 ) becomes,
$$x^2+x-6=0$$
$$\Rightarrow$$ $$x^2+3x-2x-6=0$$
$$\Rightarrow$$ $$x(x+3)-2(x+3)=0$$
$$\Rightarrow$$ $$(x+3)(x-2)=0$$
$$\Rightarrow$$ $$x=-3,2$$
When $$x<0$$, then $$|x|=-x$$, so equation ( 1 ) becomes,
$$x^2-x-6=0$$
$$\Rightarrow$$ $$x^2-3x+2x-6=0$$
$$\Rightarrow$$ $$x(x-3)+2(x-3)=0$$
$$\Rightarrow$$ $$(x-3)(x+2)=0$$
$$\Rightarrow$$ $$x=3,-2$$
$$\Rightarrow$$ Sum of the roots $$=-3+2+3-2=0$$
$$\therefore$$ Here, roots are real and equal.
Question 6
1 / -0

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

A
$$\dfrac{2\sin \theta}{(1-\cos\theta)^2+4\sin^2 \theta}$$

B
$$\dfrac{1+\cos \theta}{(1-\cos\theta)^2+4\sin^2 \theta}$$

C
$$\cfrac { \left( 1-\cos { \theta } \right) -2i\sin { \theta } }{ { \left( 1-\cos { \theta } \right) }^{ 2 }+4{ i }^{ 2 }\sin ^{ 2 }{ \theta } }$$

D
$$\dfrac{1-\cos \theta}{(1-\cos\theta)^2+4\sin^2 \theta}$$

Solution

$$\Rightarrow \cfrac { 1 }{ \left( 1-\cos { \theta } \right) +2i\sin { \theta } } =\cfrac { \left( 1-\cos { \theta } \right) -2i\sin { \theta } }{ { \left( 1-\cos { \theta } \right) }^{ 2 }-4{ i }^{ 2 }\sin ^{ 2 }{ \theta } } $$
$$=\cfrac { \left( 1-\cos { \theta } \right) -2i\sin { \theta } }{ { \left( 1-\cos { \theta } \right) }^{ 2 }+4{ i }^{ 2 }\sin ^{ 2 }{ \theta } }$$
Question 7
1 / -0

$$\dfrac{{{{\left( {1 + i} \right)}^3}}}{{2 + i}}$$ is equal to

A
$$\dfrac{2}{5} - \dfrac{6}{5}i$$

B
$$0$$

C
$$ - \dfrac{1}{5} + \dfrac{6}{5}i$$

D
$$ - \dfrac{2}{5} + \dfrac{6}{5}i$$

Solution

$$\Rightarrow \cfrac { { \left( 1+i \right) }^{ 3 } }{ 2+i } =\cfrac { { \left( 1+i \right) }^{ 2 }\left( 1+i \right) }{ 2+i } =\cfrac { \left( 1+{ i }^{ 2 }+2i \right) \left( 1+i \right) }{ \left( 2+i \right) } =\cfrac { 2i\left( 1+i \right) }{ 2+i } $$
$$\Rightarrow \cfrac { 2\left( i-1 \right) \left( 2-i \right) }{ \left( 2+i \right) \left( 2-i \right) } =\cfrac { 2\left( 2i-{ i }^{ 2 }-2+i \right) }{ 4-{ i }^{ 2 } } =\cfrac { -2 }{ 5 } +\cfrac { 6 }{ 5 } i$$
Question 8
1 / -0

The modulus of the complex number $$z=\dfrac{1-i}{3-4i}$$ is

A
$$\dfrac{5}{\sqrt{2}}$$

B
$$\dfrac{\sqrt{2}}{5}$$

C
$$\sqrt{\dfrac{2}{5}}$$

D
none of these

Solution

Now,
$$z=\dfrac{1-i}{3-4i}$$
or, $$z=\dfrac{(1-i)(3+4i)}{3^2+4^2}$$
or, $$z=\dfrac{7+i}{25}$$.
Then $$|z|=\sqrt{\left(\dfrac{7}{25}\right)^2+\left(\dfrac{1}{25}\right)^2}=\dfrac{\sqrt{50}}{25}=\dfrac{\sqrt{2}}{5}$$.
Question 9
1 / -0

If $$z=\dfrac{1+i}{\sqrt{2}}$$, then the value of $$z^{1929}$$ is

A
$$1+i$$

B
$$-1$$

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

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

Solution

Let $$z=\dfrac{1+i}{\sqrt{2}}$$

$$\Rightarrow |z|=\sqrt{\dfrac{1}{2}+\dfrac{1}{2}}=1$$

$$\Rightarrow Arg=tan^{-1}(1)=\dfrac{\pi}{4}$$

$$\therefore z^{1929}=(1\cdot e^{\dfrac{\pi i}{4}})^{1929}$$

$$=e^{1929 \times \dfrac{\pi i}{4}}$$

$$=e^{i\left(482\pi+\dfrac{\pi}{4}\right)}$$

$$=e^{i\dfrac{\pi}{4}}$$

$$=\dfrac{1}{\sqrt{2}}+i\dfrac{1}{\sqrt{2}}$$

$$\therefore z^{1929}=\dfrac{1+i}{\sqrt{2}}$$
Question 10
1 / -0

The equation $$4\sin^2x+4\sin x+a^2-3=0$$ possesses a solution if 'a' belongs to the interval.

A
$$(-1, 3)$$

B
$$(-3, 1)$$

C
$$[-2, 2]$$

D
$$R-(-2, 2)$$

Solution

We are given, the equation
$$4\sin^{2}x+4\sin x+a^{2}-3=0$$
Here, $$A=4, B=4$$ & $$C=a^{2}-3$$
now, Discriminant $$D=\sqrt{B^{2}-4 AC}$$
$$\therefore D^{2}=B^{2}-4AC; D\ge 0$$
here solution is possible so,
$$D^{2}\ge 0$$
$$\therefore B^{2}-4AC\ge 0$$
$$\therefore (4)^{2}-4(4)(a^{2}-3)\ge 0$$
$$\therefore 16-16(a^{3}-3)\ge 0$$
$$\therefore 16(1-a^{3}+3)\ge 0$$
$$\therefore 1-a^{2}+3\ge 0$$
$$\therefore a^{2}-4\le 0$$
$$\therefore (a-2)(a+2)\le 0$$
Now, From the value of a we get
the condition $$B^{2} \ge 4 AC$$
For that $$a\in [-2,2]$$ So that the condition will be true.

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Complex Numbers and Quadratic Equations Test 28

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Question 1

$$(0, 1)$$

$$(1, 0)$$

$$(0,- 1)$$

$$(1, 1)$$

Solution

Question 2

$$\sqrt{10}$$

$$\sqrt{9}$$

$$\sqrt{8}$$

$$\sqrt{7}$$

Solution

Question 3

$${\sec ^{ - 1}}2$$

$${\cot ^{ - 1}}0$$

$${\sin ^{ - 1}}\left( {\dfrac{{\sqrt 3 - 1}}{{2\sqrt 2 }}} \right)$$

$${\cos ^{ - 1}}\left( {\dfrac{{ - 1}}{2}} \right)$$

Solution

Question 4

Rational & District

Irrational & District

Real & District

Imaginary

Solution

Question 5

real and equal

real and sum $$0$$

real with sum $$1$$

real with product $$0$$

Solution

Question 6

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

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

$$\cfrac { \left( 1-\cos { \theta } \right) -2i\sin { \theta } }{ { \left( 1-\cos { \theta } \right) }^{ 2 }+4{ i }^{ 2 }\sin ^{ 2 }{ \theta } }$$

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

Solution

Question 7

$$\dfrac{2}{5} - \dfrac{6}{5}i$$

$$0$$

$$ - \dfrac{1}{5} + \dfrac{6}{5}i$$

$$ - \dfrac{2}{5} + \dfrac{6}{5}i$$

Solution

Question 8

$$\dfrac{5}{\sqrt{2}}$$

$$\dfrac{\sqrt{2}}{5}$$

$$\sqrt{\dfrac{2}{5}}$$

none of these

Solution

Question 9

$$1+i$$

$$-1$$

$$\dfrac{1+i}{2}$$

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

Solution

Question 10

$$(-1, 3)$$

$$(-3, 1)$$

$$[-2, 2]$$

$$R-(-2, 2)$$

Solution

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