Complex Numbers and Quadratic Equations Test 40

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Complex Numbers and Quadratic Equations Test 40

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Weekly Quiz Competition

Question 1
1 / -0

If $$a, b, c$$ are real distinct numbers satisfying the condition $$a + b + c = 0$$, then the roots of the quadratic equation $$3ax^2 + 5bx + 7c = 0$$ are

A
positive

B
negative

C
real and distinct

D
imaginary

Solution

Given equation is : $$3a{ x }^{ 2 }+5bx+7c=0$$
Also given is that $$a+b+c=0\dots(1)$$

Discriminant of the equation $$\Rightarrow D =b^2-4ac $$

$$= { \left( 5b \right) }^{ 2 }-4\left( 3a \right) \left( 7c \right)$$

$$ = 25{ b }^{ 2 }-84ac\dots(2)$$

Substituting the value of b from equation (1) in equation (2),we get

D $$=25{ \left( -a-c \right) }^{ 2 }-84ac$$

$$ =25{ \left( a+c \right) }^{ 2 }-84ac$$

$$= 25{ a }^{ 2 }+25{ c }^{ 2 }-34ac$$

$$= 4{ a }^{ 2 }+4{ c }^{ 2 }+8ac$$ + $$21{ a }^{ 2 }+{ 21c }^{ 2 }-42ac$$

$$= 4{ \left( a+c \right) }^{ 2 }+21{ \left( a-c \right) }^{ 2 }> 0$$

Since D>0,both the roots will be real and distinct.

Also,in the absence of information on the coefficients a,b and c in the equation we can't conclude whether both the roots will be positive or negative.

Hence,option (C) is the correct alternative.
Question 2
1 / -0

Find the modulus, argument and the principal argument of the complex numbers.
$$z=1+cos\frac {10\pi}{9}+i sin \left (\frac {10\pi}{9}\right )$$

A
Principal Arg $$z=-\frac {4\pi}{9}; |z|=2 cos \frac {4\pi}{9}; Arg z=2 k\pi -\frac {4\pi}{9} k\epsilon l$$

B
Principal Arg $$z=-\frac {10\pi}{9}; |z|=2 cos \frac {10\pi}{9}; Arg z=2 k\pi -\frac {10\pi}{9} k\epsilon l$$

C
Principal Arg $$z=-\frac {-10\pi}{9}; |z|=2 cos \frac {-10\pi}{9}; Arg z=2 k\pi -\frac {4\pi}{9} k\epsilon l$$

D
Principal Arg $$z=-\frac {-4\pi}{9}; |z|=2 cos \frac {-4\pi}{9}; Arg z=2 k\pi -\frac {4\pi}{9} k\epsilon l$$

Solution

Simplifying, we get
$$z=2\cos^{2}\left(\dfrac{5\pi}{9}\right)+2i\sin\left(\dfrac{5\pi}{9}\right)\cos \left(\dfrac{5\pi}{9}\right)$$

$$=2\cos \left(\dfrac{5\pi}{9}\right)[\cos \left(\dfrac{5\pi}{9}\right)+i\sin\left(\dfrac{5\pi}{9}\right)]$$

$$=-2\cos \left(\dfrac{4\pi}{9}\right)[-\cos \left(\dfrac{4\pi}{9}\right)+i\sin\left(\dfrac{4\pi}{9}\right)]$$

$$=2\cos \left(\dfrac{4\pi}{9}\right)[\cos \left(\dfrac{4\pi}{9}\right)-i\sin\left(\dfrac{4\pi}{9}\right)]$$
Hence
$$|z|=2\cos \left(\dfrac{4\pi}{9}\right)$$
And
Principal Arg$$\left(z\right)=-\dfrac{4\pi}{9}$$
Question 3
1 / -0

Dividing $$ f(z) $$ by $$ z - i$$ we obtain the remainder i and dividing it by $$ z + i$$ we get the remainder $$1 + i$$ then remainder upon the division of $$ f(z)$$ by $$ z^{2} + 1$$ is

A
$$\displaystyle \frac{1}{2}(z+1)+i$$

B
$$\displaystyle \frac{1}{2}(iz+1)+i$$

C
$$\displaystyle \frac{1}{2}(z-1)+i $$

D
$$\displaystyle \frac{1}{2}(z+1)+1$$

Solution
Question 4
1 / -0

If the equation $$ \displaystyle 4x^{2}+x\left ( p+1 \right )+1=0 $$ has exactly two equal roots , then one of the value of $$p$$ is

A
$$5$$

B
$$-3$$

C
$$0$$

D
$$3$$

Solution

Since the equation $$\displaystyle 4x^{2}+x(p+1)+1=0$$ has two equal roots therefore
Discriminant$$=0$$
or $$\displaystyle b^{2}-4ac=0$$
or $$\displaystyle (p+1)^{2}-4\times 4\times 1=0$$
or $$\displaystyle p^{2}+2p+1-16=0$$
or $$\displaystyle p^{2}+2p-15=0$$
or $$\displaystyle \left ( p+5 \right )\left ( p-3 \right )=0$$
$$\displaystyle \therefore p=-5$$ or $$p=3 $$
Hence, one of the values of $$p = 3$$.
Question 5
1 / -0

If $$(\sqrt 3-i)^n=2^n, n\in N$$, then $$n$$ is a multiple of

A
$$6$$

B
$$10$$

C
$$9$$

D
$$12$$

Solution

$$\sqrt3-i = 2\left(\dfrac{\sqrt{3}-i}{2}\right)$$
$$=2.e^{-i\frac{\pi}{6}}$$
Hence, $$\left(2.e^{-i\frac{\pi}{6}}\right)^{n}=2^{n}$$

$$2^{n}e^{-i\frac{n\pi}{6}}=2^{n}.e^{i2k\pi}$$, $$k\in N$$
$$\because e^{i2k\pi} = 1$$

Hence, $$-\dfrac{n\pi}{6}=2k\pi$$
$$-n=12k$$
Now $$\cos(-x)=\cos x$$
Hence, $$n=12k$$
Where $$k$$ is a natural number.
Question 6
1 / -0

Find the modulus, argument and the principal argument of the complex numbers.
$$(tan 1-i)^2$$

A
$$Modulus =\sec^21, Arg (z) = 2 n\pi +(2- \pi), Principal\ Arg (z) = (2 -\pi)$$

B
$$Modulus =\text{cosec}^21, Arg (z) = 2 n\pi -(2- \pi), Principal\ Arg (z) = (-2- \pi)$$

C
$$Modulus =\sec^21, Arg (z) = 2 n\pi -(2- \pi), Principal\ Arg (z) = (-2- \pi)$$

D
$$Modulus =\text{cosec}^21, Arg (z) = 2 n\pi +(2- \pi), Principal\ Arg (z) = (2- \pi)$$

Solution

Let
$$z=\tan1-i$$
Then
$$|z|=\sqrt{\tan^{2}1+1}=\sec1$$

$$Arg(z)=\tan^{-1}(\dfrac{-1}{\tan1})$$

$$=-\tan^{-1}(cot(1))$$

$$=-(\dfrac{\pi}{2}-\cot^{-1}(cot(1)))$$

$$=1-\dfrac{\pi}{2}$$

Hence
$$Z=(\tan1-i)^{2}=z^{2}=|z|^{2}.e^{2i.(1-\frac{\pi}{2})}$$

$$=\sec^{2}(1)e^{i.(2-\pi)}$$
Hence
$$|Z|=\sec^{2}(1)$$ and principal Arg(z)=$$(2-\pi)$$.
Question 7
1 / -0

The set of all real values of $$p$$ for which the equation $$x + 1 = \displaystyle \sqrt{px}$$ has exactly one root is

A
{0}

B
{4}

C
{0, 4}

D
{0,2}

Solution

The equation can be written as $$x^{ 2 }+(2-p)x+1$$
The equation will have exactly 1 root if the equation is a perfect square,
$$\Longrightarrow b^{ 2 }-4ac=0$$
$$\Longrightarrow (2-p)^{ 2 }-4=0$$
$$\Longrightarrow |2-p|=2$$
$$ \Longrightarrow 2-p=2$$ or $$2-p=-2$$,
$$ p=0,4$$
Question 8
1 / -0

Given that $$i = \sqrt {-1}$$, find the multiplicative inverse of $$5 - i$$.

A
$$5 + i$$

B
$$\dfrac {5 + i}{26}$$

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

D
$$\dfrac {5 + i}{24}$$

E
$$\dfrac {5 - i}{24}$$

Solution

Let the multiplicative inverse of $$5-i$$ be $$a+bi$$
Therefore we have $$(5-i)(a+bi) = 1$$
$$\Rightarrow 5a+b +i(5b-a) = 1$$
$$\Rightarrow 5a+b = 1$$ and $$5b-a = 0$$
$$\Rightarrow a=5b$$
Substitute this in $$5a+b =1$$, we get
$$b = \dfrac {1}{26}$$ and $$a = \dfrac {5}{16}$$
Therefore, multiplicative inverse is $$\dfrac {(5+i)}{26}$$.
Question 9
1 / -0

If $$(cos\theta+i\, sin \theta)\,\,( cos\,2\theta+i\,sin\,\theta)$$ ....
$$(cos\, n\theta +i\,sin\,n\theta) = 1$$, then the value of $$\theta$$ is

A
$$\dfrac{2m\pi}{n\,(n+1)}$$

B
$$4\,m\,\pi$$

C
$$\dfrac{4\,m\pi}{n\,(n+1)}$$

D
$$\dfrac{m\pi}{n\,(n+1)}$$

Solution

We have, $$(cos \theta+ i\, sin \theta) (cos \,2\theta + i \,sin 2\theta) ...$$
$$(cos \,n\theta + i \,sin\, n\theta) = 1$$
$$\therefore cos (\theta + 2\theta + 3\theta + .... + n\theta)+ i\, sin (\theta + 2\theta + 3\theta + ......... + n\theta) = 1$$
$$\Rightarrow cos\dfrac{n\,n+1}{2}\theta+i\,sin\dfrac{n\,n+1}{2}\theta=1$$
On comparing the coefficients of real and imaginary parts on both sides; we get
$$cos\dfrac{n\,(n+1)}{2}\theta=1$$
And $$sin\,\dfrac{n(n+1)}{2}\theta=0$$
$$\therefore \dfrac{n\,n+1}{2}\therefore = 2\,m\pi$$
$$\Rightarrow \theta = \dfrac{4\,m\pi}{n\,(n+1)}$$
Question 10
1 / -0

What is the product of the complex numbers $$\left( -3i+4 \right) $$ and $$\left( 3i+4 \right) $$?

A
$$1$$

B
$$7$$

C
$$25$$

D
$$-7+24i$$

E
$$7+24i$$

Solution

Consider the product of the complex numbers as shown below:
$$(-3i+4)(3i+4)=(-3i)(3i)-3i(4)+4(3i)+4(4)$$
$$=-9i^2-12i+12i+16$$
$$=-9(-1)+16=9+16$$
$$=25$$
Hence, option C is the correct answer.

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Complex Numbers and Quadratic Equations Test 40

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Question 1

positive

negative

real and distinct

imaginary

Solution

Question 2

Principal Arg $$z=-\frac {4\pi}{9}; |z|=2 cos \frac {4\pi}{9}; Arg z=2 k\pi -\frac {4\pi}{9} k\epsilon l$$

Principal Arg $$z=-\frac {10\pi}{9}; |z|=2 cos \frac {10\pi}{9}; Arg z=2 k\pi -\frac {10\pi}{9} k\epsilon l$$

Principal Arg $$z=-\frac {-10\pi}{9}; |z|=2 cos \frac {-10\pi}{9}; Arg z=2 k\pi -\frac {4\pi}{9} k\epsilon l$$

Principal Arg $$z=-\frac {-4\pi}{9}; |z|=2 cos \frac {-4\pi}{9}; Arg z=2 k\pi -\frac {4\pi}{9} k\epsilon l$$

Solution

Question 3

$$\displaystyle \frac{1}{2}(z+1)+i$$

$$\displaystyle \frac{1}{2}(iz+1)+i$$

$$\displaystyle \frac{1}{2}(z-1)+i $$

$$\displaystyle \frac{1}{2}(z+1)+1$$

Solution

Question 4

$$5$$

$$-3$$

$$0$$

$$3$$

Solution

Question 5

$$6$$

$$10$$

$$9$$

$$12$$

Solution

Question 6

$$Modulus =\sec^21, Arg (z) = 2 n\pi +(2- \pi), Principal\ Arg (z) = (2 -\pi)$$

$$Modulus =\text{cosec}^21, Arg (z) = 2 n\pi -(2- \pi), Principal\ Arg (z) = (-2- \pi)$$

$$Modulus =\sec^21, Arg (z) = 2 n\pi -(2- \pi), Principal\ Arg (z) = (-2- \pi)$$

$$Modulus =\text{cosec}^21, Arg (z) = 2 n\pi +(2- \pi), Principal\ Arg (z) = (2- \pi)$$

Solution

Question 7

{0}

{4}

{0, 4}

{0,2}

Solution

Question 8

$$5 + i$$

$$\dfrac {5 + i}{26}$$

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

$$\dfrac {5 + i}{24}$$

$$\dfrac {5 - i}{24}$$

Solution

Question 9

$$\dfrac{2m\pi}{n\,(n+1)}$$

$$4\,m\,\pi$$

$$\dfrac{4\,m\pi}{n\,(n+1)}$$

$$\dfrac{m\pi}{n\,(n+1)}$$

Solution

Question 10

$$1$$

$$7$$

$$25$$

$$-7+24i$$

$$7+24i$$

Solution

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