If $$z=\sqrt{\dfrac{1-i}{1+i}}$$ , then $$\arg \mathrm{z}=$$

$$-\displaystyle \dfrac{\pi}{4},\dfrac{3\pi}{4}$$

$$\displaystyle \dfrac{\pi}{4},\dfrac{\pi}{2}$$

$$-\displaystyle \dfrac{\pi}{4},\dfrac{\pi}{2}$$

$$\displaystyle \dfrac{3\pi}{4},\pi$$

MCQ Questions for Class 11 Engineering Maths Complex Numbers And Quadratic Equations Quiz 3

$z=2-i\sqrt{3}$ then

$z^{4}-4z^{2}+8z+35$ is :

0%
$6$
0%
$0$
0%
$1$
0%
$2$

Explanation

Given

$z = 2 - i \sqrt3,$

$\therefore$

$z - 2 = - i \sqrt3$
Squaring both sides,

$\therefore$

$z^2 - 4z + 4 = -3$

$\therefore$

$z^2 - 4z = -7$
Now

$z^4 - 4z^2 + 8z + 35 = (z^2 - 4z + 7)(z^2 + 4z + 5)$
Thus

$z^4 - 4z^2 + 8z + 35 = 0.$

$z_{1},\ z_{2}$ are roots of equation

$z^{2}-az+a^{2}=0$ , then

$|\displaystyle \frac{z_{1}}{z_{2}}|=$

0%
$a^{2}$
0%
a
0%
2a
0%
1

Explanation

$z_{1}$ and $z_{2}$ are roots of $z^{2}-az+a^{2}=0$
$\therefore z_{1},z_{2}=\dfrac{a\pm \sqrt{a^{2}-4a^{2}}}{2}$
$=a(\dfrac{1\pm i\sqrt{3}}{2})$
$\therefore z_{1} , z_{2}$ are multiple of complex cube roots of unity.
$\therefore |\dfrac{z_{1}}{z_{2}}|=|\dfrac{\omega ^{2}}{\omega }|=|\omega |=1$

$\log_{\frac{1}{2}}|\mathrm{z}-2|>\log_{\frac{1}{2}}|z|$ then

0%
$x>1$
0%
$x<1$
0%
$x<2$
0%
$x>2$

Explanation

$Let \quad z= x+iy$

$\log_{\frac{1}{2}}|\mathrm{z}-2|>\log_{\frac{1}{2}}|z|$

$\Rightarrow |\mathrm{z}-2| < |z|$

$\Rightarrow |x - 2 + iy| < |x + iy|$

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

$\Rightarrow -4x + 4 < 0$

$\Rightarrow x > 1.$

For n an integer, the argument of

$Z=\displaystyle \frac{(\sqrt{3}+i)^{4n+1}}{(1-i\sqrt{3})^{4n}}$ is

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

Explanation

$Z=\dfrac{(\dfrac{-1+\sqrt{3}i}{i})^{4n+1}}{(1-i\sqrt{3})^{4n}}$

$Z=\dfrac{(\dfrac{2\omega }{i})^{4n+1}}{(-2\omega )^{4n}}$

$=\dfrac{2\omega }{i^{4n+1}}\times (\dfrac{2\omega }{-2\omega })^{4n}$

$=\dfrac{2\omega }{i}\times 1$

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

$=\sqrt{3}+i$
$\therefore arg |z|=tan^{-1}(\dfrac{1}{\sqrt{3}})$
$=\dfrac{\pi }{6}$

The amplitude of

$1 + \cos x - i \sin x$ is

0%
$\displaystyle \frac{x}{2}$
0%
$\displaystyle x$
0%
$-\displaystyle \frac{x}{2}$
0%
$\displaystyle 2 x$

Explanation

$\displaystyle 1+\cos x- i \sin x$

$\displaystyle =2\cos^{2}\frac{x}{2}-2i \sin \frac{x}{2}\cos \frac{x}{2}$

$\displaystyle =2\cos \frac{x}{2}\left (\cos\frac{x}{2}-i \sin \frac{x}{2}\right)$

$\displaystyle =2\cos \frac{x}{2} e^{-i\frac{x}{2}}$

$\displaystyle \therefore$ amplitude

$\displaystyle=\frac{-x}{2}$

The region represented by z such that

$\left | \dfrac{\mathrm{z}-a}{z+a} \right |=1({\rm Im} (a) = 0)$ is

0%
$y=0$
0%
$x=0$
0%
$x+y=0$
0%
$x-y=0$

Explanation

Let

$z=x+iy$

$\Rightarrow \left | (x-a)+iy \right |=\left | (x+a)+iy \right |$

$\Rightarrow (x-a)^{2}+\not{ y^{2}} =(x+a)^{2}+\not{ y^{2}}$

$\Rightarrow \not{ x^{2}} +\not{ a^{2}} -2ax =\not{ x^{2}} +\not{ a^{2}} +2ax$

$\Rightarrow 0$

The principal argument of

$\displaystyle \frac{i-3}{i-1}$ is

0%
$\tan^{-1}\dfrac{1}{2}$
0%
$\tan^{-1}\dfrac{3}{2}$
0%
$\tan^{-1}\dfrac{5}{2}$
0%
$\tan^{-1}\dfrac{7}{2}$

Explanation

We have,

arg

$(i-3) = tan^{-1} \left (\dfrac{-1}{3} \right)$ &

arg

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

arg

$(i-3) -$ arg

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

The principal argument of

$\displaystyle \sqrt{2}[\cos\frac{5\pi}{3}+i\sin\frac{5\pi}{3}]$ is

0%
$\displaystyle \frac{5\pi}{3}$
0%
$\displaystyle \frac{\pi}{3}$
0%
$-\displaystyle \frac{\pi}{3}$
0%
$-\displaystyle \frac{\pi}{2}$

Explanation

$\sqrt{2}[cos(\dfrac{5\pi}{3})+isin(\dfrac{5\pi}{3})]$

$=\sqrt{2}[cos(-2\pi+\dfrac{5\pi}{3})+isin(-2\pi+\dfrac{5\pi}{3})]$

$=\sqrt{2}[cos(\dfrac{-\pi}{3})+isin(\dfrac{-\pi}{3})]$

$=\sqrt{2}.e^{-i\frac{\pi}{3}}$ .
Hence principal argument is

$-\dfrac{\pi}{3}$ .

$\left | \displaystyle \frac{1}{(1-i)^{2}}-\frac{1}{(1+i)^{2}}\right |=$

0%
4
0%
3
0%
2
0%
1

Explanation

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

$=\left | \dfrac{\not{1}+2i-\not{1}-(\not{1}-2i)-\not{1} }{(1^{2}+1^{2})^{2}}\right |$

$=\left | \dfrac{4i}{4} \right |=1$

The principal amplitude of

$(2-i)(1-2i)^{2}$ is in the interval

0%
$\left (0,\displaystyle \frac{\pi}{2} \right)$
0%
$\left (-\displaystyle \frac{\pi}{2},0 \right)$
0%
$\left (-\displaystyle \pi, -\frac{\pi}{2} \right)$
0%
$\left (-\displaystyle \frac{\pi}{2},\frac{\pi}{2} \right)$

Explanation

$(2-i)(1-2i)^2$

$= (2-i)( -3 -4i)$

$= -10 -5i$

This lies in the third quadrant. Hence, option C is correct.

The principal argument of

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

0%
$\displaystyle \frac{\pi}{3}$
0%
$\displaystyle \frac{\pi}{6}$
0%
$\displaystyle \frac{\pi}{8}$
0%
$\displaystyle \frac{\pi}{4}$

Explanation

For a complex number

$z=x+iy$ , the principal argument is given by

$tan^{-1}\left(\dfrac{y}{x} \right)$
For

$z=1+\sqrt{2}+i,$
arg

$(z)=tan^{-1}\left (\dfrac{1}{1+\sqrt{2}} \right)$

$=\dfrac{\pi }{8}$

$Z_{1},Z_{2}$ are two unimodular Complex numbers then

$\left |\displaystyle \frac{1}{Z_{1}}+\frac{1}{Z_{2}} \right|=$

0%
$1$
0%
$|Z_{1}+Z_{2}|$
0%
$|Z_{1}|+|Z_{2}|$
0%
$2$

Explanation

$\left |\dfrac{1}{z_1}+\dfrac{1}{z_2}\right|$

$=\left|\dfrac{z_1+z_2}{z_1z_2}\right|$

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

$= \dfrac{|z_1+z_2| }{1\times 1}$

$=|z_1+z_2|$

The real value of

$\theta$ for which the expression,

$\displaystyle \frac{1+i\cos\theta}{1-2i\cos\theta}$ is real number is

0%
$n\displaystyle \pi\pm\frac{\pi}{2}$
0%
$n\displaystyle \pi-\frac{\pi}{2}$
0%
$n\displaystyle \pi+\frac{\pi}{2}$
0%
$2n\displaystyle \pi\pm \cfrac {\pi}{2}$

Explanation

Let

$z=\cfrac{1+i\cos\theta}{1-2i\cos\theta}$

$=\cfrac{(1+i\cos\theta)(1+2i\cos\theta)}{(1-2i\cos\theta)(1+2i\cos\theta)}$

$=\cfrac{1-2\cos^{2}\theta+i3\cos\theta}{1+4\cos^{2}\theta}$

$=\cfrac{1-2\cos^{2}\theta}{1+4\cos^{2}\theta}+i\left (\dfrac{3\cos\theta}{1+4\cos^{2}\theta}\right)$
If the above complex number is purely real, then

$Im(z)=0$

$3\cos\theta=0$

$\theta=\cfrac{(2n\pm1)\pi}{2}$ , where

$n\epsilon N$

$\alpha,\ \beta,\ \gamma$ are modulus of the complex number

$3+4i, -5+12i,\ 1-i$ , then the increasing order for

$\alpha, \beta$ and

$\gamma$ is

0%
$\alpha,\ \gamma,\ \beta$
0%
$\alpha,\ \beta,\ \gamma$
0%
$\gamma,\ \alpha,\ \beta$
0%
can't be determined

Explanation

$\alpha =\sqrt { { 3 }^{ 2 }+{ 4 }^{ 2 } } =5$

$\beta =\sqrt{(5)^{2}+(12)^{2}}=13$

$\gamma =\sqrt{1+1^{2}}=\sqrt{2}$

$\Rightarrow \gamma < \alpha < \beta$

$\mathrm{l}\mathrm{n}$ a G.

$\mathrm{P}$ first term is

$\sqrt{3}+i$ and common ratio is

$\sqrt{3}-i$ then the modulus of the

$n^{th}$ term of the G.

$\mathrm{P}$ . is

0%
4
0%
$2^{n-1}$
0%
$2^{n}$
0%
$2^{n+1}$

Explanation

$a_{n}=(\sqrt{3+i})(\sqrt{3}-i)^{n-1}$

$\downarrow$ $\downarrow$

first ratio

$|a_{n}|=|\sqrt{3}+i|(|\sqrt{3-i}|)^{n-1}$

$[as |z^{n-1}|=(|z|)^{n-1}]$

$=(\sqrt{3+1})\times(\sqrt{3+1})^{n-1}$

$=2\times2^{n-1} = 2^n$

$z=\sqrt{\dfrac{1-i}{1+i}}$ , then

$\arg \mathrm{z}=$

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

Explanation

We have,

$\displaystyle z=\sqrt{\frac{1-i}{1+i}}=\sqrt{\frac{1-i}{1+i}\times \frac{1-i}{1-i}}=\sqrt{\frac{(1-i)^2}{1^2-i^2}}=\sqrt{\frac{(1-i)^2}{2}} [\because i^2 =-1]$

$\Rightarrow\displaystyle z =\pm {\frac{(1-i)}{\sqrt{2}}}=\dfrac{1-i}{\sqrt2},\frac{-1+i}{\sqrt2}$

$\dfrac{1-i}{\sqrt2}$ lies in fourth quadrant

$\Rightarrow\text{ar}(z)=\tan^{-1}(-1)=-\dfrac{\pi}{4}$
And

$\dfrac{-1+i}{\sqrt2}$ lies in second quadrant

$\Rightarrow\text{ar}(z)=\tan^{-1}(-1)=\dfrac{3\pi}{4}$
Hence option D is correct choice.

$\displaystyle \log_{(\frac{1}{3})}|z+1|>\log_{(\frac{1}{3})}|z-1|$ , then

0%
${\rm Re}(z)\geq 0$
0%
${\rm Re}(z)<0$
0%
${\rm Im}(z)>0$
0%
${\rm Im}(z)\leq 0$

Explanation

$\displaystyle \log(\frac{1}{3})|z+1|>\log(\frac{1}{3})|z-1|$

$\Rightarrow |z + 1| < |z - 1|$

$\Rightarrow (x^2 + 2x + 1 + y^2) < (x^2 - 2x + 1 + y^2)$

$\Rightarrow | 4x | < 0$
Thus,

$Re(z) < 0.$

$\dfrac{x+3i}{2+iy}=1-i$ , then the value of

$\left ( 5x-7y \right )^2$ is

0%
$1$
0%
$0$
0%
$2$
0%
$4$

Explanation

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

$x+3i=(1-i)(2+iy)$

$x+3i=2+iy-2i+y$

$x-y-2=i(y-5)$
Or

$(x-y-2)+i(5-y)=0$

Hence

$5-y=0$ or

$y=5$ and

$x-y-2=0$ or

$x-y=2$

Since

$y=5$

Hence

$x=y+2$ or

$x=5+2$

$x=7$

Therefore

$x=7$ and

$y=5$ .

Therefore

$(5x-7y)^{2}$

$=(5(7)-7(5))^{2}$

$=(35-35)^{2}$

$=0$

$z_{1},\ z_{2}$ are any two complex numbers then

$\left |z_{1^{+}}\sqrt{\mathrm{z}_{1}^{2}-\mathrm{z}_{2}^{2}} \right |+\left|\mathrm{z}_{1}-\sqrt{\mathrm{z}_{1}^{2}-\mathrm{z}_{2}^{2}} \right|$ is equal to

0%
$|\mathrm{z}_{1}|$
0%
$|\mathrm{z}_{2}|$
0%
$|z_{1}+\mathrm{z}_{2}|+|\mathrm{z}_{1}-\mathrm{z}_{2}|$
0%
$|\mathrm{z}_{1}+\mathrm{z}_{2}|-|\mathrm{z}_{1}-\mathrm{z}_{2}|$

Explanation

${ \left( \left| { z }_{ 1 }+\sqrt { { z }_{ 1 }^{ 2 }-{ z }_{ 2 }^{ 2 } } \right| +\left| { z }_{ 1 }-\sqrt { { z }_{ 1 }^{ 2 }-{ z }_{ 2 }^{ 2 } } \right| \right) }^{ 2 }$

$={ \left| { z }_{ 1 } \right| }^{ 2 }+{ \left| { z }_{ 1 } \right| }^{ 2 }-{ \left| { z }_{ 2 } \right| }^{ 2 }+{ \left| { z }_{ 1 } \right| }^{ 2 }+{ \left| { z }_{ 1 } \right| }^{ 2 }-{ \left| { z }_{ 2 } \right| }^{ 2 }+2{ \left| { z }_{ 1 } \right| }\left| \sqrt { { z }_{ 1 }^{ 2 }-{ z }_{ 2 }^{ 2 } } \right| +\left( -2\left| { z }_{ 1 } \right| \left| \sqrt { { z }_{ 1 }^{ 2 }-{ z }_{ 2 }^{ 2 } } \right| \right) +2\left( { \left| { z }_{ 1 } \right| }^{ 2 }-\left| { \left| { z }_{ 1 } \right| }^{ 2 }-{ \left| { z }_{ 2 } \right| }^{ 2 } \right| \right)$

$=4{ z }_{ 1 }^{ 2 }\longrightarrow \left( 1 \right)$ (On simplifying )

$\Rightarrow { \left( \left| { z }_{ 1 }+{ z }_{ 2 } \right| +\left| { z }_{ 1 }-{ z }_{ 2 } \right| \right) }^{ 2 }={ \left| { z }_{ 1 } \right| }^{ 2 }+{ \left| { z }_{ 2 } \right| }^{ 2 }+2\left| { z }_{ 1 } \right| \left| { z }_{ 2 } \right| +{ \left| { z }_{ 1 } \right| }^{ 2 }-2\left| { z }_{ 1 } \right| \left| { z }_{ 2 } \right| +\left( { \left| { z }_{ 1 } \right| }^{ 2 }-{ \left| { z }_{ 2 } \right| }^{ 2 } \right) ^{ 2 }$

$=4{ z }_{ 1 }^{ 2 }\longrightarrow \left( 2 \right)$ (On Simplifying )

$\Rightarrow (1)=(2)$

Hence, the answer is

$\left| { z }_{ 1 }+{ z }_{ 2 } \right| +\left| { z }_{ 1 }-{ z }_{ 2 } \right|.$

$z\neq 0$ , then

$\displaystyle \int_{0}^{50}\arg(-|z|)dx$ equals

0%
50
0%
not defined
0%
0
0%
$50\pi$

Explanation

$arg(-\left | z \right |)=arg(-ve\ real\ number)=\pi$
Hence

$\begin{matrix}50\\\int \\0 \end{matrix}arg(-\left | z \right |)dx=\begin{matrix}50\\\int \\0 \end{matrix}\pi\ dx$

$=50\pi$

Number of solutions of the equation

$|z|^{2}+7{z}=0$ is

0%
$1$
0%
$2$
0%
$4$
0%
$6$

Explanation

Let

$\ z=x+iy$
Then,

$\left | z \right |^{2}+7z=x^{2}+y^{2}+7(x+iy)=0$

$\Rightarrow x^{2}+y^{2}+7x=0$ &

$7y=0$

$\Rightarrow x^{2}+ 7x=0$ and

$y=0$

$\Rightarrow x=-7\ or\ x=0$ and

$y=0$
So, solutions are

$z=-7$ or

$z=0$
Hence, two solutions.

Which of the following equations have no real roots ?

0%
$x^2-2\sqrt{3}+5=0$
0%
$2x^2+6\sqrt{2}x+9=0$
0%
$x^2-2\sqrt{3}-5=0$
0%
$2x^2-6\sqrt{2}x-9=0$

Explanation

$\textbf{Step -1: Finding Discriminant of equation in option A.}$

$\text{The discriminant of any quadratic equation, }ax^2+bx+c=0, \text{ is }D=b^2-4ac.$

$\text{Discriminant of }x^2-2\sqrt3x+5=0\text{ is: }$

$D=(-2\sqrt3)^2-4\times1\times5$

$=12-20$

$=-8<0$

$\textbf{Step -2: Finding discriminant of equation in option B.}$

$\text{Discriminant of }2x^2+6\sqrt2x+9=0\text{ is: }$

$D=(6\sqrt2)^2-4\times2\times9$

$=72-72$

$=0$

$\textbf{Step -3: Finding discriminant of equation in option C.}$

$\text{Discriminant of }x^2-2\sqrt3x-5=0\text{ is: }$

$D=(-2\sqrt3)^2-4\times1\times-5$

$=12+20$

$=32>0$

$\textbf{Step -4: Finding discriminant of equation in option D.}$

$\text{Discriminant of }2x^2-6\sqrt2x-9=0\text{ is: }$

$D=(-6\sqrt2)^2-4\times2\times-9$

$=72+72$

$=144>0$

$\textbf{Step -5: Finding the correct option.}$

$\because \text{The equation in option A has discriminant less than zero.}$

$\Rightarrow \text{It has no real roots.}$

$\textbf{ Hence, option A is correct.}$

The values of

$k$ for which the equation

$2x^2 + kx + x + 8 = 0$ will have real and equal roots are

0%
$10$ and $-6$
0%
$7$ and $-9$
0%
$6$ and $-10$
0%
$-7$ and $9$

Explanation

From the given equation,

$a=2,\quad b=k+1,\quad c=8$
If an equation has real and equal roots, then,

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

${ (k+1) }^{ 2 }-4(2)(8)=0$

${ k }^{ 2 }+2k+1-64=0$

${ k }^{ 2 }+2k-63=0$
Solving for

$k$ ,

$k=\dfrac{-2\pm\sqrt{4-4.(-63)}}{2}$

${ k }=7$ and

${ k }=-9$
Option B is correct!

The equation

$2x^2 + 2(p + 1) x +p= 0$ , where

$p$ is real, always has roots that are

0%
Equal
0%
Equal in magnitude but opposite in sign
0%
Irrational
0%
Real

Explanation

The discriminant of a quadratic equation

$ax^2 + bx + c = 0$ is given by

$b^2-4ac$

$a=2,b=2(p+1)$ and

$c=p$

$[2(p+1)]^2-4(2p)\Rightarrow 4(p+1)^2-8p$

$\Rightarrow 4[(p+1)^2-2p]\Rightarrow 4[(p^2+2p+ 1)-2p]$

$\Rightarrow 4(p^2+1)$
For any real value of

$p,4(p^2+1)$ will always be positive as

$p^2$ cannot be negative for real

$p$ .
Hence, the discriminant

$b^2-4ac$ will always be positive.

When the discriminant is greater than

$0$ or is positive, then the roots of a quadratic equation will be real.

If the equation

$x^2- (2 + m) x + 1 (m^2 -4m + 4) = 0$ has coincident roots, then:

0%
$m=0$
0%
$m=6$
0%
$m=2$
0%
$m=\cfrac{2}{3}$

Explanation

If equation

$ax^{2}+bx+c=0$ , then the roots are equal (coincident) when

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

In equation

$x^{2}-(2+m)+1(m^{2}-4m+4)=0$

The a=1

$b=-(2+m)$

$c=m^{2}-4m+4$

Then

$b^{2}-4ac=\left \{ (-(2+m)) \right \}^{2}-4(1)(m^{2}-4m+4)=0$

$\Rightarrow m^{2}+4m+4-4m^{2}+16m-16=0$

$\Rightarrow -3m^{2}+20m-12=0\Rightarrow -3m^{2}+18m+2m+6=0$

$-3m(m-6)+2(m-6)=0\Rightarrow (-3m+2)(m-6)=0$

If -3m+2=0 then

$m=\frac{2}{3}$

$m-6=0 \ then \ m=6$

Let us consider a quadratic equation

$x^2+\lambda x+\lambda +1.25=0$ , where

$\lambda$ is a constant.

The value of

$\lambda$ such that the above quadratic equation has two coincident roots

0%
$\lambda =5$ or $\lambda=-1$
0%
$\lambda =1$ or $\lambda=5$
0%
$\lambda =-5$ or $\lambda=1$
0%
None of these

Explanation

The given equation is

$x^2+\lambda x+\lambda +1.25=0$

$a=1,b=\lambda,c=\lambda+1.25$

$b^2-4ac=\lambda^2-4\times 1.(\lambda+1.25)$

$=\lambda^2-4\lambda-5=(\lambda-5)(\lambda+1)$

The equation has two coincident roots if

$b^2-4ac=0$ ,

$(\lambda-5)(\lambda+1)=0$

$\Rightarrow$ Either

$\lambda-5=0,\lambda=5$

$\Rightarrow \lambda+1=0 \Rightarrow \lambda=-1$

$\therefore \lambda =5$ or

$-1$
Hence, the given equation has coincident roots for

$\lambda =5$ or

$-1$

If the equation

$(m^2+n^2)x^2-2(mp+nq)x+p^2+q^2=0$ has equal roots , then

0%
$mp = nq$
0%
$mq =np$
0%
$mn=pq$
0%
$mq=\sqrt{np}$

Explanation

For real and equal root, discriminant,

$D=0$

$b^2=4ac$

$\Rightarrow 4 (mp + nq)^2 = 4 (m^2 + n^2) (p^2 + q^2)$

$\Rightarrow m^2q^2 + n^2p^2- 2mnpq=0$

$\Rightarrow (mq-np)^2=0\Rightarrow mq-np=0$

$\therefore mq=np$

If the roots of the equation

$px^2+2qx+r=0$ and

$qx^2-2\sqrt{pr}x+q=0$ be real, then

0%
$p = q$
0%
$q^2=pr$
0%
$p^2=qr$
0%
$r^2=pq$

Explanation

Equation

$px^2 + 2qx + r = 0$ and

$qx^2-2\sqrt{pr}x+q=0$ have real roots, then

$b^2-4ac\geq0$ for both equations

From first

$4q^2 - 4pr\geq 0\Rightarrow q^2\geq pr$ ...............(1)
and from second

$4(pr)-4q^2\geq0$ (for real root)

$\Rightarrow pr\geq q^2$ ...............(2)

From (1) and (2), we get result

$q^2=pr$

Which of the following equations have no real roots ?

0%
$x^2 -2\sqrt{3}x+5=0$
0%
$-2x^2 +6\sqrt{2}+11=0$
0%
$x^2 -2\sqrt{3}x-5=0$
0%
$-2x^2 -6\sqrt{2}x-9=0$

Explanation

$\textbf{Step -1: Finding Discriminant of equation in option A.}$

$\text{The discriminant of any quadratic equation, }ax^2+bx+c=0, \text{ is }D=b^2-4ac.$

$\text{Discriminant of }x^2-2\sqrt3x+5=0\text{ is: }$

$D=(-2\sqrt3)^2-4\times1\times5$

$=12-20$

$=-8<0$

$\textbf{Step -2: Finding discriminant of equation in option B.}$

$\text{Discriminant of }-2x^2+6\sqrt2x+11=0\text{ is: }$

$D=(6\sqrt2)^2-4\times-2\times11$

$=72+88$

$=160>0$

$\textbf{Step -3: Finding discriminant of equation in option C.}$

$\text{Discriminant of }x^2-2\sqrt3x-5=0\text{ is: }$

$D=(-2\sqrt3)^2-4\times1\times-5$

$=12+20$

$=32>0$

$\textbf{Step -4: Finding discriminant of equation in option D.}$

$\text{Discriminant of }-2x^2-6\sqrt2x-9=0\text{ is: }$

$D=(-6\sqrt2)^2-4\times-2\times-9$

$=72-72$

$=0$

$\textbf{Step -5: Finding the correct option.}$

$\because \text{The equation in option A has discriminant less than zero.}$

$\Rightarrow \text{It has no real roots.}$

$\textbf{ Hence, option A is correct.}$

If the equation

$2x^2-6x+p=0$ has real and different roots, then the values of

$p$ are given by

0%
$p<\cfrac{9}{2}$
0%
$p\leq \cfrac{9}{2}$
0%
$p>\cfrac{9}{2}$
0%
$p\geq \cfrac{9}{2}$

Explanation

For having distinct real roots,

Discriminant,

$D = b^2-4ac>0\Rightarrow (-6)^2>4\cdot2\cdot p\Rightarrow 36>8p$

$\Rightarrow 8p<36\Rightarrow p<\dfrac{9}{2}$

Let us consider a quadratic equation

$x^2+3ax+2a^2=0$
If this equation has roots

$\alpha ,\beta$ and it is given that

$\alpha^2 +\beta ^2=5$ , then value of discriminant,

$D$ , for the above quadratic equation is

0%
$D>0$
0%
$D<0$
0%
$D=0$
0%
none of these

Explanation

$Ax^{2}+Bx+C=0$

Then

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

In given equation

$x^{2}+3ax+2a^{2}=0$

$A=1 ,B=3a$ and

$C=2a^{2}$

Then,

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

So,

$(3a)^2 - 4(2a^2) = 9a^2 -8a^2 = a^2$

Thus,

$a^2>0$

Then

$D> 0$

Solve the equation

$\left|z\right| = z + 1 + 2i$ .

0%
$3 - 2i$
0%
$2 - \displaystyle \frac{3}{2}i$
0%
$2 +\displaystyle \frac{3}{2}i$
0%
$\displaystyle \frac{3}{2} - 2i$

Explanation

$\left|z\right| = z + 1 +2i$

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

$\Longrightarrow \sqrt{x^2 +y^2} = x + 1$ and

$0 = 2 + y$ or

$y = -2$

$\Longrightarrow \sqrt{x^2 + 4} = x + 1$

$\Rightarrow x^2 + 4 = x^2 + 2x + 1$

$\Rightarrow 2x = 3$

$\therefore x = \dfrac{3}{2}$

$\Longrightarrow x + iy = \dfrac{3}{2} - 2i$

Ans: D

Both the roots of the given equation

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

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

Explanation

Given equation

$(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0$ can be written as

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

Now,

$D= B^2-4AC=4\left \{(a+b+c)^2-3(ab+bc+ca)\right \}$

$=4(a^2+b^2+c^2-ab-bc-ca)$

$=2(a^2+b^2-2ab+b^2+c^2-2bc+c^2+a^2-2ca)$

$=2\left \{(a-b)^2+(b-c)^2+(c-a)^2\right \} \geq 0$ as sum of squares is always greater than or equal to zero

$\Rightarrow D\ge 0$

Hence, both roots are always real .

$\sqrt {5-12i}+\sqrt {-5-12i}=z$ , then principal value of arg z can be

0%
$-\dfrac {\pi}{4}$
0%
$\dfrac {\pi}{4}$
0%
$\dfrac {3\pi}{4}$
0%
$-\dfrac {3\pi}{4}$

Explanation

${5-12i}=(3-2i)^2$

$-5-12i=(2-3i)^2$

$z=\sqrt{5-12i}+\sqrt{-5-12i}=\pm(3-2i)^{}\pm(2-3i)^{}$
Hence

$z=5-5i$ , arg

$= \displaystyle \frac{-\pi}{4}$

$z=1+i$ , arg

$= \displaystyle \frac{\pi}{4}$

$z=-1-i$ , arg

$= \displaystyle \frac{-3\pi}{4}$

$z=-5+5i$ , arg

$= \displaystyle \frac{3\pi}{4}$
Hence, all options are correct.

The nature of roots of

$x^2 + x + 1$ is

0%
real and equal
0%
real and unequal
0%
imaginary and distinct
0%
imaginary and equal

Explanation

$x^2+x+1=0$

Calculate D

$D=b^2-4ac$

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

$D=-3$

$D < 0$

So, roots are imaginary and distinct.

The greatest positive argument of complex number satisfying

$|z-4|=Re(z)$ is

0%
$\displaystyle \frac {\pi}{3}$
0%
$\displaystyle \frac {2\pi}{3}$
0%
$\displaystyle \frac {\pi}{2}$
0%
$\displaystyle \frac {\pi}{4}$

Explanation

$|z-4|=Re(z)$

$\Rightarrow \sqrt {(x-4)^2+y^2}=x$
or

$x^2-8x+16+y^2=x^2$
or

$y^2=8(x-2)$
The given relation represents the part of the parabola with focus

$(4, 0)$ lying above the

$x-axis$ and the imaginary axis as the directrix. The two tangents from directrix are at right angle. Hence, greatest positive argument of

$z$ is

$\dfrac {\pi}{4}$ .

$k+\mid k+z^2\mid =\mid z\mid^2 (k\epsilon R^-)$ , then possible argument of

$z$ is

0%
$0$
0%
$\pi$
0%
$\displaystyle \dfrac {\pi}{2}$
0%
None of these

Explanation

$|k+z^{2}|=|k|+|z^{2}|$
Hence

$k$ and

$|z^{2}|$ are collinear.
Now

$k\epsilon{R^{-}}$
Since,

$k$ and

$z^{2}$ is collinear, then

$z^{2}$ has to be purely real.
Thus,

$arg(z^{2})=(2n-1)\pi$
Now

$z=re^{i\theta}$

$z^{2}=r^{2}e^{2i\theta}=me^{2i\theta}$
Hence

$arg(z^{2})=2arg(z)$
Therefore,

$arg(z)=\dfrac{1}{2}arg(z^{2})$

$=\dfrac{(2n-1)\pi}{2}$
Putting

$n=1$ , we get

$arg(z)=\dfrac{\pi}{2}$
Hence, option 'C' is correct.

$b^2-4ac\geq 0$ , then the root of quadratic equation

$ax^2+bx+c=0$ is

0%
$\frac {b}{2a}\pm \frac {\sqrt {b^2-4a}}{2a}$
0%
$-\frac {b}{2a}\pm \frac {\sqrt {b^2+4ac}}{2a}$
0%
$\frac {b}{2a}\pm \frac {\sqrt {b^2+4ac}}{2a}$
0%
$-\frac {b}{2a}\pm \frac {\sqrt {b^2-4ac}}{2a}$

The principal value of arg z where

$z = 1 + \cos \displaystyle \frac{6 \pi}{5} + i \sin\frac{6 \pi}{5}$ is given by

0%
$\displaystyle \frac{3 \pi}{5}$
0%
$- \displaystyle \frac{ \pi}{5}$
0%
$- \displaystyle \frac{3 \pi}{5}$
0%
$\displaystyle \frac{ \pi}{5}$

Explanation

$1+cos\dfrac{6\pi}{5}+isin(\dfrac{6\pi}{5})$

$=2cos^{2}\dfrac{3\pi}{5}+i2sin\dfrac{3\pi}{5}.cos\dfrac{3\pi}{5}$

$=2cos(\dfrac{3\pi}{5})[cos\dfrac{3\pi}{5}+isin\dfrac{3\pi}{5}]$

$=2cos(\dfrac{3\pi}{5})[cos\dfrac{3\pi}{5}+isin\dfrac{3\pi}{5}]$ ....Since

$\dfrac{3\pi}{5}$ is in the first quadrant.

Find the complex numbers z which simultaneously satisfy the equation

$\displaystyle \left | \frac{z - 12}{z - 8 i} \right | = \frac{5}{3}$ and

$\displaystyle \left | \frac{z - 4}{z - 8} \right | = 1$ .

0%
6 + 8 i or 6 + 17 i
0%
6 + 8 i or 6 - 17 i
0%
6 - 8 i or 6 + 17 i
0%
6 - 8 i or 6 - 17 i

Explanation

Put

$z=x+iy$

$\displaystyle \left| \frac { z-4 }{ z-8 } \right| =1\Rightarrow \left| \frac { x-4+iy }{ x-8+iy } \right| =1\\ \Rightarrow { \left( x-4 \right) }^{ 3 }+{ y }^{ 2 }={ \left( x-8 \right) }^{ 2 }+{ y }^{ 2 }\\ \Rightarrow x=6$

With

$\displaystyle x=6,\left| \frac { z-12 }{ z-8i } \right| =\frac { 5 }{ 3 }$

$\displaystyle \Rightarrow \left| \frac { -6+iy }{ 6+i\left( y-8 \right) } \right| =\frac { 5 }{ 3 } \\ \Rightarrow 9\left( 36+{ y }^{ 2 } \right) =25\left[ 36+\left( y-8 \right) 2 \right] \\ \Rightarrow { y }^{ 2 }-25y+136=0\\ \Rightarrow y=17,8$
Hence the required numbers are

$z=6+17i,6+8i$

$i{ z }^{ 3 }+{ z }^{ 2 }-z+i=0$ , then

0%
$\left| z \right| <1$
0%
$\left| z \right| >1$
0%
$\left| z \right| =1$
0%
$\left| z \right| =0$

Explanation

Given

$i{ z }^{ 3 }+{ z }^{ 2 }-z+i=0\Rightarrow i{ z }^{ 2 }\left( z-i \right) -\left( z-i \right) =0$

$\Rightarrow \left( z-i \right) \left( i{ z }^{ 2 }-1 \right) =0\Rightarrow z=i$ or

$\displaystyle { z }^{ 2 }=\frac { 1 }{ i } =-i$

Now

$z=i\Rightarrow \left| z \right| =\left| i \right| =1$

and

${ z }^{ 2 }=-i\Rightarrow \left| { z }^{ 2 } \right| =\left| -i \right| \Rightarrow \left| { z }^{ 2 } \right| =1\Rightarrow \left| z \right| =1$

Thus, in both cases

$\left| z \right| =1$

Number of roots of the equation

$z^{10} - z^5 - 992 = 0$ where real parts are negative is

0%
$3$
0%
$4$
0%
$5$
0%
$6$

Explanation

We have,

$z^{10} - z^5 - 992 = 0$
Put

$z^5=u$

$\Rightarrow u^2-u-992=0$

$\Rightarrow u = \dfrac{1\pm\sqrt{1+ 4\times 992}}{2}$

$\quad =\dfrac{1\pm 63}{2}=32, -31$

$\Rightarrow z^5 =32, (-1)^531$

$\Rightarrow z=(32)^{1/5}, -(31)^5$
Hence number of roots of

$z$ having negative real parts is

$5$
corresponding to

$z^5 =-31$

For any integer

$n$ , the argument of

$\displaystyle z=\frac { { \left( \sqrt { 3 } +i \right) }^{ 4n+1 } }{ { \left( 1-i\sqrt { 3 } \right) }^{ 4n } }$ is

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

Explanation

We have,

$\displaystyle z=\frac { { \left( \sqrt { 3 } +i \right) }^{ 4n+1 } }{ { \left( 1-i\sqrt { 3 } \right) }^{ 4n } } =\frac { { \left( 2{ e }^{ i\frac { \pi }{ 6 } } \right) }^{ 4n+1 } }{ { \left( 2{ e }^{ -i\frac { \pi }{ 3 } } \right) }^{ 4n } }$

$\displaystyle=\frac { { 2 }^{ 4n+1 }.{ e }^{ i\left( 4n+1 \right) \frac { \pi }{ 6 } } }{ { 2 }^{ 4n }.{ e }^{ -i4n\frac { \pi }{ 3 } } }$

$\displaystyle =2.{ e }^{ i\left( 2n+1 \right) \frac { \pi }{ 6 } }\quad$

$\displaystyle=2.{ e }^{ 2n\pi i }.{ e }^{ \frac { \pi i }{ 6 } }=2.{ e }^{ \frac { \pi i }{ 6 } }\quad \quad \quad \left( \because { e }^{ 2n\pi i }=1 \right)$

$\displaystyle \therefore arg(z)=\frac { \pi }{ 6 }$

$z = x + iy$ and

$x^2 + y^2 = 16$ , then the range of

0%
[0, 4]
0%
[0,2]
0%
[2, 4]
0%
none of these

Explanation

$z=x+iy$
And

$|z|^{2}=16$

$|z|=4$
If z is purely real,

$x=\pm4$ and

$||x|-|y||=4$ ...(i)
If z is purely imaginary

$y=\pm4i$ and

$||x|-|y||=4$ ...(ii)
Now if

$|x|=|y|$
Thus

$2x^{2}=2y^{2}=16$

$x=y=\pm2\sqrt{2}$
Under such case

$||x|-|y||=0$
Hence the range of values of

$||x|-|y||$ is

$[0,4]$

For

$\left|z - 1\right| = 1$ , find tan

$[arg\dfrac{((z -1)}{(2 - 2 \frac{i}{z}}))].$

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

Explanation

Let

$z=x+iy$
Hence,

$|z-1|=1$

$(x-1)^{2}+y^{2}=1$
Hence by using parametric equation of circle.

$x-1=cos\theta$

$x=1+cos\theta$ and

$y=sin\theta$
Then,

$z=(1+cos\theta)+isin\theta$
Let

$\theta=\frac{\pi}{2}$
Then,

$z=1+i$ ...(i)
Hence,

$z-1=i$ and

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

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

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

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

$=1-i$ ...(ii)
Hence,

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

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

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

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

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

$=\dfrac{1}{\sqrt{2}}.e^{i\frac{3\pi}{4}}$
Hence,

$tan(\frac{3\pi}{4})$

$=-1$ .

Find the greatest and the least value of

$\left|z_1 + z_2\right|$ if

$z_1 = 24 + 7i$ and

$\left|z_2\right| = 6.$

0%
least value is 25, greatest value is 31
0%
least value is 19, greatest value is 31
0%
least value is 19, greatest value is 25
0%
least value is 13, greatest value is 25

Explanation

$\left|z_1 + z_2\right|$ is 19 and the greatest value is 31.

Ans: C

$(w - \overline{w}z)/(1-z)$ is purely real where

$w = \alpha + i\beta, \beta \neq 0$ and

$z \neq 1$ , then set of the values of

$z$ is

0%
${z : \left|z\right| = 1}$
0%
${z : z = \overline{z}}$
0%
${z : z \neq 1}$
0%
${z : \left|z\right| = 1, z \neq 1}$

Explanation

$\dfrac{(\omega - \overline{\omega} z)}{1-z}$ is purely real and

$z\ne 1$

$\dfrac{(\omega - \overline{\omega} z)}{1-z} = \dfrac{\overline{\omega -\overline{\omega} z}}{(\overline{1-z})}$

$\dfrac{(\omega - \overline{\omega} z)}{1-z} = \dfrac{\overline {\omega} - \omega\overline{z}}{1-\overline z }$

$(\omega - \overline{\omega} z)(1- \overline{z}) = (1-z)(\overline{\omega}-\omega\overline z)$

$\omega -\omega\overline z-\overline{\omega}z + \overline{\omega} |z|^2 = \overline{\omega}+\omega|z|^2 -\overline{\omega}z-\omega\overline z$

$\omega - \overline{\omega} = |z|^2 (\omega - \overline{\omega})$

$\therefore |z|^2 = 1$ (as

$Im(\omega) = \beta \ne 0$ )

$\Rightarrow |z| =1$ But

$z\ne 1$

$\therefore$ Set of values of

$z$ is

$z:|z| =1,z\ne1$

Complex number

$z$ satisfy the equation

$\left|z - (4/z)\right| = 2$ . Locus of z if

$\left|z - z_1\right| = \left|z - z_2\right|$ , where

$z_1$ and

$z_2$ are complex numbers with the greatest and the least moduli, is

0%
line parallel to the real axis
0%
line parallel to the imaginary axis
0%
line having a positive slope
0%
line having a negative slope

Explanation

Squaring the given equation we have,

$\left(z-\dfrac4z\right)\left(\bar{z}-\dfrac{4}{\bar{z}}\right) = 4$

$\therefore |z|^2 - 4\left(\dfrac{z}{\bar{z}}+\dfrac{\bar{z}}{{z}}\right)+\dfrac{16}{|z|^2} = 4$

Let,

$z = re^{i \theta}$

$\therefore r^2 - 4\left(e^{i2\theta}+e^{-i2\theta}\right)+\dfrac{16}{r^2} = 4$

$\therefore r^2 - 8\left(\cos2\theta\right)+\dfrac{16}{r^2} = 4$

$\therefore r^2 +\dfrac{16}{r^2} = 4(1 + 2\left(\cos2\theta\right))\ldots\ldots$ (1)

For greatest or least modulus as

$r = f(\theta)$ ,

$\dfrac{dr}{d\theta} = 0$ .

This gives

$\sin2\theta = 0$

$\therefore \cos2\theta = 1\ldots\ldots$ (

$\cos2\theta$ cannot be

$-1$ as it will not satisfy equation (1))

$\therefore r^2 +\dfrac{16}{r^2} = 12$

Solving for

$r$ , we get

$r = \sqrt5\pm1$

$r_{max} = \sqrt5+1$ and

$r_{min} = \sqrt5-1$

These numbers lie on real axis as

$2\theta = 2n\pi$ ,

$\theta = n\pi$

$|z-z_1| = |z-z_2|$ is locus of line equidistant from

$z_1, z_2$ which is perpendicular bisector. Hence, it is parallel to imaginary axis.

0%
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
0%
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
0%
Assertion is correct but Reason is incorrect
0%
Assertion is incorrect and Reason is correct

Explanation

$\displaystyle \left| { z }_{ 1 }{ +z }_{ 2 } \right| =\left| \frac { 1 }{ { z }_{ 1 } } +\frac { 1 }{ { z }_{ 2 } } \right| \quad$

$\displaystyle \quad \Longrightarrow \left| { z }_{ 1 }{ +z }_{ 2 } \right| =\left| \frac { { z }_{ 1 }+{ z }_{ 2 } }{ { z }_{ 1 }{ z }_{ 2 } } \right|$

$\quad \because \quad { z }_{ 1 }\neq { -z }_{ 2 }$

$\quad \therefore \quad \left| { z }_{ 1 }{ z }_{ 2 } \right| =1$
Therefore

${ z }_{ 1 }{ z }_{ 2 }$ is unimodular.
But

${ z }_{ 1 }$ &

${ z }_{ 2 }$ are not unimodular.

Ans: c

$(\sqrt{8} + i)^{50} = 3^{49}(a + ib)$ , then find the value of

$a^2 + b^2$ .

0%
$(a^2 + b^2) = 9$
0%
$(a^2 + b^2) = 27$
0%
$(a^2 + b^2) = 3$
0%
$(a^2 + b^2) = 1$

Explanation

Given that

$(\sqrt{8} + i)^{50} = 3^{49}(a + ib)$
Taking modulus and squaring both sides, we get

$(8 + 1)^{50} = 3^{98}(a^2 + b^2)$
or

$9^{50}=3^{98}(a^2 + b^2)$
or

$3^{100} = 3^{98} (a^2 + b^2)$
or

$(a^2 + b^2) = 9$

Ans: A

CBSE Questions for Class 11 Engineering Maths Complex Numbers And Quadratic Equations Quiz 3 - 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 3 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

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