If $$(x+iy)(2-3i)=4+i$$ then (x, y) =

$$\left ( 1,\dfrac{1}{13} \right )$$

MCQ Questions for Class 11 Engineering Maths Complex Numbers And Quadratic Equations Quiz 1

Express

$\dfrac{1}{(1 - cos \theta + 2 i sin \theta)}$ in the form

$x + iy$

0%
$\left(\displaystyle \frac{1}{5 + 3 cos \theta}\right) + \left(\displaystyle \frac{2 cot \theta/2}{5 + 3 cos \theta}\right)i$
0%
$\left(\displaystyle \frac{1}{5 - 3 cos \theta}\right) + \left(\displaystyle \frac{-2 cot \theta/2}{5 - 3 cos \theta}\right)i$
0%
$\left(\displaystyle \frac{1}{5 + 3 cos \theta}\right) + \left(\displaystyle \frac{-2 cot \theta/2}{5 + 3 cos \theta}\right)i$
0%
$\left(\displaystyle \frac{1}{5 - 3 cos \theta}\right) + \left(\displaystyle \frac{2 cot \theta/2}{5 - 3 cos \theta}\right)i$

Explanation

To Express

$\dfrac{1}{(1-cos\theta +2isin\theta )}$ in the form

$x+iy$
We have ,

$\dfrac{1}{(1-cos\theta +2isin\theta )}$

$= \dfrac{1}{(1-cos\theta) +2isin\theta } \times \dfrac{(1-cos\theta) - 2isin\theta }{(1-cos\theta ) - 2isin\theta }$

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

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

$= \dfrac{(1-cos\theta )}{(1-cos\theta )( 3cos\theta + 5)} - \dfrac{2isin\theta }{(1-cos\theta )( 3cos\theta + 5)}$

$= \dfrac{1}{( 3cos\theta + 5)} - \dfrac{2isin\dfrac{\theta }{2}cos\dfrac{\theta }{2} }{2sin^{2}\dfrac{\theta }{2}( 3cos\theta + 5)}$

$= \dfrac{1}{( 3cos\theta + 5)} - \dfrac{2cot\dfrac{\theta }{2} }{( 3cos\theta + 5)}i$

$= \dfrac{1}{( 5 + 3cos\theta )} + \dfrac{(-2cot{\theta }/{2} )}{( 5 + 3cos\theta )}i$

Hence , Option C

$z = x + iy$ and

$\omega = \dfrac{(1 -iz)}{(z-i)}$ , then

$\left|\omega\right| = 1$ implies that in the complex plane

0%
z lies on the imaginary axis
0%
z lies on the real axis
0%
z lies on the unit circle
0%
none of these

Explanation

Given

$w=\dfrac { 1-iz }{ z-i }$ and

$\left| w \right| =1$

$\Rightarrow \left| \dfrac { 1-iz }{ z-i } \right| =1$ ...(1)

Substitute

$z=x+iy$ in equation (1)

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

$\Rightarrow \left| 1+y-ix \right| =\left| x+i\left( y-1 \right) \right| \\ \Rightarrow { \left( 1+y \right) }^{ 2 }+{ x }^{ 2 }={ x }^{ 2 }+{ \left( y-1 \right) }^{ 2 }\\ \Rightarrow y=0$

Therefore z lies on the real axis.

Ans: B

$a, b$ and

$c$ are real numbers then the roots of the equation

$(x - a)(x - b) + (x - b)(x - c) + (x - c)(x - a) = 0$ are always

0%
Real
0%
Imaginary
0%
Positive
0%
Negative

Explanation

Given equation is

$(x - a)(x - b) + (x - b)(x - c) + (x - c)(x - a) = 0$

$\Rightarrow 3x^{2} - 2(b + a + c) x + ab + bc + ca = 0$

Now, here

$A = 3, B = -2(a + b + c)$

$C = ab + bc + ca$

Therefore,

$D = \sqrt {B^{2} - 4AC}$

$= \sqrt {(-2(a + b + c))^{2} - 4(3)(ab + bc + ca)}$

$= \sqrt {4(a + b +c)^{2} - 12(ab + bc + ca)}$

$= 2\sqrt {a^{2} + b^{2} + c^{2} - ab - bc - ca}$

$= 2\sqrt {\dfrac {1}{2}\left \{(a - b)^{2} - (b - c)^{2} + (c - a)^{2} \right \}} \geq 0$

This is always

$\geq 0$ , we have real roots for the equation

$(x - a)(x - b) + (x - b)(x - c) + (x - c)(x - a) = 0$

$(x+iy)(2-3i)=4+i$ then (x, y) =

0%
$\left ( 1,\dfrac{1}{13} \right )$
0%
$\left ( -\dfrac{5}{13},\dfrac{14}{13} \right )$
0%
$\left ( \dfrac{5}{13},\dfrac{14}{13} \right )$
0%
$\left ( -\dfrac{5}{13},-\dfrac{14}{13} \right )$

Explanation

$(x+iy)(2-3i)=4+i$

$2x-(3x)i+(2y)i-3yi^{2}=4+i$

$\underbrace{2x+3y}_{Real }+\underbrace{(2y-3x)}_{Imaginary }i=4+i$

Comparing the real & imaginary parts,

$2x+3y=4$ --------------------------(1)

$2y-3x=1$ ----------------------------(2)
Solving eq(1) & eq(2),

$4x+6y=8$

$-9x+6y=3$

$13x=5\Rightarrow x=\dfrac{5}{13}$

$y=\dfrac{14}{13}$

$\therefore (x,y)=\left (\dfrac{5}{13},\dfrac{14}{13} \right )$

$z =3+5i$ , then

$z^3+z+198=$

0%
$3 - 15i$
0%
$-3 - 15i$
0%
$-3 + 15i$
0%
$3 + 15i$

Explanation

$z=3+5i$

$z^{3}=(3+5i)^{3}$

$=3^{3}+3.3^{2}(5i)+3.3(5i)^{2}+(5i)^{3}$

$=27-125i+135i-225$

$=-225+27+(135-125)i$

$=-198+10i$

$\therefore z^{3}+z+198$

$=-198+10i+3+5i+198$

$=3+15i$

$z=2-3i$ then

$z^2-4z+13=$

0%
$0$
0%
$1$
0%
$2$
0%
$3$

Explanation

$z=2-3i$

$z^{2}=2^{2}-3^{2}-12i$

$=-5-12i$

$\therefore z^{2}-4z+13$

$=(-5-12i)-4(2-3i)+13$

$=-5-12i-8+12i +13$

$=-13+13$

$=0$

The complex number

$\displaystyle \frac{1+2i}{1-i}$ lies in the quadrant :

0%
I
0%
II
0%
III
0%
IV

Explanation

Let

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

$\Rightarrow z =\dfrac{(1+2i)}{1-i}\times \dfrac{1+i}{1+i}$

$= \dfrac{1+2i+i+2i^2}{1-i^2}$

$=\dfrac{1+3i-2}{1+1}$ ............

$[\because i^2 = -1]$

$\Rightarrow z =\dfrac{-1+3i}{2}$

$=-\dfrac{1}{2}+\dfrac{3}{2}i$

$=x+iy$
Clearly

$x<0$ and

$y>0$
Hence

$z$ lies in

$\text{II}$ quadrant

$\sqrt{-3}\sqrt{-75}=$

0%
$15$
0%
$15i$
0%
$-15$
0%
$-15i$

Explanation

$\sqrt{-3}\times\sqrt{-75}=\sqrt{3\times(-1)}\sqrt{75\times(-1)}$

$=\sqrt{3}i\times\sqrt{75}i$

$=\sqrt{225}i^{2}$

$= - 15$

The sum of two complex numbers

$a + ib$ and

$c +id$ is a real number if

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$a + c = 0$
0%
$b + d = 0$
0%
$a + b= 0$
0%
$b + c = 0$

Explanation

It is given that

$z_{1}=a+ib$ and

$z_{2}=c+id$

Then

$z_{1}+z_{2}=(a+c)+i(b+d)$

Now

$(z_{1}+z_{2})$ is purely real.

Then the imaginary part has to be

$0$ .

Hence

$b+d=0$ .

The locus of complex number z such that z is purely real and real part is equal to - 2 is

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Negative y-axis
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Negative x-axis
0%
The point (-2, 0)
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The point (2, 0)

Explanation

$z=x+iy$

$(x,y)$

$z$ is purely real and the real part equals

$-2$

$\therefore y=0$ &

$x=-2$

$z=-2$
Hence, this would be represented by the point (-2,0) on the Argand Plane.

$\dfrac{1}{i-1}+\dfrac{1}{i+1}$ is

0%
positive rational number
0%
purely imaginary
0%
positive Integer
0%
negative integer

Explanation

Let

$Z= \dfrac{1}{i-1}+\dfrac{1}{i+1}$

$=\dfrac{i+1+i-1}{(i-1)(i+1)}$

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

$=\dfrac{2i}{-2}$

$\therefore Z=-i$

The argument of every complex number is

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Double valued
0%
Single valued
0%
Many valued
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Triple valued

Explanation

$z=x+iy$
amplitude

$= tan^{-1}\frac{y}{x}$

$\Rightarrow$ amplitude

$=\theta \pm 2k\pi$ where

$\theta \epsilon \left [ -\pi ,\pi \right ]$

$\forall k\epsilon R$
since

$k\epsilon R$

$\Rightarrow$ Amplitude of any complex number is many valued.

The sum of two complex numbers

$a + ib$ and

$c+ id$ is purely imaginary if

0%
$a + c = 0$
0%
$a + d = 0$
0%
$b + d = 0$
0%
$b + c = 0$

Explanation

It is given that

$z_{1}=a+ib$ and

$z_{2}=c+id$

$z_{1}+z_{2}=(a+c)+i(b+d)$

$z_{1}+z_{2}$ is purely imaginary. (Given)
Then the real part has to be

$0$ .
Hence

$a+c=0$ .

For

$a < 0$ , arg

$(ia) =$

0%
$\dfrac{\pi }{2}$
0%
$-\dfrac{\pi }{2}$
0%
$\pi$
0%
$-\pi$

Explanation

Let

$z= ia$

$a <0$ . Hence

$z$ must lie on the negative imaginary axis ,
Hence,

$arg(z) = -\dfrac{\pi}{2}$

The principal value of the argument of

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

0%
$\dfrac{\pi }{6}$
0%
$\dfrac{3\pi }{6}$
0%
$\dfrac{5\pi }{6}$
0%
$\dfrac{7\pi }{6}$

Explanation

Given,

$z=-\sqrt 3+i$

We have

$| z| = \sqrt{\sqrt{3}^2 +1^2} =2$

Therefore,

$\cos \theta = \dfrac{-\sqrt{3} }{2}$

and

$\sin \theta = \dfrac{1}{2}$

Hence,

$\theta =\dfrac{5\pi}{6}$

Amplitude of

$\dfrac{1+i}{1-i}$ is :

0%
$0$
0%
$\pi$
0%
$\dfrac{\pi }{2}$
0%
$-\pi$

Explanation

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

$=\dfrac{(1+i)(1+i)}{(1-i)(1+i)}$

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

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

$=i$

therefore amplitude

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

Which of the following equations has two distinct real roots ?

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$2x^2-3\sqrt 2 x+\dfrac 94=0$
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$x^{2}+x-5=0$
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$x^{2}+3x+2\sqrt{2}=0$
0%
$5x^{2}-3x+1=0$

Explanation

An equation is said to have two distinct and real roots if the discriminant

$b^2-4ac>0$

Case (i): For equation:

$2x^2-3\sqrt 2 x+\dfrac 94=0$ .

Here

$a=2, b=-3\sqrt 2, c=\dfrac 94$

The discrimant is

$(-3\sqrt 2)^2-4(2)\left(\dfrac 94\right)=18-18=0$

Hence this equation has equal real roots

Case (ii): For equation:

$x^2+x-5=0$ .

Here

$a=1, b=1, c=-5$

The discrimant is

$1^2-4(1)(-5)=1+20=21>0$

Hence this equation has two distinct real roots

Case (iii): For equation:

$x^2+3x+2\sqrt 2=0$ .

Here

$a=1, b=3, c=2\sqrt 2$

The discrimant is

$3^2-4(1)(2\sqrt2)=9-8\sqrt 2<0$

Hence this equation has no real roots

Case (iv): For equation:

$5x^2-3x+1=0$ .

Here

$a=5, b=-3, c=1$

The discrimant is

$(-3)^2-4(5)(1)=9-20<0$

Hence this equation has no real roots

A quadratic equation

$ax^2 + bx+c=0$ has two distinct real roots, if

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$a=0$
0%
$b^2-4ac = 0$
0%
$b^2-4ac < 0$
0%
$b^2-4ac > 0$

Explanation

$a=0$ , it becomes linear equation.
If

${ b }^{ 2 }-4ac=0$ , then there will be real and equal roots.
If

${ b }^{ 2 }-4ac<0$ , then the roots will be unreal.
Only if

${ b }^{ 2 }-4ac>0$ , we will get two real distinct roots.
Option D is correct!

For

$a > 0$ , arg

$(ia) =$

0%
$\dfrac{\pi }{2}$
0%
$-\dfrac{\pi }{2}$
0%
$\pi$
0%
$-\pi$

Explanation

$z= 0 + ia$

$\therefore arg(z) = arg (ia) = \tan^{-1} \cfrac a0$

$\cfrac {\pi}{2}$ ...(since a is greater than or equal to zero)

The modulus of

$\sqrt{2}i-\sqrt{-2}i$ is:

0%
2
0%
$\sqrt{2}$
0%
0
0%
$2\sqrt{2}$

Explanation

Let

$z=\sqrt{2}i-\sqrt{-2}i$

$\Rightarrow z=\sqrt{2}i-\sqrt{2}i^2$

$\Rightarrow z=\sqrt{2}+\sqrt{2}i$
Now,

$|z|=\sqrt{2+2}=2$

The roots of the equation

$3x^{2} - 4x + 3 = 0$ are :

0%
real and unequal
0%
real and equal
0%
imaginary
0%
none of these

Explanation

Given equation is

$3x^2-4x+3=0$

To find, the nature of the roots of the equation

An equation is said to have

(i) two distinct and real roots if the discriminant

$b^2-4ac>0$

(ii) equal real roots if

$b^2-4ac=0$

(iii) no real roots or imaginary roots if

$b^2-4ac<0$

In the given equation

$a=3, b=-4, c=3$

Hence the discriminant is

$(-4)^2-4(3)(3)=16-36=-20<0$

Therefore the roots of the given equation are imaginary in nature.

For

$a<0$ , arg

$a=$

0%
$\dfrac{\pi }{2}$
0%
$\dfrac{-\pi }{2}$
0%
$\pi$
0%
$-\pi$

Explanation

It is given that

$a<0$
Thus, 'a' is purely real.
Hence

$arg(a)$ will be

$0$ or,

$\pi$
But it is given that

$a<0$
Hence

$cos(arg(a))=-1$
Or

$arg(a)=\pi$ .

If the square of

$(a + ib)$ is real, then

$ab=$

0%
$0$
0%
$1$
0%
$-1$
0%
$2$

Explanation

$(a+ib)^{2}=a^{2}-b^{2}+2iab$ is given to be real

$\Rightarrow ab=0$

Find the argument of

$-1 - i\sqrt{3}$

0%
$\theta= -2\pi/3$
0%
$\theta= 2\pi/3$
0%
$\theta= -4\pi/3$
0%
$\theta= 4\pi/3$

Explanation

Let

$z = -1 -i\sqrt{3}$

Then

$\alpha = \tan^{-1}\left|b/a\right| = \tan^{-1} \left|\sqrt{3}/1\right| = \pi/3$
Here,

$z$ is in the third quadrant.

Therefore, argument is

$\theta= -(\pi - \alpha) = -(\pi - \pi/3) = -2\pi/3$

Ans: A

The roots of

$x^{2}-x+1=0$ are:

0%
Real and equal
0%
Real and not equal
0%
Imaginary
0%
Reciprocals

Explanation

Given equation is

$x^2-x+1=0$

We know

$D=b^{2}-4ac$

Here

$a=1, b=-1, c=1$
Therefore,

$D=(-1)^{2}-4(1)(1)$

$=1-4$

$=-3< 0$
Thus roots are imaginary.

Nature of the roots of the quadratic equation

$2x^{2}-2\sqrt{6}x+3=0$ is:

0%
Real, irrational, unequal
0%
Real, rational, equal
0%
Real, rational, unequal
0%
Complex

Explanation

(B)

$b^{2}-4ac=(-2\sqrt{6})^{2}-4(2)(3)$

$=24-24$

$=0$

$\therefore$ Real, rational, equal

Determine the nature of roots of the equation

$x^2 + 2x\sqrt{3}+3=0$ .

0%
Real and distinct
0%
Non-real and distinct
0%
Real and equal
0%
Non-real and equal

Explanation

The nature of the roots can be determined from the discriminant

$b^2-4ac$

$\therefore b^2-4ac=(2\sqrt {3})^2-(4\times 1\times 3)$

$\Rightarrow b^2-4ac=12-12$

$\Rightarrow b^2-4ac=0$

$\because b^2-4ac=0$
There are two real and equal roots.

Find the value of

$x$ of the equation

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

0%
$1$
0%
$2$
0%
$0$
0%
none of these

Explanation

Then

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

$\Rightarrow |{(1-i)}|^{x} =|2|^{x}$

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

$\Rightarrow { \left( \sqrt { 2 } \right) }^{ x }={ 2 }^{ x }$

$\Rightarrow \dfrac{x}{2}=x$

$\Rightarrow 2x=x\Rightarrow x=0$

Hence, option C is correct.

If the discriminant of a quadratic equation is negative, then its roots are:

0%
unequal
0%
equal
0%
inverse
0%
imaginary

Explanation

If the discriminant

$b^2 - 4ac$ of a quadratic equation is negative, then its roots are imaginary.

Solve

$\displaystyle \left ( 1-i \right )x+\left ( 1+i \right )y= 1-3i,$

0%
$\displaystyle x= -1, y= 2.$
0%
$\displaystyle x= 2, y= -1.$
0%
$\displaystyle x= 2, y= 1.$
0%
$\displaystyle x= 1, y= 2.$

Explanation

$\displaystyle \left ( 1-i \right )x+\left ( 1+i \right )y= 1-3i.$
Equating real and imaginary parts, we get

$\displaystyle x+y= 1$ and

$\displaystyle -x+y= -3.$
Adding both equations we get

$y=-1$

Substituting this value of

$y$ in any of the 2 equations, we get

$x=2$

$\therefore \displaystyle x= 2, y= -1.$

Ans: B

The roots of

$4x^{2}-2x+8=0$ are:

0%
Real and equal
0%
Rational and not equal
0%
Irrational
0%
Not real

Explanation

Given equation is

$4x^2-2x+8=0$

We know value of discriminant

$D=b^{2}-4ac$

Here

$a=4, b=-2, c=8$
Therefore,

$D=(-2)^{2}-4(4)(8)$

$=4-128$

$=-124< 0$
Hence, roots are not real.

Evaluate :

$\sqrt{-25} + 3 \sqrt{-4} +2 \sqrt{-9}$

0%
$-17i$
0%
$5i$
0%
$17i$
0%
$6i$

Explanation

$\sqrt{-25} + 3 \sqrt{-4} +2 \sqrt{-9}$

$=5\sqrt{-1}+6\sqrt{-1}+6\sqrt{-1}$ we know,

$\sqrt{-1}=i$

$= 5i + 6i + 6i = 17i$

$x^{2}-2px+8p-15=0$ has equal roots, then

$p=$

0%
$3$ or $-5$
0%
$3$ or $5$
0%
$-3$ or $5$
0%
$-3$ or $-5$

Explanation

Given equation is

$x^2-2px+8p-15=0$

Here

$a=1, b=-2p, c=8p-15$

We know for equal roots,

$b^{2}-4ac=0$
Therefore,

$(-2p)^{2}-4(1)(8p-15)=0$

$\Rightarrow 4p^{2}-32p+60=0$

$\Rightarrow p^{2}-8p+15=0$

$\Rightarrow (p-5)(p-3)=0$
i.e.,

$p=5$ or

$3$

Determine the values of

$p$ for which the quadratic equation

$2x^2 + px + 8 = 0$ has equal roots.

0%
$p=\pm 64$
0%
$p=\pm 8$
0%
$p=\pm 4$
0%
$p=\pm 16$

Explanation

$ax^2+bx+c=0,$
Discriminant

$D$

$=b^2-4ac$ ,
D

$=0$ , for the roots to be real and equal

$2x^{2}+px+8=0$

Thus, we have

$a=2 ,b=p$ and

$c=8$

Then

$D$

$=b^2-4ac=0$

$\therefore p^{2}-4\times 2\times 8=0\Rightarrow p^{2}-64=0$

$\Rightarrow p=\pm 8$ .

Find the values of

$k$ for the following quadratic equation, so that they have two real and equal roots:

$2x^2 + k x + 3 = 0$

0%
$k = \pm 2\sqrt 3$
0%
$k = \pm 2\sqrt 6$
0%
$k = \pm \sqrt 6$
0%
$k = \pm \sqrt 3$

Explanation

$ax^2+bx+c=0$

Discriminant

$D=b^2-4ac$

$D=0$ , so they have equal and real roots.

Then,

$2x^{2}+kx+3=0$

$a=2$ ,

$b=k$ and

$c=3$

$D=k^{2}-4\times 2\times 3=0$

$\Rightarrow k^{2}-24=0$

$\Rightarrow k=\pm \sqrt{24}$

$\Rightarrow k=\pm 2\sqrt{6}$

$\displaystyle \frac{\displaystyle i^{4n + 3} + (-i)^{8n - 3}}{\displaystyle(i)^{12 n- 1} - i^{2 - 16 n}}, n \varepsilon N$ is equal to

0%
1 + i
0%
2i
0%
-2i
0%
-1 - i

Explanation

Given expression

$= \displaystyle \frac{\displaystyle i^3 + (-1) (i)^{-3}}{(-i)^{-1} - (i)^2} = \frac{\displaystyle -i - i}{\displaystyle i + 1}$

$=\displaystyle \frac{\displaystyle -2i}{1 + i} \times \frac{1 - i}{1 - i} = \frac{\displaystyle -2i - 2}{\displaystyle 1 + 1} = - 1 - i$

$i^2 + i^4 + i^6 + ........+ i^{2n}$ is

0%
Positive
0%
Negative
0%
Zero
0%
Cannot be determined

Explanation

$1+i^2+i^4+........+i^{2n}$

$=1-1+1-1+1-1......$

Hence it can b positive or negative depending on the value of

$2n$

Hence It cannot be determined from the given data.

Find the modulus and the principal value of the argument of the number

$1-i$

0%
$\displaystyle \sqrt{2},\pi/4$
0%
$\displaystyle \sqrt{2},-\pi/4$
0%
$\displaystyle \sqrt{2},-\pi/3$
0%
$\displaystyle \sqrt{2},3\pi/4$

Explanation

Let

$\displaystyle 1=r\cos \theta ,-1=r\sin \theta$

Squaring and adding these relations, we get

$\displaystyle r^{2}\left ( \cos ^{2}\theta +\sin ^{2}\theta \right )=1^{2}+\left ( -1 \right )^{2}=2$ .

Since,

$cos^2\theta+sin^2\theta=1, \Rightarrow r=\pm\sqrt{2}$

$r$ cannot be negative

Hence.

$r=\sqrt{2}$

Then

$\displaystyle \cos \theta =\dfrac{1}{\sqrt{2}},\sin \theta =\dfrac{-1}{\sqrt{2}},$

The value of

$\displaystyle \theta$ between-

$\displaystyle \pi$ and

$\displaystyle \pi$ which satisfies these equations is

$\displaystyle -\left ( \pi /4 \right )$ Thus

$\displaystyle \left | 1-i \right |=r=\sqrt{2}$ and

$arg (1-i)=-\displaystyle \left ( \pi /4 \right ).$

Ans: B

$i^2 = - 1$ , then the value of

$\displaystyle \sum^{200}_{n = 1} i^n$ is

0%
50
0%
-50
0%
0
0%
100

Explanation

Given,

$i^2=-1$

$\displaystyle \sum_{n = 1}^{200} i^n = i + i^2 + i^3 + ......... + i^{200} = \displaystyle \frac{i (1 - i^{200})}{1 - i}$ (since G. P.)

$= \displaystyle \frac{\displaystyle i (1 - 1)}{\displaystyle 1 - i} = 0$

If i =

$\sqrt {-1}, then 1 + i^2 + i^3 -i^6 + i^8$ is equal to -

0%
2- i
0%
1
0%
-3
0%
-1

Explanation

Given,

$i= \sqrt {-1}$

$\therefore 1+i^{ 2 }+i^{ 3 }-i^{ 6 }+i^{ 8 }=1-1-i+1+1$

$=2-i$

Check whether

$2x^2 - 3x + 5 = 0$ has real roots or no.

0%
The equation has real roots.
0%
The equation has no real roots.
0%
Data insufficient
0%
None of these

Explanation

Here the quadratic equation is

$2x^2 - 3x + 5 = 0$
Comparing it with

$ax^2+bx+c=0$ , we get

$a=2, b=-3, c=5$
Therefore, discriminant,

$D=b^2-4ac$

$(-3)^2-4\times 2\times 5$

$=9-40$

$=-31$
Here,

$D<0$

Therefore the equation has no real roots.

(

$i^{10}+1) (i^9 + 1)(i^8 +1).......(i+1)$ equal to

0%
-1
0%
1
0%
0
0%
i

Explanation

$(i^{10}+1) (i^9 + 1)(i^8 +1).......(i+1)$

$=(i^{10}+1) (i^9 + 1)(i^8 +1).......(i^3+1)(i^2+1)(i+1)$

$=(i^{10}+1) (i^9 + 1)(i^8 +1).......(i^3+1)(-1+1)(i+1)[\because i^2=-1]$

$=(i^{10}+1) (i^9 + 1)(i^8 +1).......(i^3+1)(0)(i+1) =0$

$i^2$

$= -1$ , then find the odd one out of the following expressions.

0%
$-i^2$
0%
$(-i)^2$
0%
$i^4$
0%
$(-i)^4$
0%
$-i^6$

Explanation

$i$ is an imagiary number which has a value of

$i=\sqrt{-1}$ . Then

$-i^2=-1*-1=1$

${(-i)}^2=i^2=-1$

$i^4={(i^2)}^2={(-1)}^2=1$

${(-i)}^4={(i^2)}^2={(-1)}^2=1$

$-i^6=-{(i^2)}^3=-1\times {(-1)}^3=-1\times (-1)=1$

All are

$1$ except option

$B$ which is

$-1$ .

When

$(3-2i)$ is subtracted from

$(4 + 7i)$ , then the result is

0%
$1 + 5i$
0%
$1 + 9i$
0%
$7 + 5i$
0%
$7 + 9i$

Explanation

We need to subtract

$(3-2i)$ from

$(4+7i)$

$\therefore 4 + 7i - (3 - 2i)$

$=4 + 7i - 3 + 2i$

$= 1 + 9i$

The value of k for which polynomial

$x^{2} - kx + 4$ has equal zeroes is

0%
$4$
0%
$2$
0%
$-4$
0%
$-2$

Explanation

$x^{2}-kx+4$ has equal roots

$\Rightarrow D = 0$

$\Rightarrow k^{2}-4\times 4\times 1\Rightarrow k^{2} = 16 \Rightarrow \boxed{k = \pm 4}$

1104178_517074_ans_e4c1a10f8af04a3dbe2ea102299f36a2.jpg

If the discriminant of a quadratic equation is negative, then its roots are

0%
Unequal
0%
Equal
0%
Inverses
0%
Imaginary

Explanation

$x = \dfrac{-b\pm \sqrt{D}}{2a}$ if D is

$-$ ve

$\Rightarrow$ imaginary Roots
1097630_516995_ans_139402a3e3b3476abd08468647e3ae9d.jpg

If the equation

$(1 + m^{2}) x^{2} + 2mcx + (c^{2} - a^{2}) = 0$ has equal roots, then

$c^{2} =$

0%
$a^{2} (1 + m^{2})$
0%
$a (1 + m^{2})$
0%
$a^{4} (1 - m^{2})$
0%
$a^{2} (1 - m^{2})$

Explanation

As the eq

$^{n}$ has equal roots

$\Rightarrow D = 0$

$\Rightarrow (2mc)^{2}-4(1+m^{2})(c^{2}-a^{2}) = 0$

$\Rightarrow 4-m^{2}c^{2}-4[c^{2}-a^{2}+m^{2}c^{2}-m^{2}a^{2}] = 0$

$\Rightarrow c^{2} = m^{2}a^{2}+a^{2}\Rightarrow \boxed{c^{2} = a^{2}(1+m^{2})}$

1115086_516967_ans_b5abf02279864e97885d5bec0b227f73.jpg

Amplitude of

$\displaystyle \frac{1 +\sqrt 3i}{ \sqrt3 + i}$ is

0%
$\displaystyle \frac {\pi}{3}$
0%
$\displaystyle \frac {\pi}{2}$
0%
0
0%
$\displaystyle \frac {\pi}{6}$

Explanation

$\displaystyle Z= \frac{1 +\sqrt 3i}{ \sqrt3 + i}$

$\displaystyle\Rightarrow \left( \frac { 1+\sqrt { 3 } i }{ \sqrt { 3 } +i } \right) \left( \frac { \sqrt { 3 } -i }{ \sqrt { 3 } -i } \right) =\frac { 2\sqrt { 3 } +2i }{ 4 }$

$\therefore$ Amplitude of

$Z$ is

$\tan ^{ -1 }\left({\displaystyle \frac { 1 }{ \sqrt3 } } \right)=\displaystyle\frac{\pi}{6}$
Hence, option D.

For

$i=\sqrt{-1}$ , what is the sum

$\left(7+3i\right) + \left(-8+9i\right)$ ?

0%
$-1+12i$
0%
$-1-6i$
0%
$15+12i$
0%
$15-6i$

Explanation

Given:

$(7+3i)+(-8+9i)$

$= (7+(-8))+(3i+9i)$

$=(7-8)+12i$

$=-1+12i$

Option A is correct.

If the equation

$\displaystyle x^{2}-bx+1=0$ does not possess real roots then

0%
$-3 < b < 3$
0%
$-2 < b < 2$
0%
$b > 2$
0%
$b < -2$

Explanation

Given equation is

$x^{2}-bx+1=0$
If the equation doesnot possess real roots,then

$b^2-4ax<0$

$=>(-b)^2-4(1)(1)<0$

$=>b^2<4$

$=>b<\pm 2$

$\therefore -2<b<2$

CBSE Questions for Class 11 Engineering Maths Complex Numbers And Quadratic Equations Quiz 1 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Maths Complex Numbers And Quadratic Equations Quiz 1 - MCQExams.com

0:0:1

Practice Class 11 Engineering Maths Quiz Questions and Answers

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