MCQ Questions for Class 11 Engineering Maths Complex Numbers And Quadratic Equations Quiz 7

If '

$\omega$ ' is a complex cube root of unity,then

$\omega ^{ \begin{pmatrix} \frac { 1 }{ 3 } & +\frac { 2 }{ 9 } +\frac { 4 }{ 27 } ...\infty \end{pmatrix} }+\omega^{ \begin{pmatrix} \frac { 1 }{ 2 } & +\frac { 3 }{ 8 } +\frac { 9 }{ 32 } ...\infty \end{pmatrix} }=$

0%
$1$
0%
$-1$
0%
$\omega$
0%
$i$

Explanation

$\dfrac{1}{3}+\dfrac{2}{9}+\dfrac{4}{27}+......+\infty$ (infinte G.P.)

$S_\omega = \dfrac{a}{1-r} =1$

$\dfrac{1}{2}+\dfrac{3}{8}+\dfrac{9}{32}+......+\infty$ (infinte G.P.)

$S_\omega = \dfrac{a}{1-r} =2$

$\therefore\, \omega^1+\omega^2=-1$

Find a complex number z satisfying the equation

$z+\sqrt{2}|z+1|+i=0$

0%
$2-i$
0%
$-2-i$
0%
$\sqrt{2}-i$
0%
None of these

Explanation

Let

$z=x+iy$ . Then,

$z+\sqrt{2}|z+1|+i=0$

$x+iy+\sqrt{2}|(x+1)+iy|+i=0$

$x+\sqrt{2}\sqrt{(x+1)^2+y^2}+(y+1)i=0$

$x+\sqrt{2(x+1)^2+2y^2}+(y+1)i=0$

$x+\sqrt{2(x+1)^2+2}=0$ and

$y=-1$

$\sqrt{2(x+1)^2+2}=-x$ and

$y=-1$

$2(x+1)^2+2=x^2$ and

$y=-1$

$x^2+4x+4=0$ and

$y=-1$

$(x+2)^2=0$ and

$y=-1$

$x=-2$ and

$y=-1$

Hence,

$z=x+iy=-2-i$

Consider the quadratic equation

$nx^{2} + 7\sqrt {n}x + n = 0$ , where

$n$ is a positive integer. Which of the following statements are necessarily correct?
I. For any

$n$ , the roots are distinct.
II. There are infinitely many values of

$n$ for which both roots are real.
III. The product of the roots is necessarily an integer.

0%
III only
0%
I and III only
0%
II and III only
0%
I, II and III

Explanation

Let the discriminant of the quadratic equation

$ax^{2} + bx + c$ , where

$a, b$ and

$c$ are real, be given by

$D = b^{2} - 4ac$ . Roots of the given equation are

i) real and distinct if

$D > 0$

ii) real and equal if

$D = 0$

iii) imaginary and distinct if

$D < 0$

Product of the roots is

$\dfrac{c}{a}$ .

For the given problem,

$D = 49n-4n^{2}$

I) roots are distinct if

$D \neq 0$

$\Rightarrow 49n \neq 4n^{2}$

$\Rightarrow n \neq \dfrac{49}{4}$

$n$ is a positive integer, so for all n, the roots are distinct.

II) Both roots are real

$\Rightarrow D \geq 0$

$\Rightarrow 49n\geq 4n^{2}$

$\Rightarrow n\leq \dfrac{49}{4}$

So, such values of

$n$ are not infinite.

III) Product of roots

$= \dfrac{n}{n} = 1$ , which is an integer.

${z}_{1}=-3+5i;{z}_{2}=-5-3i$ and

$z$ is a complex number lying on the line segment joining

${z}_{1}$ and

${z}_{2}$ , then

$arg(z)$ can be:

0%
$-\cfrac { 3\pi }{ 4 }$
0%
$-\cfrac { \pi }{ 4 }$
0%
$\cfrac { \pi }{ 6 }$
0%
$\cfrac { 5\pi }{ 6 }$

$\dfrac {lz_{2}}{mz_{1}}$ is purely imaginary number, then

$\left |\dfrac {\lambda z_{1} + \mu z_{2}}{\lambda z_{1} - \mu z_{2}}\right |$ is equal to

0%
$\dfrac {l}{m}$
0%
$\dfrac {\lambda}{\mu}$
0%
$\dfrac {-\lambda}{\mu}$
0%
$1$

Explanation

$\cfrac { { lz }_{ 2 } }{ m{ z }_{ 1 } } =ki\\ \cfrac { { z }_{ 2 } }{ { z }_{ 1 } } =\cfrac { mki }{ l } \\ \implies |\cfrac { { \lambda z }_{ 1 }+\mu { z }_{ 2 } }{ { \lambda z }_{ 1 }-\mu { z }_{ 2 } } |\\ \implies |\cfrac { \lambda +\mu { z }_{ 2 }/{ z }_{ 1 } }{ \lambda -\mu { z }_{ 2 }/{ z }_{ 1 } } |\\ \implies |\cfrac { \lambda +\mu \cfrac { mki }{ l } }{ \lambda -\mu \cfrac { mki }{ l } } |\\ \implies|\cfrac { \lambda l+\mu mki }{ \lambda l-\mu mki } |$

$\implies|\cfrac{re^{i\theta}}{re^{-i\theta}}|$

$\implies |e^{2i\theta}|$

$\implies 1$

The complex number

$\dfrac{1+2i}{1-i}$ lies in which quadrant of the compiles plan

0%
First
0%
Second
0%
Third
0%
Fourth

Explanation

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

Multiply the coyugate of

$(1-i)$ is numerator

$\times$ denominator

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

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

$=\dfrac{1+3i-4}{1+1}$

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

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

$\therefore$ It is of the form

$-a+3i (-a, b)$

$\therefore \left(\dfrac{-3}{2}, \dfrac{3}{2}\right)$

$\therefore 2^{nd}$ Quandrant

If in applying the quardratic formula to a quadratic equation

$f(x) = ax^2 + bx + c = 0$ , it happens that

$c = b^2/4a$ , then the graph of

$y = f(x)$ will certainly:

0%
have a maximum
0%
have a minimum
0%
tangent to the $x$ -axis
0%
be tangent to the $y$ -axis
0%
lie in one quadrant only

Explanation

The condition

$c = b^2/4a$ implies equal real coefficients, i.e., the curve touches the

$x$ -axis at exactly one point and, therefore, the graph is tangent to the

$x$ -axis.

The real and imaginary part of the complex number

$1 + \sqrt {i}$ where

$i = \sqrt {-1}$ are

0%
$1 - \dfrac {1}{\sqrt {2}}$ and $-\dfrac {1}{\sqrt {2}}$ respectively
0%
$1 - \dfrac {1}{\sqrt {2}}$ and $\dfrac {1}{\sqrt {2}}$ respectively
0%
$1 + \dfrac {1}{\sqrt {2}}$ and $\dfrac {1}{\sqrt {2}}$ respectively
0%
$1 + \dfrac {1}{\sqrt {2}}$ and $-\dfrac {1}{\sqrt {2}}$ respectively

Explanation

$\sqrt { i } ={ cis\left( \dfrac { \pi }{ 2 } \right) }^{ 0.5 }$

From De moivres theorem,

$\sqrt { i } ={ cis\left( \dfrac { \pi }{ 2 } \right) }^{ 0.5 }=\pm cis\left( 0.5\frac { \pi }{ 2 } \right) =\dfrac { 1 }{ \sqrt { 2 } } +i\dfrac { 1 }{ \sqrt { 2 } }$ ,

$\therefore$ the real and imaginary parts are

$1+\dfrac { 1 }{ \sqrt { 2 } } ,\dfrac { 1 }{ \sqrt { 2 } }$ and

$1-\dfrac { 1 }{ \sqrt { 2 } } ,-\dfrac { 1 }{ \sqrt { 2 } }$

The given quadratic equations have real roots and the roots are equal and that is

$\dfrac{1}{\sqrt2}$ :

$2x^2 \, - \, 2\sqrt{2}x \, + \, 1 \, = \, 0$

0%
True
0%
False

Explanation

$2x^2-2\sqrt{2}x+1=0$

$\Rightarrow$

$2x^2-\sqrt{2}x-\sqrt{2}x+1=0$

$\Rightarrow$

$\sqrt{2}x(\sqrt{2}x-1)-1(\sqrt{2}x-1)=0$

$\Rightarrow$

$(\sqrt{2}x-1)(\sqrt{2}x-1)=0$

$\Rightarrow$

$\sqrt{2}x-1=0$ and

$\sqrt{2}x-1=0$

$\Rightarrow$

$x=\dfrac{1}{\sqrt{2}}$ and

$x=\dfrac{1}{\sqrt{2}}$

$\Rightarrow$ Hence, we can see roots iven in question is correct.

The given quadratic equation have real roots and roots are

$\dfrac{\sqrt5}{3}, \, -\sqrt5$ :

$3x^2 \, + \, 2\sqrt{5x} \, - \, 5 \, = \, 0$

0%
True
0%
False

Explanation

$3x^2+2\sqrt{5}x-5=0$

$\Rightarrow$

$3x^2+3\sqrt{5}x-\sqrt{5}x-5=0$

$\Rightarrow$

$3x(x+\sqrt{5})-\sqrt{5}(x+\sqrt{5})=0$

$\Rightarrow$

$(x+\sqrt{5})(3x-\sqrt{5})=0$

$\Rightarrow$

$x+\sqrt{5}=0$ and

$3x-\sqrt{5}=0$

$\therefore$

$x=-\sqrt{5}$ and

$x=\dfrac{\sqrt{5}}{3}$

$\therefore$ We can see roots given in question are correct.

The roots of the following quadratic equation are Real and equal.

$3x^2-4\sqrt3x+4=0$

0%
True
0%
False

Explanation

$\Rightarrow$ The given quadratic equation is

$3x^2-4\sqrt{3}x+4=0$ , comparing it with

$ax^2+bx+c=0$

$\Rightarrow$ We get,

$a=3,\,b=-4\sqrt{3},\,c=4$

$\Rightarrow$ Discriminant

$=b^2-4ac$

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

$=48-48$

$=0$

$\therefore$ Discriminant

$=0$

$\therefore$ The roots of given equation are real and equal.

The given quadratic equations have real roots and the roots are

$-\sqrt2, \, \dfrac{-5}{\sqrt2}$ :

$\sqrt{2}x^2 \, + \, 7x \, + \, 5\sqrt{2} \, = \, 0$

0%
True
0%
False

Explanation

$\sqrt{2}x^2+7x+5\sqrt{2}=0$

$\Rightarrow$

$\sqrt{2}x^2+5x+2x+5\sqrt{2}=0$

$\Rightarrow$

$x(\sqrt{2}x+5)+\sqrt{2}(\sqrt{2}x+5)=0$

$\Rightarrow$

$(x+\sqrt{2})(\sqrt{2}x+5)=0$

$\Rightarrow$

$x+\sqrt{2}=0$ and

$\sqrt{2}x+5=0$

$\Rightarrow$

$x=-\sqrt{2}$ and

$x=\dfrac{-5}{\sqrt{2}}$

$\therefore$ Here, we can see roots are real and roots are

$-\sqrt{2},\,\dfrac{-5}{\sqrt{2}}$ .

In the following, determine whether the given quadratic equations have real roots and if so, find the roots :

$16x^2$ = 24x + 1

0%
$\text{not real}$
0%
$real,\dfrac{3 \, \pm \, \sqrt{10}}{4}$
0%
$\dfrac{3 \, \pm \, \sqrt{10}}{2}$
0%
$\dfrac{3 \, \pm \, \sqrt{13}}{4}$

Explanation

$16x^2=24x+1$

$\Rightarrow$

$16x^2-24x-1=0$

$\Rightarrow$

$a=16,\,b=-24,\,c=-1$

$\Rightarrow$

$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

$=\dfrac{-(-24)\pm\sqrt{(24)^2-4(16)(-1)}}{2(16)}$

$=\dfrac{24\pm\sqrt{576+64}}{32}$

$=\dfrac{24\pm\sqrt{640}}{32}$

$=\dfrac{24\pm8\sqrt{10}}{32}$

$=\dfrac{3\pm\sqrt{10}}{4}$

$\therefore$

$x=\dfrac{3\pm\sqrt{10}}{4}$

$\therefore$ We can see roots we have got is real.

The given quadratic equations have real roots and the roots are equal and it is

$1$ :

$x^2$ - 2x + 1 = 0

0%
True
0%
False

Explanation

$x^2-2x+1=0$

$\Rightarrow$

$x^2-x-x+1=0$

$\Rightarrow$

$x(x-1)-1(x-1)=0$

$\Rightarrow$

$(x-1)(x-1)=0$

$\Rightarrow$

$x-1=0$ and

$x-1=0$

$\Rightarrow$

$x=1$ and

$x=1$

$\therefore$ Here, we can see roots are real and equal.

The roots of the following quadratic equation Real and equal.

$3x^2$ - 2

$\sqrt{6}x$ + 2 = 0

0%
True
0%
False

Explanation

Step 1: Find the coefficients $\boldsymbol{a,\ b, \text{and}\ c}$

Compare the given equation with $ax^2+bx+c=0$ , we get

$a=3,\ b=-2\sqrt{6},\ c=2$

Step 2: Find discriminant of the given quadratic equation

$D=b^2-4ac=(-2\sqrt{6})^2-4(3)(2)=12-12=0$

Hence the roots of given quadratic equation are real and equal.

Hence the given statement is true.

The discrimination of the equation

$x^2 + 2x\sqrt3 + 3 = 0$ is zero. Hence, its roots are:

0%
Real and Equal
0%
Rational and Equal
0%
Rational and Unequal
0%
Irrational and Unequal
0%
Imaginary

Explanation

${ x }^{ 2 }+2x\sqrt { 3 } +3=0\\ \Longrightarrow { (x+\sqrt { 3 } ) }^{ 2 }=0\\ \Longrightarrow x=-\sqrt { 3 } ,-\sqrt { 3 }$

Roots are real and equal.

The roots of the following quadratic equation are Not real
2

$(a^2 + b^2)x^2$ + 2(a + b) x + 1 = 0

0%
True
0%
False

Explanation

$2(a^2+b^2)x^2+2(a+b)x+1=0$

$\Rightarrow$ Here,

$A=2(a^2+b^2),\,B=2(a+b),\,C=1$

$\Rightarrow$

$B^2-4AC$

$\Rightarrow$

$[2(a+b)]^2-4[2(a^2+b^2)](1)$

$\Rightarrow$

$4(a^2+2ab+b^2)-4(2a^2+2b^2)$

$\Rightarrow$

$4a^2+8ab+4b^2-8a^2-8b^2$

$\Rightarrow$

$-4a^2-4b^2+8ab$

$\Rightarrow$

$-4(a^2+b^2-2ab)$

$\Rightarrow$

$-4(a-b)^2$

$\Rightarrow$ Here

$Discriminant$ is negative.

$\Rightarrow$

$D<0$

$\therefore$ The equation has no real roots

The complex number

$x+iy$ whose modulus is unity,

$y\neq 0$ , can be represented as

$x+iy=\dfrac { a+i }{ a-i }$ , where

$a$ is real number.

0%
True
0%
False

Explanation

$\cfrac{a+i}{a-i}$

$=\cfrac{(a+i)^2}{a^2+1}=\cfrac{a^2-1+2ai}{a^2+1}$

$\cfrac{a^2-1}{a^2+1}+\cfrac{2ai}{a^2+1}=z$ (Let)

$|z|=\sqrt{\cfrac{(a^2-1)^2}{(a^2+1)^2}+(\cfrac{2a}{a^2+1})^2}$

$\implies \cfrac{\sqrt{(a^2+1)^2}}{a^2+1}=1$

Hence true.

$\mid{z_1}\mid=2$ ,

$\mid{z_2}\mid=3$ ,

$\mid{z_3}\mid=4$ and

$\mid{z_1+z_2+z_3}\mid=2$ , then the value of

$\mid{4z_2z_3+9z_3z_1+16z_1z_2}\mid$ .

0%
24
0%
48
0%
96
0%
120

Explanation

$z_1\bar{z_1}=|z_1|^2=1$

$,z_2\bar{z_2}=|z_2|^2=9$

$,z_3\bar{z_3}=|z_3|^2=6$

$|4z_2z_3+9z_3z_1+16z_1z_2|$

$\implies |\bar{z_1}z_1z_2z_3$

$+\bar{z_2}z_1z_2z_3$

$+\bar{z_3}z_1z_2z_3|$

$\implies |z_1||z_2||z_3||\bar{z_1}+\bar{z_2}+\bar{z_3}|$

$\implies 24\times$

$|\bar{z_1}+\bar{z_2}+\bar{z_3}|$

$\implies 48$

Evaluate:

${ \left( \dfrac { cos\dfrac { \pi }{ 8 } -isin\dfrac { \pi }{ 8 } }{ cos\dfrac { \pi }{ 8 } +isin\dfrac { \pi }{ 8 } } \right) }^{ 4 }$

0%
$1$
0%
$-1$
0%
$2$
0%
$\dfrac { 1 }{ 2 }$

Explanation

We know that

$e^{i\theta}=cos\theta+isin\theta$ and

$e^{-1\theta}=cos\theta-isin\theta$

${ \left( \dfrac { cos\dfrac { \pi }{ 8 } -isin\dfrac { \pi }{ 8 } }{ cos\dfrac { \pi }{ 8 } +isin\dfrac { \pi }{ 8 } } \right) }^{ 4 }$

$=\left(\dfrac{e^{-i(\pi/8)}}{e^{i(\pi/8)}}\right)^4=(e^{-i(\pi/4)})^4=e^{-i\pi}=cos\pi-isin\pi=-1$

Let

$z_1$ and

$z_2$ are two complex numbers such that

$(1-i)z_1=2z_2$ and

$arg(z_1z_2)=\dfrac{\pi}{2}$ then

$arg(z_2)$ is equals to:

0%
$\dfrac{3 \pi}{8}$
0%
$\dfrac{\pi}{8}$
0%
$\dfrac{5 \pi}{8}$
0%
$\dfrac{-7 \pi}{8}$

Explanation

$(1-i)z_1=2z_2$

$\cfrac{z_2}{z_1}=\cfrac{1}{2}-\cfrac{i}{2}$

Let

$z_1=r_1e^{i\theta_1}\ and\ z_=r_1r^{\theta_2}$

$arg(\cfrac{z_2}{z_1})=\tan^{-1}\cfrac{-1/2}{1/2}=-\pi/4$

$\theta_2-\theta_1=-\pi/4$ and

$arg(z_1z_2)=\pi/2$ (given)

$\implies \theta_2-\theta_1=-\pi/4$ and

$\theta_2 +\theta_1=\pi/2$

$\implies 2\theta_2=\pi/4,\theta_2=\pi/8$

$\implies arg(z_2)=\pi/8$

Let $z$ be a complex number such that $\left| z+\dfrac { 1 }{ z } \right| =2$ .

If $\left| z \right| ={ r }_{ 1 }$ and $\left| \dfrac { 1 }{ z } \right| =$ ${r}_{2}$ for $\arg z=\dfrac { \pi }{ 4 }$ then

$\left| { r }_{ 1 }-{ r }_{ 2 } \right| =$

0%
$\dfrac { 1 }{ \sqrt { 2 } }$
0%
$1$
0%
$\sqrt { 2 }$
0%
$2$

Explanation

$\left| z+\dfrac { 1 }{ z } \right| ^{ 2 }={ r }^{ 2 }+\dfrac { 1 }{ { r }^{ 2 } } +2r.\dfrac { 1 }{ r } \cos { \left( \theta +\theta \right) } =4$
or

${ r }^{ 2 }+\dfrac { 1 }{ { r }^{ 2 } } =4-2\cos { 2\theta }$

$\therefore\quad \left( r-\dfrac { 1 }{ r } \right) ^{ 2 }=2\left( 1-\cos { 2\theta } \right) =4\sin { ^{ 2 }\theta }$

$\left| { r }_{ 1 }-{ r }_{ 2 } \right| =\left| r-\dfrac { 1 }{ r } \right| =2\sin { \theta } =2\sin { \dfrac { \pi }{ 4 } } =\sqrt { 2 }$

${z}_{1}=1+2i,\ {z}_{2}=2+3i,\ {z}_{3}=3+4i$ , then

${z}_{1},\ {z}_{2}$ and

${z}_{3}$ are collinear.

0%
True
0%
False

Explanation

Line formula from

$z_3$ and

$z_1$

$(y-y_1)=\cfrac{y_3-y_1}{x_3-x_1}$

$\times(x-x_1)$

$\implies (y-2)=\cfrac{(4-2)(x-1)}{3-1}$

$\implies y-2=x-1\\ \implies y=x+1$

$z_2=(2,3)$

$y=x+1$ satisfies

$z_2$ .

Hence

$z_1,z_2$ and

$z_3$ are collinear.

${z_1}$ and

${z_2}$ are two non-zero complex number such that

$\left| {{{{z_1}} \over {{z_2}}}} \right|$ = 2 and

$\arg \left( {{z_1}{z_2}} \right) = {{3\pi } \over 2}$ , then

${{\overline {{z_1}} } \over {{z_2}}}$ is equal to

0%
2i
0%
-2
0%
-2i
0%
2

Explanation

$Let\quad { z }_{ 1 }=2r{ e }^{ i{ \theta }_{ 1 } },{ z }_{ 2 }=r{ e }^{ i{ \theta }_{ 2 } }\\ arg\left( { z }_{ 1 }{ z }_{ 2 } \right) ={ \theta }_{ 1 }+{ \theta }_{ 2 }=\cfrac { 3\pi }{ 2 } \\ \bar { { z }_{ 1 } } =2r{ e }^{ -i{ \theta }_{ 1 } }\\ \cfrac { \bar { { z }_{ 1 } } }{ { z }_{ 2 } } =\cfrac { 2r{ e }^{ -i{ \theta }_{ 1 } } }{ r{ e }^{ i{ \theta }_{ 2 } } } =2r{ e }^{ -i({ \theta }_{ 1 }+{ \theta }_{ 2 }) }=2r{ e }^{ -i\cfrac { 3\pi }{ 2 } }=2i$

When will the quadratic equation

$ax^2+bx+c=0$ have Real Roots?

0%
$b^2- 4ac \ge 0$
0%
$b^2-4ac \le 0$
0%
$b^2-4ac < 0$
0%
None of the above.

Explanation

We learnt that the quadratic equation

$ax^2+bx+c=0$ will have Real Roots when

$b^2-4ac0$ Option

$a$ is correct.

If the value of ' $b^2-4ac$ 'is equal to zero, the quadratic equation $ax^2+bx+c=0$ will have

0%
Two real roots which are equal
0%
Two Distinct Real Roots.
0%
No Real Roots.
0%
No Roots or Solutions.

Explanation

$ax^{2}+bx+c=0$

For roots,

$D \equiv b^{2}-4ac$

and given that

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

$\Rightarrow D=0$

SO, equation will have only one root as,

$x=\cfrac{-b\pm\sqrt{b^{2}-4ac}}{2(a)}$

$x=\cfrac{-b\pm 0}{2a}$

$x=\cfrac{-b}{2a}$ Only one real root.

If z satisfies

$\left| {z - 1} \right| < \left| {z + 3} \right|$ then

$w = 2z + 3 - i$ , ( where

$w = 2z + 3 - i$ ) satisfies:

0%
$\left| {w - 5 - i} \right| < \left| {w + 3i} \right|$
0%
$\left| {w - 5} \right| < \left| {w + 3} \right|$
0%
$\left( {iw} \right) > 1$
0%
$\left| {\arg \left( {w - 1} \right)} \right| < {\pi \over 2}$

If the equation

${ x }^{ 2 }+nx+n=0,n\epsilon I$ , has integral roots then

${ n }^{ 2 }-4n$ can assume

0%
no integral value
0%
one integral value
0%
two integral value
0%
three integral value

Explanation

${ x }^{ 2 }+nx+n=0,n\epsilon 1$ 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

$p=0$

Real part of

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

0%
$-1/5$
0%
$1/5$
0%
$1/10$
0%
$-1/10$

Explanation

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

$-\dfrac{1}{5}$ .

$\dfrac{2z_1}{3z_2}$ is a purely imaginary number,then

$\left|\dfrac{z_1-z_2}{z_1+z_2}\right|=$

0%
$3/2$
0%
$1$
0%
$2/3$
0%
$4/9$

Explanation

$\dfrac{2z_1}{3z_2}=k'i$

$\dfrac{z_1}{z_2}=\dfrac{3k'i}{2}$

$\dfrac{z_1}{z_2}=ki$ where

$\dfrac{3k'}{2}=k$

$z_1=kiz_2$

Thus

$\left|\dfrac{z_1-z_2}{z_1+z_2}\right|=\left|\dfrac{kiz_2-z_2}{kiz_2+z_2}\right|$

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

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

$=\dfrac{\sqrt{k^2+1^2}}{\sqrt{k^2+1^2}}=1$

The value of

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

0%
$- 1 - i$
0%
$- 1 + i$
0%
$1 - i$
0%
$1 + i$

Explanation

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,

$a={ i }^{ 0 }=1$

$r=i$

$n={ 102 }$

${ i }^{ 102 }t=\cfrac { 1\left( { i }^{ 102 }-1 \right) }{ i-1 } =\cfrac { { i }^{ 2 }-1 }{ i-1 } =\cfrac { -2\left( i+1 \right) }{ \left( i-1 \right) \left( i+1 \right) }$

${ i }^{ 102 }t=i+1\Rightarrow { i }^{ 2 }t=i+1$

$t=-i-1$

Find the real number

$x$ if

$(x-2i)(1+i)$ is purely imaginary.

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

Explanation

$(x-2i)(1+i)$

$= x+ix -2i-2i^2$

$= (x+2) -i(x-2)$ is purely imaginary if (x+2)=0

$\therefore x=-2$

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

0%
$2i\log (x-i)-i\log (x^2+1)$
0%
$2i\log (x-i)+i\log (x^2+1)$
0%
$2i\log (x+i)-3i\log (x^2+1)$
0%
$2i\log (x-i)-i\log (x^2+i)$

Explanation

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

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

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

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

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

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

$i^2= -1$ , then

$1+ i^2+ i^4 +i^6+i^8 +.............to ( 2n +1)$ terms is equal to

0%
$0$
0%
$1$
0%
$3i$
0%
$4i$

Explanation

$1+ i^2 + i^4 + ......(2n+1)terms$

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

$i^4 = 1$

$=1$

If roots of equation

$2{x}^{2}+bx+c=0;b,c\in R$ , are real & distinct then the roots of equation

$2{cx}^{2}+(b-4c)x+2c-b+1=0$ are

0%
imaginary
0%
equal
0%
real and distinct
0%
cant say

Explanation

$1$ st Equation is

$2x^2 + bx + c = 0$

$\to$ root are real distinct

$\therefore D > 0$

$b^2 - 4ac>0$

$b^2 - 8c > 0$ ........ (1)

$2$ nd equation is

$2cx^2 + (b-4c)x+2c-b+1=0$

To find the nature of roots, calculate

$=b^2 - 4ac$

$D = (b-4c)^2 - 4(2c)(2c-b+1)$

$D = b^2-8c$

From (1)

It is +ve, so, roots are real and distinct.

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

0%
${\sec ^{ - 1}}2$
0%
${\cot ^{ - 1}}0$
0%
${\sin ^{ - 1}}\left( {\dfrac{{\sqrt 3 - 1}}{{2\sqrt 2 }}} \right)$
0%
${\cos ^{ - 1}}\left( {\dfrac{{ - 1}}{2}} \right)$

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:

0%
$\dfrac{1}{9}$
0%
$\dfrac{2}{9}$
0%
$\dfrac{3}{9}$
0%
$\dfrac{4}{9}$

Explanation

$x^2+4x+c=0$

For roots to be real

$0\ge 0$

$16-4c\ge c$

$4\ge c$

$\therefore \{1,2,3,4\}$

$\therefore$ Probability

$=\cfrac{4}{9}$

$a,b,c$ are integers and

${ b }^{ 2 }=4\left( ac+{ 5d }^{ 2 } \right)$ ,

$\in$ , then roots of the quadratic equation

${ ax }^{ 2 }+bx+c=0$ are

0%
Irrational
0%
Rational and different
0%
Complex conjugate
0%
Rational and equal

$a+ ib= \sum_{k=1}^{101} i^k$ , then

$(a, b)$ equals

0%
$(0, 1)$
0%
$(1, 0)$
0%
$(0,- 1)$
0%
$(1, 1)$

Explanation

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,

$\left( a, b \right) = \left( 0, 1 \right)$ .

$z = -3- i,$ find

$|z|$ .

0%
$\sqrt{10}$
0%
$\sqrt{9}$
0%
$\sqrt{8}$
0%
$\sqrt{7}$

Explanation

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

The modulus of the complex number

$z=\dfrac{1-i}{3-4i}$ is

0%
$\dfrac{5}{\sqrt{2}}$
0%
$\dfrac{\sqrt{2}}{5}$
0%
$\sqrt{\dfrac{2}{5}}$
0%
none of these

Explanation

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

$|z|=\sqrt{\left(\dfrac{7}{25}\right)^2+\left(\dfrac{1}{25}\right)^2}=\dfrac{\sqrt{50}}{25}=\dfrac{\sqrt{2}}{5}$ .

(A) Which of the following quadratic polynomials has zeros
-9 and 9

0%
$x-81=0$
0%
${x^2} - 9=0$
0%
${x^2} - 64=0$
0%
${x^2} - 81=0$

Let z be a complex number such that

$z\in c\ R$ and

$\dfrac{1+z+z^2}{1-z+z^2}\in R$ , then

$|z|=3$ .

0%
True
0%
False

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

$z^{1929}$ is

0%
$1+i$
0%
$-1$
0%
$\dfrac{1+i}{2}$
0%
$\dfrac{1+i}{\sqrt{2}}$

Explanation

Let

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

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

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

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

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

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

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

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

$\therefore z^{1929}=\dfrac{1+i}{\sqrt{2}}$

The equation

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

0%
$(-1, 3)$
0%
$(-3, 1)$
0%
$[-2, 2]$
0%
$R-(-2, 2)$

Explanation

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.

1425708_1058027_ans_ad2e9646f17a4c59a5b88b3f6e8e79af.png

$|Z|=2,|z_{2}|=3,|z_{3}=4|$ and

$|z_{1}+z_{2}+z_{3}|=5$ then

$|4z_{2}z_{3}+9z_{3}z_{1}+16z_{1}z_{2}|=$

0%
$20$
0%
$24$
0%
$48$
0%
$120$

Explanation

Solution:- (D) 120

$\left| {z}_{1} \right| = 2, \left| {z}_{2} \right| = 3, \left| {z}_{3} \right| = 4, \left| {z}_{1} + {z}_{2} + {z}_{3} \right| = 5$

$\left| 4 {z}_{2} {z}_{3} + 9 {z}_{3} {z}_{1} + 16 {z}_{1} {z}_{2} \right| = \cfrac{\left| {z}_{1} \right| \left| {z}_{2} \right| \left| {z}_{3} \right|}{\left| {z}_{1} \right| \left| {z}_{2} \right| \left| {z}_{3} \right|} \left| 4 {z}_{2} {z}_{3} + 9 {z}_{3} {z}_{1} + 16 {z}_{1} {z}_{2} \right|$

$= \left| {z}_{1} \right| \left| {z}_{2} \right| \left| {z}_{3} \right| \left| \cfrac{4}{{z}_{1}} + \cfrac{9}{{z}_{2}} + \cfrac{16}{{z}_{3}} \right|$

$= \left( 2 \times 3 \times 4 \right) \left| \cfrac{4 \bar{{z}_{1}}}{{\left| {z}_{1} \right|}^{2}} + \cfrac{9 \bar{{z}_{2}}}{{\left| {z}_{2} \right|}^{2}} + \cfrac{16 \bar{{z}_{3}}}{{\left| {z}_{3} \right|}^{2}}\right|$

$= 24 \left| \cfrac{4 \bar{{z}_{1}}}{{2}^{2}} + \cfrac{9 \bar{{z}_{2}}}{{3}^{2}} + \cfrac{16 \bar{{z}_{3}}}{{4}^{2}} \right|$

$= 24 \left| \cfrac{4 \bar{{z}_{1}}}{4} + \cfrac{9 \bar{{z}_{2}}}{9} + \cfrac{16 \bar{{z}_{3}}}{16} \right|$

$= 24 \left| \bar{{z}_{1}} + \bar{{z}_{2}} + \bar{{z}_{3}} \right|$

$= 24 \left| \overline{{z}_{1} + {z}_{2} + {z}_{3}} \right|$

$= 24 \times 5$

$= 120$

Hence,

$\left| 4 {z}_{2} {z}_{3} + 9 {z}_{3} {z}_{1} + 16 {z}_{1} {z}_{2} \right| = 120$ .

$\dfrac{x - 3}{3 + i} + \dfrac{y - 3}{3 - i} = i$ where

$x , y \in R$ then

0%
$x = 2$ & $y = -8$
0%
$x = -2$ & $y = 8$
0%
$x = -2$ & $y = -6$
0%
$x = 2$ & $y = 8$

Explanation

$\dfrac{x-3}{3+i}+\dfrac{y-3}{3-i}=i$

$\dfrac{(x-3)(3-i)}{10}+\dfrac{(y-3)(3+i)}{10}=i$

$3(x-3)+3(y-3)-i((x-3+(3-y))=10i$

$3(x+y-6)-i(x-y)=10i$

compare real and imaginary point

$x+y=6$

$y-x=10$

$\Rightarrow y=8, x=-2$

The complex number

$\dfrac{1 + 2i}{1 - i}$ lies in which quadrant of the complex plane.

0%
First
0%
Second
0%
Third
0%
Fourth

Explanation

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

$\Rightarrow \dfrac{1+2i}{1-i}\times \dfrac{1+i}{1+i}$

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

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

$\therefore$ It lies in

$2^{nd}$ Quadrant.

Given

$\left| z \right| =4$ and

$Argz=\dfrac{5z}{6}$ , then

$z$ is

0%
$2\sqrt{3}+2i$
0%
$2\sqrt{3}-2i$
0%
$-2\sqrt{3}+2i$
0%
$-\sqrt{3}+i$

Explanation

Given that

$\Rightarrow |z|=4$ and

$Arg \space z=\dfrac{5\pi}{6}$

$\Rightarrow$ Let

$z=x+iy$

$\Rightarrow |z|=4$

$\Rightarrow \sqrt{x^2+y^2}=4$

$\Rightarrow {x^2+y^2}=16$ ...

$(1)$

$\Rightarrow Arg \space z=\dfrac{5\pi}{6}$

$\Rightarrow \tan ^{-1}(\dfrac{y}{x})=\dfrac{5\pi}{6}$

$\Rightarrow \dfrac{y}{x}=\tan (\dfrac{5\pi}{6})$

$\Rightarrow \dfrac{y}{x}=-\dfrac{1}{\sqrt 3}$

$\Rightarrow x^2=3y^2$

Substituting this in

$(1)$ we get,

$\Rightarrow x^2+\dfrac{x^2}{3}=16$

$\Rightarrow \dfrac{4x^2}{3}=16$

$\Rightarrow x^2=12$

$\Rightarrow x=\pm 2\sqrt 3$

$\Rightarrow y=\mp 2$

$\Rightarrow \theta$ lies in

$2^{nd} Quadrant$

Hence,

$\Rightarrow z=-2\sqrt 3+2i$

$a < b < c < d$ , then the roots of the equation

$(x-a)(x-c)+2(x-b)(x-d)=0$ are?

0%
Real and distinct
0%
Real and equal
0%
Imaginary
0%
None of these

CBSE Questions for Class 11 Engineering Maths Complex Numbers And Quadratic Equations Quiz 7 - 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 7 - MCQExams.com

0:0:1

Practice Class 11 Engineering Maths Quiz Questions and Answers

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