MCQ Questions for Class 11 Engineering Maths Complex Numbers And Quadratic Equations Quiz 4

Let z be a complex number such that the imaginary part of z is nonzero and a =

$z^2 + z + 1$ is real. Then a cannot take the value

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

Explanation

Let

$z=\alpha+i\beta$ . Then

$Im(z^2+z+1)=2\alpha\beta+\beta=0$

$\Rightarrow \alpha=-\dfrac{1}{2}$ since

$\beta\neq0$ . Then

$a=\alpha^2-\beta^2+\alpha+1$

$a=\dfrac{3}{4}-\beta^{2}$

$a\neq\dfrac{3}{4}$ .

$[\because \beta\neq0]$

Complex number

$z$ satisfy the equation

$\left|z - (4/z)\right| = 2$ . Then the value of

$arg(z_1/z_2)$ , where

$z_1$ and

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

0%
2 $\pi$
0%
$\pi$
0%
$\dfrac {\pi}{2}$
0%
none of these

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

$2\theta = 2n\pi$ ,

$\theta = n\pi$

$\therefore arg\left(\dfrac{z_1}{z_2}\right)= \pm \pi$

Find the argument of

$sin \alpha + i(1 - cos \alpha), 0 < \alpha < \pi$

0%
$\displaystyle \frac{\alpha}{2}$
0%
$\displaystyle \frac{\alpha}{4}$
0%
$2\alpha$
0%
$\alpha$

Explanation

$z = sin \alpha + i (1 - cos \alpha)$

$\Longrightarrow amp(z) = tan^{-1}\left(\dfrac{1 - cos \alpha}{sin \alpha}\right) (\therefore$ z lies in the first quadrant)

$= tan^{-1}\left(\dfrac{2 sin^2 \dfrac{\alpha}{2}}{2 sin \dfrac{\alpha}{2} cos \dfrac{\alpha}{2}}\right) = tan^{-1} tan \left(\dfrac{\alpha}{2}\right) = \dfrac{\alpha}{2}$

Ans: A

Find the modulus and the principal argument of the complex number

$(\tan 1 - i)^2$

0%
$|z|=(\tan1)^2+1$ ,z lies in 4rd quadrant, $arg(z)=2-\pi/2$
0%
$|z|=(\tan1)^2+1$ ,z lies in 4rd quadrant, $arg(z)=2-\pi$
0%
$|z|=(\tan1)^2+1$ ,z lies in 3rd quadrant, $arg(z)=2-\pi/2$
0%
$|z|=(\tan1)^2+1$ ,z lies in 3rd quadrant, $arg(z)=2-\pi$

Explanation

$z = (\tan 1 - i)^2 = (\tan^2 1- 1) - (2 \tan 1)i$

$\left|z\right| = \sqrt{(\tan^21 - 1)^2 + 4 \tan^2 1}$

$= \sqrt{(\tan^21 + 1)^2}$

$= \tan^2 1 + 1$

Since,

$\tan^2 1 - 1 > 0$ and

$-2\tan 1 < 0$ .

$\Rightarrow z$ lies in the fourth quadrant

$\Rightarrow arg(z) = -\pi + \tan^{-1} \left|\dfrac{2 \tan 1}{1 - \tan^2 1}\right|$

$= -\pi + \tan^{-1} \left|\tan 2\right|$

$=2 - \pi$

Find the argument of

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

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

Explanation

Given,

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

Rationalizing the denominator, we get

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

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

$=\dfrac{2\sqrt{3}+2i}{4}$

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

$=\cos30^{0}+i\sin30^{0}$

Hence,

$Arg(Z)=30^{0}$

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

Find the minimum value of

$|z-1|$ if

0%
$\left|z - 1\right| \ge 0$
0%
$\left|z - 1\right| \ge 1$
0%
$\left|z - 1\right| \ge 2$
0%
$\left|z - 1\right| \ge 3$

Explanation

We know that,

$|z_1+z_2|⩽||z_1|+|z_2||$

Therefore,

$\Rightarrow \left|z - 1\right| \ge 1$

Ans: B

If z is a complex number, then find the minimum value of

0%
$E = 1$
0%
$E = 2$
0%
$E = 3$
0%
$E = 4$

Explanation

$\left|z_1 + z_2 + z_3\right|$ .

0%
the greatest value is 6.
0%
the greatest value is 7.
0%
the greatest value is 9.
0%
the greatest value is 12.

Explanation

Given,

$\left|z_1 + z_2 + z_3\right|$

Find the modulus and the principal argument of the complex number

$\displaystyle \frac{i - 1}{i\left(1 - cos \frac{2\pi}{5}\right) + sin \frac{2\pi}{5}}$

0%
$\displaystyle \frac{cosec(\frac{\pi}{5})}{\sqrt2}$ , $\frac{9\pi}{20}$
0%
$\displaystyle \frac{sin(\frac{\pi}{5})}{\sqrt2}$ , $\displaystyle \frac{11\pi}{20}$
0%
$\displaystyle \frac{cosec(\frac{\pi}{5})}{\sqrt2}$ , $\displaystyle \frac{11\pi}{20}$
0%
$\displaystyle \frac{sin(\frac{\pi}{5})}{\sqrt2}$ , $\displaystyle \frac{9\pi}{20}$

Explanation

$z\displaystyle = \frac{i - 1}{i\left(1 - cos \frac{2\pi}{5}\right) + sin \frac{2\pi}{5}}$

$\displaystyle = \frac{i-1}{i2 sin^2 \frac{\pi}{5} + 2 sin \frac{\pi}{5} cos \frac{\pi}{5}}$

$\displaystyle = \frac{i-1}{\left(2 sin \frac{\pi}{5}\right) \left(cos \frac{\pi}{5} + i sin \frac{\pi}{5}\right)}$

$\displaystyle \left|z\right|= \frac{\left|i-1\right|}{\left(2 sin\frac{\pi}{5}\right) \left|\left(cos \frac{\pi}{5} + i sin \frac{\pi}{5}\right)\right|}$

$\displaystyle = \frac{\sqrt{2}}{\left(2 sin \frac{\pi}{5}\right)\left|\left(cos \frac{\pi}{5} + i sin \frac{\pi}{5}\right)\right|}$

$\displaystyle = \frac{1}{\sqrt{2}} cosec \frac{\pi}{5}$

$\displaystyle arg z = arg \left[\frac{i-1}{\left(2 sin \frac{\pi}{5}\right)\left(cos \frac{\pi}{5} + i sin \frac{\pi}{5}\right)}\right]$

$\displaystyle = arg (-1 + i) - arg\left(2 sin \frac{\pi}{5}\right) - arg \left(cos\frac{\pi}{5} + i sin \frac{\pi}{5}\right)$

$\displaystyle = \frac{3\pi}{4} - 0 - \frac{\pi}{5} = \frac{11\pi}{20}$

Ans: B

$\displaystyle\ z=1+i\ \tan \alpha$ , where

$\displaystyle\ \pi < \alpha < \frac{3\pi }{2}$ is

$|z|$ is equal to

0%
$\displaystyle\ \sec \alpha$
0%
$\displaystyle\ -\sec \alpha$
0%
$\displaystyle\ cosec\alpha$
0%
none of these

Explanation

$z=1+i\tan\alpha$

$\Rightarrow |z| =\sqrt{1+\tan^2\alpha}=\sqrt{\sec^2\alpha}=|\sec\alpha|$
Now we know that, modulus of any complex number is always non-zero
and it is given that, alpha lies in third quadrant in which

$\sec\alpha <0$
Hence

$|z|=-\sec\alpha>0$
Hence option 'B' is correct choice.

Let

$\displaystyle\ z= \frac{\cos \theta +i\sin \theta }{\cos \theta -i\sin \theta }$ ,

$\displaystyle\ \frac{\pi}{4}< 0< \frac{\pi}{2}$ . Then arg z is

0%
$\displaystyle\ 2\theta$
0%
$\displaystyle\ 2\theta-\pi$
0%
$\displaystyle\ \pi +2\theta$
0%
None of these

Explanation

Converting to Euler's form, we get

$z=\dfrac{e^{i\theta}}{e^{-i\theta}}$

$=e^{i\theta-(-i\theta)}$

$=e^{i(2\theta)}$

$=cos2\theta+isin2\theta$

$\Rightarrow arg(z)=2\theta$
Hence, option 'A' is correct.

The value of the sum

$\displaystyle\ \sum _{n=1}^{13}\left ( i^{n}+i^{n+1} \right )$ , where

$\displaystyle\ i=\sqrt{-1}$

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

Explanation

$\sum (i^n+i^{n+1})=i+2(i^2+i^3+i^4.........i^{13})+i^{14}$

$=i+2(0)-1$ .........................(

$i^2=-1,i^3=-i,i^4=1$ )

$=i-1$

$\displaystyle \displaystyle\ |z_{1}-1|<1, |z_{2}-2|<2, |z_{3}-3|<3$ then

$\displaystyle\ |z_{1}+z_{2}+z_{3}|$

0%
$\displaystyle\$ is less than 6
0%
$\displaystyle\$ is more than 3
0%
$\displaystyle\$ is less than 12
0%
$\displaystyle\$ lies between 6 and 12

Explanation

Given the conditions in the question, the maximum values of

$z_1, z_2,z_3$ are 2, 4 and 6 respectively.
Thus, the maximum value of

$|z_1 + z_2 + z_3|$ becomes 12 when all the values are maximum simultaneously.
So,

$|z_1 + z_2 + z_3|$ is less than 12.

The equation

$\displaystyle x^{2}-6x+8+\lambda (x^{2}-4x+3)=0,\lambda \in R$ , has

0%
real and unequal roots for all $\lambda$
0%
real roots for $\lambda < 0$ only
0%
real roots for $\lambda > 0$ only
0%
real and unequl roots for $\lambda =0$ only

Explanation

$x^{ 2 }-6x+8+\lambda (x^{ 2 }-4x+3)=0$

$\Rightarrow \left( 1+\lambda \right) { x }^{ 2 }-\left( 6+4\lambda \right) x+8+3\lambda =0$
Now,

$D=4\left[ { \left( 3+2\lambda \right) }^{ 2 }-\left( 1+\lambda \right) \left( 8+3\lambda \right) \right] =4\left( { \lambda }^{ 2 }+\lambda +1 \right)$

Therefore,

$D=4\left( { \lambda }^{ 2 }+\lambda +1 \right) >0$ for all

$\lambda \in R$ .
Therefore, given equation has real and distinct roots.

Ans: A

${ a }^{ 3 }+{ b }^{ 3 }+{ c }^{ 3 }-3abc=0$ then the roots of the equation

$\left( { a }^{ 2 }-bc \right) { x }^{ 2 }+2\left( { b }^{ 2 }-ac \right) x+{ c }^{ 2 }-ab=0$ are

0%
imaginary
0%
real and unequal
0%
real and equal
0%
Cannot say

Explanation

Given quadratic equation is

$\left(a^2-bc\right)x^2+2\left(b^2-ac\right)x+\left(c^2-ab\right)=0$ and

$a^{ 3 }+b^{ 3 }+c^{ 3 }=3abc$

$D=4\left( b^{ 2 }-ac \right) ^{ 2 }-4\left( a^{ 2 }-bc \right) \left( c^{ 2 }-ab \right) \\ \Rightarrow D=4b\left( a^{ 3 }+b^{ 3 }+c^{ 3 }-3abc \right) = 0$

$\therefore$ Roots are real and equal.

$\displaystyle\ z=\frac{\sqrt{3}+i}{\sqrt{3}-i}$ then the fundamental amplitude of z is

0%
$\displaystyle\ -\frac{\pi}{3}$
0%
$\displaystyle\ \frac{\pi}{3}$
0%
$\displaystyle\ \frac{\pi}{6}$
0%
None of these

Explanation

Given,

$\displaystyle\ z=\frac{\sqrt{3}+i}{\sqrt{3}-i}$

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

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

$\Rightarrow \displaystyle z=\frac{2+2\sqrt{3}i}{4}, \because i^2=-1$

$\Rightarrow \displaystyle z=\frac{1+\sqrt{3}i}{2}$

$\therefore \text{amp(z)}=\tan^{-1}\sqrt{3}=\cfrac{\pi}{3}$

$l,m$ are real

$l\neq m$ then the roots of the equation

$\displaystyle (l-m)x^{2}-5(l+m)x-2(l-m)=0$ are

0%
real and equal
0%
non real complex
0%
real and unequal
0%
none of these

Explanation

$(l-m)x^{ 2 }-5(l+m)x-2(l-m)=0$

$D=25{ \left( l+m \right) }^{ 2 }+8{ \left( l-m \right) }^{ 2 }>0$
Therefore, the roots are real and unequal.

Ans: C

In the Argand's plane, the locus of

$z (\neq 1)$ such that

$arg\displaystyle \left\{\frac{3}{2}\left(\frac{2z^2-5z+3}{3z^2 -z-2}\right)\right\} = \frac{2\pi}{3}$ is

0%
a hyperbola with the directrices at $z = -3/2$ and $z= -2/3$ .
0%
an ellipse with the directrices at $z = 3/2$ and $z= 2/3$ .
0%
a segment of a circle subtending angle $\dfrac {2\pi}{3}$ on arc between points $z = -3/2$ and $z= 2/3$ lying below real axis.
0%
a segment of a circle subtending angle $\dfrac {2\pi}{3}$ on arc between points $z = 3/2$ and $z= -2/3$ lying above real axis.

Explanation

$arg\displaystyle \left\{\frac{3}{2}\left(\frac{2z^2-5z+3}{3z^2 -z-2}\right)\right\} = \frac{2\pi}{3}$
or

$arg\displaystyle \left\{\frac{3}{2} \frac{(z-1)(2z-3)}{(z-1)(3z+2)}\right\} = \frac{2\pi}{3}$
or

$arg\displaystyle \left\{\frac{3}{2} \left(\frac{2z-3}{3z+2}\right)\right\} = \frac{2\pi}{3}$
or

$\displaystyle arg\left(\frac{z-3/2}{z+2/3}\right) = \frac{2\pi}{3}$
Hence, z discribes the segment of a circle through z = 3/2 and z= -2/3 at which the chord subtends an angle

$2\pi/3$

Ans: D

Find the principal argument of the complex number

$sin \displaystyle \frac{6\pi}{5} + i\left(1 + cos \displaystyle \frac{6\pi}{5}\right)$ .

0%
$arg(z)=\displaystyle \frac{9\pi}{10}, \left|z\right| = -2 cos \displaystyle \frac{3\pi}{5}$
0%
$arg(z)=\displaystyle \frac{\pi}{10}, \left|z\right| = -2 cos \displaystyle \frac{3\pi}{5}$
0%
$arg(z)=\displaystyle \frac{9\pi}{10}, \left|z\right| = 2 cos \displaystyle \frac{3\pi}{5}$
0%
$arg(z)=\displaystyle \frac{9\pi}{10}, \left|z\right| = -2 cos \displaystyle \frac{2\pi}{5}$

Explanation

$z \displaystyle= sin \frac{6\pi}{5} + i\left(1 + cos \frac{6\pi}{5}\right)$

$\displaystyle= 2 cos \frac{3\pi}{5}\left(sin \frac{3\pi}{5} + i cos \frac{3\pi}{5}\right)$

$\displaystyle= 2 cos \frac{3\pi}{5}\left(cos \left(\frac{\pi}{2} - \frac{3\pi}{5}\right) + i sin \left(\frac{\pi}{2} - \frac{3\pi}{5}\right)\right)$

$\displaystyle= 2 cos \frac{3\pi}{5} \left[cos\left(\frac{-\pi}{10}\right) + i sin \left(\frac{\pi}{10}\right) \right]$

$\displaystyle= -2 cos \frac{3\pi}{5}\left[- cos \frac{\pi}{10} + i sin \frac{\pi}{10}\right]$

$\displaystyle= -2 cos \frac{3\pi}{5}\left[cos \left(\pi - \frac{\pi}{10}\right) + i sin \left(\pi - \frac{\pi}{10}\right)\right]$

$\displaystyle= -2 cos \frac{3\pi}{5}\left[cos \frac{9\pi}{10} + i sin \frac{9\pi}{10}\right]$

$\displaystyle \therefore arg (Z) = \frac{9\pi}{10}$ ;
here

$\displaystyle \left|z\right| = -2 cos \frac{3\pi}{5}$

Ans: A

$\displaystyle\ z_{1}\neq-z_{2}$ and

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

0%
at least one of $\displaystyle\ z_{1}, z_{2}$ is unimodular
0%
both $\displaystyle\ z_{1}, z_{2}$ are unimodular
0%
$\displaystyle\ z_{1}. z_{2} = 1$
0%
None of these

Explanation

Let

$z_{1}=x_{1}+iy_{1}$

$z_{2}=x_{2}+iy_{2}$
Now

$z_{1}+z_{2}=(x_{1}+x_{2})+i(y_{1}+y_{2})$

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

$=\bar{z_{1}}+\bar{z_{2}}$

$=(x_{1}+x_{2})-i(y_{1}+y_{2})$ .
Now

$|z_{1}+z_{2}|$

$=\sqrt{(x_{2}+x_{1})^2+(y_{2}+y_1)^{2}}$
And

$|\dfrac{1}{z_{1}}+\dfrac{1}{z_{2}}|$

$=|(x_{1}+x_{2})-i(y_{1}+y_{2})|$

$=\sqrt{(x_{2}+x_{1})^2+(y_{2}+y_1)^{2}}$
Hence

$|z_{1}+z_{2}|=|\dfrac{1}{z_{1}}+\dfrac{1}{z_{2}}|$ for all values of

$z_{1}$ and

$z_{2}$ .

$\displaystyle u_{i}=1-\frac{1}{i}$ then

$\displaystyle u_{2}\cdot u_{3}\cdot ... \cdot u_{n}$ is equal to

0%
$\displaystyle \frac{1}{n}$
0%
$\displaystyle \frac{1}{n!}$
0%
$1$
0%
none of these

Explanation

$u_i = 1-\cfrac{1}{i}=\cfrac{i-1}{i}$ , here

$i$ is not a complex number

$\displaystyle \therefore u_2.u_3...........u_{n-1}.u_n = \frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}.....\frac{n-2}{n-1}.\frac{n-1}{n} = \frac{1}{n}$

$\displaystyle z= 1+i\sqrt{3}$ ,then

$\displaystyle z^{6}$ equals

0%
32
0%
-32
0%
64
0%
None of these

Explanation

$\displaystyle z= 1+i\sqrt{3}$

$\displaystyle z^{6}= (1+\sqrt{3})^{6}$ using Binomial Theorem.

$\displaystyle = ^{6}\:C_{0}+^{6}\:C_{1}\:(i\sqrt{3})+^{6}\:C_{2}(i\sqrt{3})^{2}+^{6}\:C_{3}(i\sqrt{3})^{3}+^{6}\:C_{4}(i\sqrt{3})^{4}+^{6}\:C_{5}(i\sqrt{3})^{5}+^{6}\:C_{6}\:(i\sqrt{3})^{6}\:$

$\displaystyle = ^{6}\:C_{0}+^{6}C_{2}(3)+^{6}C_{4}(9)-^{6}C_{6}(27)+i\sqrt{3}(^{6}C_{1}-3^{6}C_{3}+9^{6}C_{5})$

$\displaystyle = (1-15\times 3+135-27)+i\sqrt{3}(60-60)$

$\displaystyle = (136-72)+i\sqrt{3}(0)= 64$

Find the value of

$k$ for which given equation has real and equal roots.

$(k\,-\,12)\,x^2\,+\,2\,(k\,-\,12)\,x\,+\,2\,=\,0$

0%
$k\,=\,12$
0%
$k\,=\,13$
0%
$k\,=\,14$
0%
$k\,=\,15$

Explanation

$(k - 12)x^2 - 2(k - 12)x + 2 = 0$
Since, the roots are real and equal. The discriminant is zero.

$D = b^2 - 4ac = 0$

$(-2(k - 12))^2 - 4(k-12)(2) = 0$

$4 ( k - 12)^2 -8k + 96 = 0$

$4k^2 - 96k + 576 - 8k + 96 = 0$

$4k^2 - 104k + 672 = 0$

$k^2 - 26k + 168 = 0$

$k^2 - 14k - 12k + 168 = 0$

$(k - 14)(k - 12) = 0$

$k = 12, 14$
Neglect

$k = 12$ because the equation will not be quadratic if we substitute this value of

$k$ .
Hence,

$k = 14$

Determine the nature of the roots of the following equations from their discriminants.

$y^2\,-\,4y\,-\,1\,=\,0$

0%
real and equal
0%
real and unequal
0%
complex
0%
Cannot be determined

Explanation

Given equation is,

$y^2 - 4y - 1 = 0$
We know,

$D = b^2 - 4ac$

$a=1, b=-4, c=-1$

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

$\therefore D = 20$
Since,

$D > 0$ hence, the roots are real and unequal.

$\text{arg(bi)}, \left( b>0 \right)$ is

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

Explanation

We know

$\text{arg(a+ib)}=\tan^{-1}\dfrac{b}{a}$

$\therefore \text{arg(ib)}=\tan^{-1}\dfrac{b}{0}=\tan^{-1}(+\infty),\left[\because b>0\right]=\cfrac{\pi}{2}$

The locus of

$\displaystyle z= x+iy$ which satisfying the inequality

$\displaystyle \log_{1/2}\left | z-1 \right |> \log_{1/2}\left | z-i \right |$ is given by

0%
$\displaystyle x+y< 0$
0%
$\displaystyle x-y> 0$
0%
$\displaystyle x-y< 0$
0%
$\displaystyle x+y> 0$

Explanation

$log_{0.5}|z-1|>log_{0.5}|z-i|$

$-log_{2}|z-1|>-log_{2}|z-i|$

$log_{2}|z-1|<log_{2}|z-i|$
Hence

$|z-1|<|z-i|$
Now
Let

$z=x+iy$
Hence by squaring both sides

$(x-1)^{2}+y^{2}<x^{2}+(y-1)^{2}$

$(2y-1)<(2x-1)$

$2(y-x)<0$

$y-x<0$
Or

$x-y>0$

$\left| { z }_{ 1 }+{ z }_{ 2 } \right| =\left| { z }_{ 1 } \right| +\left| { z }_{ 2 } \right|$ is possible if

0%
${ z }_{ 2 }=\overline { { z }_{ 1 } }$
0%
$\displaystyle { z }_{ 2 }=\frac { 1 }{ { z }_{ 1 } }$
0%
$arg{ z }_{ 1 }=arg{ z }_{ 2 }$
0%
$\left| { z }_{ 1 } \right| =\left| { z }_{ 2 } \right|$

Explanation

Given

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

Squaring,

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

$\Rightarrow \left| { z }_{ 1 } \right| ^{ 2 }+\left| { z }_{ 2 } \right| ^{ 2 }+2Re\left| { z }_{ 1 } \right| \left| { z }_{ 2 } \right| =\left| { z }_{ 1 } \right| ^{ 2 }+\left| { z }_{ 2 } \right| ^{ 2 }+2\left| { z }_{ 1 } \right| \left| { z }_{ 2 } \right|$

$\Rightarrow 2Re\left| { z }_{ 1 } \right| \left| \overline { { z }_{ 2 } } \right| =2\left| { z }_{ 1 } \right| \left| { z }_{ 2 } \right|$

$\Rightarrow \left| { z }_{ 1 } \right| \left| \overline { { z }_{ 2 } } \right| \cos { \left( { \theta }_{ 1 }-{ \theta }_{ 2 } \right) } =\left| { z }_{ 1 } \right| \left| { z }_{ 2 } \right|$

$\Rightarrow \cos { \left( { \theta }_{ 1 }-{ \theta }_{ 2 } \right) } =1\Rightarrow { \theta }_{ 1 }-{ \theta }_{ 2 }=0\Rightarrow { \theta }_{ 1 }={ \theta }_{ 2 }$

$\Rightarrow arg{ z }_{ 1 }=arg{ z }_{ 2 }$

If arg

$\displaystyle z < 0$ then arg

$\displaystyle \left (-z \right )-$ arg z is equal to

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

Explanation

Let

$z=e^{-i\theta}$

$=cos(\theta)-isin(\theta)$

$arg(z)=-\theta$

$arg(z)<0$ .
Now

$-z=-cos(\theta)+isin(\theta)$

$arg(-z)=\pi-\theta$
Hence

$arg(-z)-arg(z)$

$=\pi-\theta-(-\theta)$

$=\pi$
Hence, option 'A' is correct.

$\displaystyle z=1+i\cot\alpha,-\frac{\pi}{2}<\alpha<0,$ then

$|z|$ is equal to

0%
$cosec\alpha$
0%
$-cosec\alpha$
0%
$cosec\alpha$ or $-cosec\alpha$
0%
none of these

Explanation

$z=1+i\cot\alpha$

$\Rightarrow |z|=\sqrt{1+\cot^2\alpha}=\sqrt{co\sec^2\alpha}$

$=|co\sec\alpha|=-co\sec\alpha,$ as

$-\pi/2<\alpha<0$

A quadratic equation with rational coefficients can have

0%
both roots equal and irrational
0%
one root rational and other irrational
0%
one root real and other imaginary
0%
none of these

Explanation

$\textbf{Step 1: If a, b, c,}$

$\boldsymbol{\in Q \ \text{for} \ ax^2+bx+c=0}$ then

$(1)$ If

$D$

$\text{ is perfect square of a rational number then roots are rational }$

$(2)$ If

$D$

$\text{is not a perfect square of a rational number then roots are irrational }$

$\displaystyle \left [ \left ( \cos \theta +i \sin \theta \right )\left ( \cos \theta -i\sin \theta \right ) \right ]^{-1}$

0%
$\displaystyle i$
0%
$\displaystyle 1$
0%
$\displaystyle -i$
0%
$\displaystyle -1$

Explanation

$\displaystyle \left [ \left ( \cos \theta +i \sin \theta \right )\left ( \cos \theta -i\sin \theta \right ) \right ]^{-1}$

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

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

$\displaystyle \left ( e^{i\theta } .e^{-i\theta }\right )^{-1}=1$

Ans: B

The inequality

$|z-4| < |z-2|$ represents the region given by

0%
$Re(z) > 1$
0%
$Re(z) < 2$
0%
$Re(z) > 0$
0%
None of these

Explanation

Let,

$z=x+iy$

Putting in the given inequality,

$|(x-4)-iy|<|(x-2)+iy|$

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

squaring both sides,

$\Rightarrow (x-4)^2+y^2<(x-2)^2+y^2$

$\Rightarrow -8x+16<-4x+4$

$\Rightarrow 4x>12\Rightarrow x>3\Rightarrow \text{Re(z)}>3$

Solve:

$\displaystyle \left ( x+iy \right )\left ( 2-3i \right )= 4+i$

0%
$\displaystyle x= \left ( 8/13 \right ), y= -\left ( 14/13 \right ).$
0%
$\displaystyle x= \left ( 5/13 \right ), y= \left ( 14/13 \right ).$
0%
$\displaystyle x= -\left ( 14/13 \right ), y= \left ( 5/13 \right ).$
0%
$\displaystyle x= \left ( 14/13 \right ), y=- \left ( 8/13 \right ).$

Explanation

Given,

$\displaystyle \left ( x+iy \right )\left ( 2-3i \right )= 4+i$

$\displaystyle \left ( 2x+3y \right )+i\left ( -3x+2y \right )= 4+i.$

Equating real and imaginary parts,

$\displaystyle 2x+3y= 4, -3x+2y= 1.$

Solving these, we get

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

Ans: B

Solve :

$(2-\sqrt{-100})(1+\sqrt{-36})$

0%
$62+2i$
0%
$52+2i$
0%
$-52+2i$
0%
$-88+2i$

Explanation

$(2-\sqrt{-100})(1+\sqrt{-36})$

$=(2-10i)(1+6i)$

$=2+12i-10i+60$

$=62+2i$

Find arg(

$\displaystyle 1+\sqrt{2}+i$ )

0%
$\displaystyle \pi /16.$
0%
$\displaystyle \pi /8.$
0%
$\displaystyle \pi /12.$
0%
$\displaystyle \pi /10.$

Explanation

Let

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

now

$|Z|=\sqrt{4+2\sqrt{2}}$

$\therefore Z=\sqrt{4+2\sqrt{2}}(\dfrac{1+\sqrt{2}}{\sqrt{4+2\sqrt{2}}}+\dfrac{i}{\sqrt{4+2\sqrt{2}}})$

Hence

$tan\theta$

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

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

$=\sqrt{2}-1$

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

Determine the nature of roots of the following equation from the discriminant:

$y^{2}\, -\, 5y\, +\, 11\, =\, 0$

0%
Real and equal
0%
Real and unequal
0%
nonreal.
0%
None of these

Explanation

Discriminant of the equation

$y^2 - 5y + 11 =0$

$D = b^2 - 4ac$

$D = (-5)^2 - 4\times1\times11$

$= 25 - 44$

$= -19$
Since,

$D < 0$ thus, roots are not real.

Find the nature of the roots of the following quadratic equation. If the real roots exist, find them.

$\displaystyle 2x^2 - 3x + 5 =0$

0%
$x=0$ and $x=-2$
0%
$x=3, x=-6$
0%
No real root.
0%
None of these

Explanation

Given,

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

Comparing with the standard quadratic equation is

$ax^2+bx+c=0$

Here,

$a=2$ ,

$b=-3$ ,

$c=5$

Now,

$D=b^2 - 4ac$

$= 9 - 4 \times 2 \times 5$

$=-31$

Here, the discriminant is negative.

Thus the quadratic question does not have any real roots.

Find the value of

$p$ for which the quadratic equation

$x^2 + p(4x + p - 1) + 2 = 0$ has equal roots ?

0%
$\displaystyle -1, \frac{2}{3}$
0%
$3 , 5$
0%
$\displaystyle -1, \frac{4}{3}$
0%
$\displaystyle \frac{3}{4}, 2$

Explanation

Given equation is

$x^2 + p(4x + p - 1) + 2 = 0$ .
Comparing it with the

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

$a=1, b=4p, c=(p^2-p+2)$
Therefore,

$D=b^2-4ac$

$x^2 + p(4x + p - 1) + 2 = 0$ .

$\Rightarrow x^2+4xp+p^2-p+2=0$

$\Rightarrow (4p)^2-4(p^2-p+2)=0$

$\Rightarrow 16p^2-4p^2+4p-8=0$

$\Rightarrow 12p^2+4p-8=0$

$\Rightarrow 3p^2+p-2=0$

$\Rightarrow 3p^2+3-2p-2=0$

$\Rightarrow 3p(p+1)-2(p+1)=0$

$\Rightarrow (p+1)(3p-2)=0$

$\Rightarrow p=-1$ and

$p=\dfrac {2}{3}$

Find the values of

$k$ for each of the following quadratic equation, so that they have two real equal roots.

$\displaystyle 2x^2 + kx + 3 =0$

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

Explanation

Given,

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

Comparing with the standard quadratic equation is $ax^2+bx+c=0$

We get,

$a=2,b=k,c=3$

Given that the quadratic equation has two real equal roots.

Thus, the discriminant is

$=0$

Here,

$D = b^2 - 4ac =0$

$\Rightarrow k^2-4\times 2\times 3=0$

$\Rightarrow k^2-24=0$

$\Rightarrow k^2=24$

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

$\Rightarrow k=\pm\sqrt{2\times 2\times 2\times 3}$

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

The nature of the roots of the equation

$x^2 - 5x + 7 = 0$ is

0%
No real roots
0%
1 real root and 1 imaginary
0%
Real and unequal
0%
Real and equal

Explanation

Given equation is

$x^{2}-5x+7=0$
Hence,

$a=1,b=-5,c=7$
Therefore discriminant

$=D$

$=b^{2}-4ac$

$=25-4(7)$

$=25-28$

$=-3$
Hence,

$D<0$
Therefore, there are no real roots of this equation.

The equation

$x^2 - px + q = 0\ p, q \in R$ has no real roots if

0%
$p^2 > 4q$
0%
$p^2 < 4q$
0%
$p^2 = 4q$
0%
None of these

Explanation

Nature of the roots for a quadratic equation

$a^2+bx+c=0$ can be determined by its Discriminant,

$D=b^2-4ac$
Here,

$D=p^2-4q$
Since, the roots are not real
Therefore,

$D<0$

$= p^2-4q<0$

$=p^2<4q$

Find the roots of the following quadratic equation by using the quadratic formula

$4x^2 + 3x + 5 =0$

0%
$\displaystyle \frac { 3\pm \sqrt { 71 } i }{ 8 }$
0%
$\displaystyle \frac { 3\pm \sqrt { 89 } i }{ 8 }$
0%
$\displaystyle \frac { -3\pm \sqrt { 71 } i }{ 8 }$
0%
$\displaystyle \frac { -3\pm \sqrt { 89 } i }{ 8 }$

Explanation

Comparing

$4x^2 + 3x + 5 =0$ with

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

$a=4, b=3, c=5$
Now the roots are

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

$=\dfrac { -3\pm \sqrt { (-3)^{ 2 }-4\times 4\times 5 } }{ 2\times 4 }$

$=\dfrac { -3\pm \sqrt { 9-80 } }{ 8 }$

$=\dfrac { -3\pm \sqrt { 71 } }{ 8 }$

Thus, the quadratic equation does not have real roots.

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

$a, b, c$ are real, has real roots if

0%
$a, b, c$ are integers
0%
$b^2> 3ac$
0%
$ac > 0$
0%
$c = 0$

Explanation

Given equation

${a}{x}^{2} + bx + c =0$

Determinant of equation should be positive for its roots to be real i.e,

${b} ^{2} - 4ac >0$ but from the options we can see that only option D is right.

Because it reduces determinant to

$b^2$ which is always positive.

$p, q, r$ are real and

$\displaystyle p\neq q,$ then roots of the equation

$\displaystyle \left ( p-q \right )x^{2}+5\left ( p+q \right )x-2\left ( p-q \right )=0$ are

0%
Real and equal
0%
Complex
0%
Real and unequal
0%
None of these

Explanation

Discriminant

$=b^{2}-4ac$

$=25(p+q)^{2}+8(p-q)^{2}$
Hence,

$D\geq 0$
Hence, roots are real.
Also

$p$ and

$q$ are real.

$\therefore D\neq0$
Hence, roots are not equal.

The roots of

$a^2x^2 + abx = b^2, a \neq 0, b \neq 0$ are:

0%
Equal
0%
Non- real
0%
Unequal
0%
None of these

Explanation

Here, $a^2x^2 + abx = b^2, a \neq 0, b \neq 0$
$\Rightarrow a^2x^2 + abx-b^2=0$
Now, roots of the quadratic equation $=\dfrac { -ab\pm \sqrt { { (ab) }^{ 2 }-4{ a }^{ 2 }(-{ b }^{ 2 }) } }{ 2{ a }^{ 2 } }$
$=\dfrac { -ab\pm \sqrt { 5{ a }^{ 2 }{ b }^{ 2 } } }{ 2{ a }^{ 2 } }$
$=\dfrac { -ab\pm \sqrt { 5 } ab }{ 2{ a }^{ 2 } }$
$=\dfrac { -b\pm \sqrt { 5 } b }{ 2{ a } }$
Thus, the roots are real and unequal.

$-2$ is a root of the quadratic equation

$x^2 + px + 2 = 0$ and the quadratic equation

$2x^2 + px+ k = 0$ has equal roots, find the value of

$k$ .

0%
$\displaystyle k = \frac{8}{9}$
0%
$\displaystyle k = -\frac{8}{9}$
0%
$\displaystyle k = -\frac{9}{8}$
0%
$\displaystyle k = \frac{9}{8}$

Explanation

Given,

$-2$ is one of the root of first equation

$x^2 + px + 2 = 0$

Let's take

$\alpha$ as other root of the equation.

Now, product of the roots is given by

$(-2).\alpha = 2$

$\longrightarrow$

$\alpha = -1$ and from sum of roots we get

$-2 + (-1) = -p$

$\longrightarrow$

$p = 3$

Now in the second equation roots are equal, take it as

$\beta$ .

From sum of roots we get

$2.\beta = \dfrac{-p}{2}$

$\longrightarrow$

$\beta = \dfrac{-3}{4}$ and from product of the roots we get

${\beta}^{2} = \dfrac{k}{2}$

$\Rightarrow$

$k = 2.{\beta}^{2}$

$\Rightarrow$

$k = \frac{9}{8}$

$z = re^{i\theta}$ , then the value of

$|e^{iz}|$ is equal to

0%
$e^{rcos\theta}$
0%
$e^{-rcos\theta}$
0%
$e^{rsin\theta}$
0%
$e^{-rsin\theta}$

Explanation

Ginen,

$z=r{ e }^{ i\theta }$

To find

$\left| { e }^{ iz } \right|$

solution,

for given,

$z=r{ e }^{ i\theta }\\ z=r(\cos\theta +i\sin\theta )\left\{ { e }^{ i\theta }=(\cos\theta +i\sin\theta ) \right\} \\ iz=ir(\cos\theta +i\sin\theta )\\ iz=r(i\cos\theta +{ i }^{ 2 }\sin\theta )\\ iz=r(-\sin\theta +i\cos\theta )\\ iz=-r\sin\theta +ir\cos\theta \\ { e }^{ iz }={ e }^{ -r\sin\theta +ir\cos\theta }\\ \left| { e }^{ iz } \right| =\left| { e }^{ -r\sin\theta } \right| \left| { e }^{ ir\cos\theta } \right|$

We know that

$\left| { e }^{ iz } \right| =1\\ \therefore \left| { e }^{ iz } \right| =\left| { e }^{ -r\sin\theta } \right| \longrightarrow \{ always\quad positive\} \\ { e }^{ iz }={ e }^{ -r\sin\theta }$

Find the discriminant of the equation and the nature of roots. Also find the roots.

$6x^2 + x - 2 = 0$

0%
$\displaystyle \frac{1}{5}, \frac{-2}{3}$
0%
$\displaystyle \frac{1}{2}, \frac{-2}{3}$
0%
$\displaystyle \frac{1}{3}, \frac{-7}{3}$
0%
$D=49$ , Real and distinct roots: $\displaystyle \frac{1}{2}, \frac{-2}{3}$

Explanation

Nature of the roots of a quadratic equation is determined by its discriminant

$D=b^2-4ac$

Comparing

$6x^2 + x - 2 = 0$ with

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

$a=6, b=1, c=-2$

Therefore,

$D=b^2-4ac$

$=1^2-4\times 6\times (-2)$

$=1+48$

$=49$

$>0$

Therefore, the roots are real.

Therefore roots are

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

$=\cfrac { -(1)\pm \sqrt { (1)^ 2-4\times 6\times (-2) } }{ 2\times 6 }$

$=\cfrac { -1\pm \sqrt { 49 } }{ 12 }$

$=\cfrac { -1\pm 7 }{ 12 }$

Therefore,

$x=\cfrac { -1+ 7 }{ 12 }$ and

$x=\cfrac { -1- 7 }{ 12 }$

$x=\cfrac { 6 }{ 12 }$ and

$x=\cfrac { -8 }{ 12 }$

$x= \cfrac { 1 }{ 2 }$ and

$x=\cfrac { -2 }{ 3 }$

Find the modulus and amplitude of

$-2i$

0%
$|z|=2; amp(z)=-\dfrac {3\pi}{2}$
0%
$|z|=2i; amp(z)=\dfrac {\pi}{2}$
0%
$|z|=2; amp(z)=\dfrac {\pi}{2}$
0%
$|z|=2; amp(z)=-\dfrac {\pi}{2}$

Explanation

$z=-2i$

$|z|=2$

$=2(-i)$

$=2e^{i\frac{-\pi}{2}}$

$=|z|(e^{i\arg(z)})$

Hence

$|z|=2$ and

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

Find the modulus and amplitude of

$-2 + 2 \sqrt 3i$

0%
$|z|=2\sqrt [ 2 ]{ 2 } ; amp(z)=\dfrac {\pi}{3}$
0%
$|z|=2\sqrt [ 2 ]{ 2 } ; amp(z)=\dfrac {2\pi}{3}$
0%
$|z|=4; amp(z)=\dfrac {2\pi}{3}$
0%
$|z|=4; amp(z)=\dfrac {\pi}{3}$

Explanation

Given

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

Hence,

$|z|=2\sqrt{1+3}\ \Rightarrow 2\sqrt{4}$

$\Rightarrow 2\times2=4$
And let

$\alpha$ be the argument of the complex number.
Then

$tan\alpha=-\sqrt{3}$

Now

$Re(z)<0$ and

$Im(z)>0$

$\therefore$ the number lies in the II quadrant
Hence,

$\alpha=\pi-\dfrac{\pi}{3}$

$=\dfrac{2\pi}{3}$

CBSE Questions for Class 11 Engineering Maths Complex Numbers And Quadratic Equations Quiz 4 - MCQExams.com

0:0:1

Practice Class 11 Engineering Maths Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Maths Complex Numbers And Quadratic Equations Quiz 4 - MCQExams.com

0:0:1

Practice Class 11 Engineering Maths Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.