If $$z$$ satisfies $$|z - 2+ 2i| \le 1$$

$$|z|_{least} = 2\sqrt{2} - 1$$

$$|z|_{least} = 2\sqrt{2} + 1$$

MCQ Questions for Class 11 Commerce Applied Mathematics Number Theory Quiz 9

What is the modulus of following complex number:

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

0%
4
0%
5
0%
2
0%
3

$a+ib=\dfrac { c+1 }{ c-1 }$ , where

$c$ is real number, then

${ a }^{ 2 }+{ b }^{ 2 }=1$ and

$\dfrac { b }{ a } =\dfrac { 2c }{ { c }^{ 2 }-1 }$

0%
True
0%
False

Which of the following is a pair of twin-prime number ?

0%
$19 , 21$
0%
$43 , 47$
0%
$59 , 61$
0%
$73 , 79$

Explanation

The even numbers between

$68$ and

$90$ is

$70 , 72 , 74 , 76 , 78 , 80 , 82 , 84 , 86 , 88 , = 10$ numbers

$z=(3+7i)(p+iq)$ , where

$p,q\in I-\left\{ 0 \right\}$ , is a purely imaginary, then minimum value of

${ \left| z \right| }^{ 2 }$ is

0%
$0$
0%
$58$
0%
$\cfrac{3364}{3}$
0%
$3364$

Explanation

$z=(3+7i)(p+iq) \quad \dots (1)$

$z=(3p-7q)+i(3q+7p)$

Now,

$z$ is purely imaginary

$3p-7q=0$

$\cfrac{p}{q}=\cfrac{7}{3}$

$\cfrac{p}{q}+i=\cfrac{7}{3}+i$

Therefore,

$\cfrac{(p+qi)}{q}=\cfrac{7+3i}{3}$

$p+iq=7+3i \quad \dots (2)$

Substitute

$(2)$ in

$(1)$

$z=21+9i+49i-21$

Thus,

$z=58i$

$\left| { z }^{ 2 } \right| =3364$

$z=\dfrac{\sqrt{3}}{2}+\dfrac{1}{2}i$ then

$z\bar{z}$ is

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

Explanation

Conjugate of

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

$\therefore\,\bar{z}=\dfrac{\sqrt{3}-i}{2}$

Now,

$z\bar{z}=\left(\dfrac{\sqrt{3}+i}{2}\right)\left(\dfrac{\sqrt{3}-i}{2}\right)=\dfrac{3-{i}^{2}}{4}=\dfrac{3+1}{4}=\dfrac{4}{4}=1$

Let

$A = \left\{ {z \in c:\left| z \right|} \right. = 2\left. 5 \right\}$ and

$B = \left\{ {z \in c:\left| {z + 5 + 12i} \right|} \right. = 4$ . Then the minimum value of

$\left| {z - w} \right|$ for

$Z \in A$ and

$\omega \in B$ is :

0%
7
0%
8
0%
9
0%
6

Solve

$i^{57}+\dfrac{1}{i^{125}}$

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

Explanation

We know that

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

${i}^{3}={i}^{2}.i=-i$

${i}^{4}={i}^{2}\times {i}^{2}=-1\times -1=1$

Thus,

${i}^{57}={\left({i}^{4}\right)}^{14}\times i={1}^{14}\times i=i$

${i}^{125}={\left({i}^{4}\right)}^{31}\times i={1}^{4}\times i=i$

$\dfrac{1}{{i}^{125}}=\dfrac{1}{i}$

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

Now

${i}^{57}+\dfrac{1}{{i}^{125}}=i+\left(-i\right)=i-i=0$

The value of the sum

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

$i=\sqrt{-1}$ , is?

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

Explanation

$= \displaystyle\sum_{n=1}^{13}(i^{n}+i^{n+1})$

$=\displaystyle \sum_{n=1}^{13}(l^{n}.(i+1))$

$= (i+1)\displaystyle\sum_{n=1}^{13}(i^{n})$

$= (i+1)[i+i^{2}+i^{3}+i^{4}+i^{5}+i^{6}+i^{7}+i^{8}+i^{9}+$

$i^{10}+i^{11}+i^{12}+i^{13}]$

$= (i+1)[i+(-1)+(-i)+1+i+(-1)+(-i)+1+1$

$+(-1)+(-i)+1+i]$

$= (i+1)(i)$

$=l^{2}+i$

$= -1+i$

$= i-1$

${z_1}$ and

${z_2}$ be complex numbers such that

${z_1} \ne {z_2}$ and

$\left| {{z_1}} \right| = \left| {{z_2}} \right|$ . If

${z_1}$ has positive real part and

${z_2}$ has negative imarinary part, then

$\frac{{\left( {{z_1} + {z_2}} \right)}}{{\left( {{z_1} - {z_2}} \right)}}$ may be

0%
Purely imaginary
0%
Real and positive
0%
Real and negative
0%
zero

$z_{1} and z_{2} are on straight line$

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

0%
$\left| z _ { 1 } + z _ { 2 } \right|$
0%
$\left| z _ { 1 } - z _ { 2 } \right|$
0%
$\left| z _ { 1 } \right| + \left| z _ { 2 } \right|$
0%
$\left| z _ { 1 } \right| - \left| z _ { 2 } \right|$

$|z_{1}+z_{2}|=|z_{1}-z_{2}|$ , then the different in the amplitudes of

$z_{1}$ and

$z_{2}$ is

0%
$\dfrac {\pi}{4}$
0%
$\dfrac {\pi}{3}$
0%
$\dfrac {\pi}{2}$
0%
$0$

Explanation

$|z_{1}+ z_{2}|= |z_{1}- z_{2}|$ then we can say that the long of diagonals of the parallelogram is equal hence the parallelogram is a rectangle so angle between

$z_{1}$ &

$z_{2}$ is

$90$ degree

For any two complex numbers

$z_{1}$ and

$z_{2}$ , then

$Re(z_{1}z_{2})=Rez_{1} Rez_{2}-\ln z_{1} \ln z_{2}$

0%
True
0%
False

The value of

$2x^{4}+5x^{3}+7x^{2}-x+41$ , when

$x=-2-\sqrt{3i}$ is:

0%
-4
0%
4
0%
-6
0%
6

Explanation

$2x^{4}+5x^{3}+4x^{2}-x+41$ ---(1)

when

$x = -2-\sqrt{3}i = -(2+\sqrt{3}i)$ ---(2)

$x^{2} = (2+\sqrt{3}i)^{2}$

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

$=4-3+4\sqrt{3}!$

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

$x^{2} = 1+4\sqrt{3}i$ ---(3)

$x^{3} = x^{2}.x$

$= (1+4\sqrt{3}i)[-(2+\sqrt{3}i)]$

$= -(2+\sqrt{3}i+8\sqrt{3}i+12!^{2})$

$= -(2-12+9\sqrt{3}i)$

$x^{3} = 10-9\sqrt{3}i$ ---(4)

$x^{4} = (x^{2})^{2}$

$= (1+4\sqrt{3}i)^{2}$

$= 1+48!^{2}+8i\sqrt{3}$

$= 1-48+8i\sqrt{3}$

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

$x^{4} = -47+8!^{3}$ ---(5)

From (1),(2),(3),(4) and (5)

$2(-47+8i\sqrt{3})+5(10-9\sqrt{3}i)+7(1+4\sqrt{3}i)-(-2\sqrt{3}i)+41$

$-94+16i\sqrt{3}+50-45\sqrt{3}i+7+28\sqrt{3}i+2+\sqrt{3}i+41$

$= 6$ d)

1112264_1200633_ans_9aa025ae23a8492fb71450047d74f29c.jpg

If the six solutions of

$x^6 = -64$ are written in the form

$a + bi$ , where

$a$ and

$b$ are real, then the product those solution with

$a < 0$ , is

0%
$4$
0%
$8$
0%
$16$
0%
$64$

$z_{1}$ and

$z_{2}$ two non-zero complex number such that

$|z_{1}+z_{2}|=|z_{1}|+|z_{2}|$ , then

$arg z_{1}-arg z_{2}$ is equal to

0%
$-p$
0%
$p/2$
0%
$-p/2$
0%
$0$

Explanation

$\Rightarrow z_1 \;\&\; z_2$ are collinear and in same direction.

$\Rightarrow arg \left(z_1\right)-arg\left(z_2\right)=0.$

Hence, the answer is

$0.$

${ z }_{ 1 }$ and

${ z }_{ 2 }$ are two complex number such that Im

$({ z }_{ 1 }={ z }_{ 2 })$ = 0= Im

$({ z }_{ 1 }{ z }_{ 2 })$ , then

0%
${ z }_{ 1 }={ z }_{ 2 }$
0%
${ z }_{ 1 }={ \overline { z } }_{ 2 }$
0%
${ z }_{ 1 }={ -z }_{ 2 }$
0%
${ z }_{ 1 }=-{ \overline { z } }_{ 2 }$

Explanation

As per given condition:

$Img(z_1=z_2)=0=Im(z_1z_2)$

$\Rightarrow Im(z_1z_2)=0$

Product of imaginary parts is zero, implies that the direction of each number is opposite in direction.

$\therefore z_1=\bar{z_2}$

$z$ is a complex number. If

$a = | x | + | y |$ and

$b = \sqrt { 2 } | x + i y |$ then which of the following is
true

0%
$a \leq b$
0%
$a > b$
0%
none of these
0%
$a - b + 2$

$z=(3+4i)^6+(3-4i)^6,$ where

$i=\sqrt { -1 },$ then

$Im(z)$ equals to

0%
$-6$
0%
$0$
0%
$6$
0%
None of these

Explanation

$z={ \left( 3+4i \right) }^{ 6 }+{ \left( 3-4i \right) }^{ 6 }$

$={ \left[ 5\left( { e }^{ i\theta } \right) \right] }^{ 6 }+{ \left[ 5\left( { e }^{ -i\theta } \right) \right] }^{ 6 }$ [where

$\theta ={ tan }^{ -1 }\left( 4/3 \right)$ ]

$={ 5 }^{ 6 }\left[ { e }^{ i6\theta }+{ e }^{ -i6\theta } \right]$

$={ 5 }^{ 6 }\left[ 2Re\left( { e }^{ i6\theta } \right) \right]$

$\therefore$ In

$\left( z \right) =0$ [B]

Number of complex numbers

$z$ such that

$|z|=1$ and

$\left|\dfrac {z}{z}+\dfrac {\bar {z}}{z}\right|=1$ is

0%
$4$
0%
$1$
0%
$8$
0%
$more\ then\ 8$

Explanation

Since

$z$ is equidistant from

$1,-1,i$ , we can say that

$z$ is center of circle with

$1,-1,i$ lying on it's circumference and so there is one one circle possible passing through

$3$ points. So, the answer is

$1$ .

If a complex number z and

$z+\dfrac { 1 }{ z }$ have same argument then-

0%
z must be purely real
0%
z must be purely imaginary
0%
z cannot be imaginary
0%
z must be raal

Explanation

we have,

z is a complex number.

Then,

Let $z=r\cos \theta +i\sin \theta$

Now, we know that,

$\overline{z}=r\left( \cos \theta -i\sin \theta \right)$

So,

$z\overline{z}={{r}^{2}}$

Then,

$\dfrac{1}{z}=\dfrac{\overline{z}}{z\overline{z}}=\dfrac{\overline{z}}{{{r}^{2}}}$

Now,

According to given question,

$z+\dfrac{1}{z}=\left| z+\dfrac{1}{z} \right|=1\,\,\,\left( \text{z}\,\text{and}\,\text{z+}\dfrac{\text{1}}{\text{z}}\,\text{same argument} \right)$

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

$=\left| \left( r+\dfrac{1}{r} \right)\cos \theta +i\left( r-\dfrac{1}{r} \right)\sin \theta \right|=1$

Now, squaring both side and we get,

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

So,

$\left( {{r}^{2}}+\dfrac{1}{{{r}^{2}}}+2 \right){{\cos }^{2}}\theta =1$

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

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

Now,

$1-2\cos 2\theta =0$

$\Rightarrow \cos 2\theta =\dfrac{1}{2}$

$\Rightarrow \cos 2\theta =cos\dfrac{\pi }{3}$

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

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

Then, Z is purely imaginary.

Hence, this is the answer.

Let P

$\left( x \right) ={ x }^{ 3 }-6{ x }^{ 2 }+Bx+C$ has 1+5i as a zero and B,C real number, then value of (B+C) is

0%
-70
0%
70
0%
24
0%
138

Explanation

Given

$1+5{i}$ is the root of

$P(x)=x^3-6{x^2}+B{x}+C$

$\implies 1-5{i}$ is also the root of

$P(x)$

Let the third root be

$h$

$\implies$ sum of roots is

$6$

$\implies 1+5{i}+1-5{i}+h=6\implies h=4$

$B=(1+5{i}+1-5{i})(4)+(1+5{i})(1-5 i)=8+26=34$

$C=-(1+5{i})(1-5{i})(4)=-104$

$B+C=34-104=-70$

A value of

$\theta$ for which

$\dfrac { 2+3isin\theta }{ 1-2isin\theta }$ is purely imaginary, is:

0%
${ sin }^{ -1 }\left( \dfrac { 1 }{ \sqrt { 3 } } \right)$
0%
$\dfrac { \pi }{ 3 }$
0%
$cos^{-1}\sqrt-1$
0%
$None of these$

Explanation

We have,

$\dfrac{2+3i\sin \theta }{1-2i\sin \theta }$

On rationalize and we get,

$\dfrac{2+3i\sin \theta }{1-2i\sin \theta }\times \dfrac{1+2i\sin \theta }{1+2i\sin \theta }$

$\Rightarrow \dfrac{2+3i\sin \theta +4i\sin \theta +6{{i}^{2}}{{\sin }^{2}}\theta }{{{1}^{2}}-4{{i}^{2}}{{\sin }^{2}}\theta }$

$\Rightarrow \dfrac{2-{{\sin }^{2}}\theta +7i\sin \theta }{1+4{{\sin }^{2}}\theta }$

$\Rightarrow \dfrac{1+1-{{\sin }^{2}}\theta +7i\sin \theta }{1+4{{\sin }^{2}}\theta }$

$\Rightarrow \dfrac{1+{{\cos }^{2}}\theta +7i\sin \theta }{1+4{{\sin }^{2}}\theta }$

$\Rightarrow \dfrac{1+{{\cos }^{2}}\theta }{1+4{{\sin }^{2}}\theta }+\dfrac{7i\sin \theta }{1+4{{\sin }^{2}}\theta }$

$\Rightarrow \dfrac{1+{{\cos }^{2}}\theta }{1+4{{\sin }^{2}}\theta }+i\dfrac{7\sin \theta }{1+4{{\sin }^{2}}\theta }$

Now,

Then purely imaginary is

Real part of equal to zero.

$\dfrac{1+{{\cos }^{2}}\theta }{1+4{{\sin }^{2}}\theta }=0$

$\Rightarrow 1+{{\cos }^{2}}\theta =0$

$\Rightarrow {{\cos }^{2}}\theta =-1$

$\Rightarrow \cos \theta =\sqrt{-1}$

$\Rightarrow \theta ={{\cos }^{-1}}\sqrt{-1}$

Hence, this is the answer.

Let

$\left| z _ { i } \right| = i , i = 1,2,3,4$ and

$\left| 16 z _ { 1 } z _ { 2 } z _ { 3 } + 9 z _ { 1 } z _ { 2 } z _ { 4 } + 4 z _ { 1 } z _ { 3 } z _ { 4 } + z _ { 2 } z _ { 3 } z _ { 4 } \right| = 48 ,$ then the value of

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

0%
$1$
0%
$2$
0%
$4$
0%
$8$

Explanation

$\begin{array}{l} \left| { { z_{ i } } } \right| =i\, \, \, \, i=1,2,3,4 \\ \left| { { z_{ 1 } } } \right| =1 \\ \left| { { z_{ 2 } } } \right| =2 \\ \left| { { z_{ 3 } } } \right| =3 \\ \left| { { z_{ 4 } } } \right| =4 \\ \left| { 16{ z_{ 1 } }{ z_{ 2 } }{ z_{ 3 } }+9{ z_{ 1 } }{ z_{ 2 } }{ z_{ 3 } }+4{ z_{ 1 } }{ z_{ 2 } }{ z_{ 3 } }+{ z_{ 2 } }{ z_{ 3 } }{ z_{ 4 } } } \right| =48 \\ \left| { { z_{ 1 } }{ z_{ 2 } }{ z_{ 3 } }{ z_{ 4 } } } \right| \left| { \overline { { z_{ 4 } } } +\overline { { z_{ 3 } } } +\overline { { z_{ 2 } } } +\overline { { z_{ 1 } } } } \right| =48 \\ 1\times 2\times 3\times 4\times \left| { { z_{ 1 } }+{ z_{ 2 } }+{ z_{ 3 } }+{ z_{ 4 } } } \right| =48 \\ \Rightarrow \left| { { z_{ 1 } }+{ z_{ 2 } }+{ z_{ 3 } }+{ z_{ 4 } } } \right| =2 \\ =\left| { \frac { 1 }{ { { z_{ 1 } } } } +\frac { 4 }{ { { z_{ 2 } } } } +\frac { 9 }{ { { z_{ 3 } } } } +\frac { { 16 } }{ { { z_{ 4 } } } } } \right| .\left| { \frac { { { z_{ 1 } }{ { \overline { z } }_{ 1 } } } }{ { { z_{ 1 } } } } +\frac { { { z_{ 2 } }{ { \overline { z } }_{ 2 } } } }{ { { z_{ 2 } } } } +\frac { { { z_{ 3 } }{ { \overline { z } }_{ 3 } } } }{ { { z_{ 3 } } } } +\frac { { { z_{ 4 } }{ { \overline { z } }_{ 4 } } } }{ { { z_{ 4 } } } } } \right| \\ =\left| { { z_{ 1 } }+{ z_{ 2 } }+{ z_{ 3 } }+{ z_{ 4 } } } \right| =2 \\ Option\, \, B\, \, is\, \, correct\, \, answer. \end{array}$

$z$ satisfies

$|z - 2+ 2i| \le 1$

0%
$|z|_{least} = 2\sqrt{2} - 1$
0%
$|z|_{least} = 2\sqrt{2} + 1$
0%
$|z - 1| \le 1$
0%
None of these

$\dfrac { z+1 }{ z+i }$ is purely imaginary, then z lies on a

0%
straight lone
0%
circle
0%
Circle with radius 1
0%
circle passing through (1, 1).

Explanation

$\dfrac { z+1 }{ z+i }$ is purely imaginary

$\Rightarrow Re\left( \dfrac { z+1 }{ z+i } \right) =0$

Say

$z=x+iy$

then,

$Re\left( \dfrac { x+iy+1 }{ x+iy+i } \right) =0$

$\Rightarrow Re\left( \dfrac { \left( x+1 \right) +iy }{ x+i\left( y+1 \right) } \right) =0$

$\Rightarrow Re\left[ \dfrac { \left[ \left( x+1 \right) +iy \right] \left[ x-i\left( y+1 \right) \right] }{ { x }^{ 2 }+{ \left( y+1 \right) }^{ 2 } } \right] =0$

$\Rightarrow Re\left[ \dfrac { x\left( x+1 \right) +y\left( y+1 \right) +i\left( xy-\left( x+1 \right) \left( y+1 \right) \right) }{ { x }^{ 2 }+{ \left( y+1 \right) }^{ 2 } } \right] =0$

$\therefore$

$\dfrac { x\left( x+1 \right) +y\left( y+1 \right) }{ { x }^{ 2 }+{ \left( y+1 \right) }^{ 2 } } =0$

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

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

$\Rightarrow { x }^{ 2 }+{ y }^{ 2 }+2.\dfrac{1}{2}x+2.\dfrac{1}{2}y=0$

The above is a circle with centre

$\left( \dfrac { -1 }{ 2 } ,\dfrac { -1 }{ 2 } \right)$ and radius

$\sqrt { \dfrac { 1 }{ 4 } +\dfrac { 1 }{ 4 } } =\sqrt { \dfrac { 1 }{ 2 } }$ units

Option B.

Purely imaginary then find the sum of statement i

$a,b$

0%
$\dfrac {5\pi}{6}$
0%
$\pi$
0%
$\dfrac {3\pi}{4}$
0%
$\dfrac {2\pi}{3}$

$\alpha$ and

$\beta$ are the roots of

${ 4x }^{ 2 }-16x+c=0,$ c>0 such that

$1<\alpha <2<\beta <3$ , then the no.of integer values of c is

0%
$17$
0%
$14$
0%
$18$
0%
$15$

Explanation

Given,

$\alpha$ and

$\beta$ are roots of

$4x^2-16x+x=0$

where

$a=4$

$b=-16$

$x > 0$

now,

$\alpha\beta =\dfrac{a}{a}$

$\Rightarrow \alpha\beta =\dfrac{c}{4}$

$\Rightarrow \beta =\dfrac{c}{4\alpha}$ ………(i)

we have

$2 < \beta < 3$

$\Rightarrow 2 < \dfrac{c}{4\alpha} < 3$ [from

$(1)$ ]

$\Rightarrow 8 < \dfrac{c}{\alpha} < 12$

$\Rightarrow 8\alpha < c < 12\alpha$ ………(ii)

and

$1 < \alpha < 2$

$\Rightarrow \alpha > 1\Rightarrow 8\alpha > 8$ ……….(iii)

and

$\alpha < 2$

$\Rightarrow 12\alpha < 24$ ………..(iv)

Substituting (iii) and (iv) in (ii) we get

$8 < 8\alpha < c < 12\alpha < 24$

$[\alpha =9, 10, 11, 12, 13, 14, 15, 16, 13, 18, 19, 20, 21, 22, 23]$

$\Rightarrow 8 < c < 24$

Therefore number of integer values of c is

$15$ .

1196924_1318670_ans_ab0c4184d5f446658aeacf0a4a2d6ac4.JPG

$\sqrt { \left( \log _ { 3 } \tan x \right) }$ is real for:

0%
$n \pi + \pi / 4 \leq x < n \pi + \pi / 2$
0%
$n \pi < x < n \pi + \pi / 2$
0%
$n \pi \pm \pi / 4 \leq x < n \pi \pm \pi / 2$
0%
None of these.

All even numbers are prime numbers.

0%
True
0%
False

Explanation

False,

$4$ is an even number but not prime numbers.

Which of the following is not a composite

number?

0%
$2 \times 3 \times 5\times 13\times 17 + 13$
0%
$7\times 6\times 5\times 4\times 3\times 2\times 1 + 5$
0%
$17\times 41\times 43\times 61 + 43$
0%
$2 \times 3\times 43 + 13$

$\left| {z - 3 + 2i} \right|\, \le 4$ then the difference between the greatest value and the least value of

$\left| z \right|$ is :

0%
$2\sqrt {13}$
0%
$8$
0%
$4 + \sqrt {13}$
0%
$\sqrt {13}$

Explanation

As we know that

$\left| z - {z}_{0} \right| = c$ represents equation of circle.

Here,

${z}_{0} = 3 - 2i$

$c = 4$

Maximum value of

$z = c + \left| z \right|$

Minimum value of

$z = c - \left| z \right|$

For

$\left| z - 3 + 2i \right| \le 4$

Maximum value

$= 4 + \sqrt{{3}^{2} + {\left( -2 \right)}^{2}} = 4 + \sqrt{13}$

Miniimum value

$= 4 - \sqrt{{3}^{2} + {\left( -2 \right)}^{2}} = 4 - \sqrt{13}$

Difference between maximum and minimum value

$= \left( 4 + \sqrt{13} \right) - \left( 4 - \sqrt{13} \right) \\ = 4 + \sqrt{13} - 4 + \sqrt{13} \\ = 2 \sqrt{13}$

Hence, this is the answer.

Find the value of

${x}^{3}+7{x}^{2}-x+16$ , when

$x=1+2i$

0%
$-11+24i$
0%
$-17+24i$
0%
$-17-24i$
0%
$-1+24i$

Explanation

Given function

$f(x)=x^3+7x^2-x+16$

We know that

$(a+b)^3=a^2+3ab(a+b)+b^3\\(a+b)^2=a^2+b^2+2ab$

When

$x=1+2$

$f(1+2i)=(1+2i)^2+(1+2i)^2-(1+2i)+16\\=1+8i^3+3\times2i\times(1+2i)+7\times(1+4i+4i^2)-(1+2i)+16\\=1+8i^3+6i+12i^2+7+28i+28i^2-1-2i+16\\=1-8i+6i-12+7+28i-28-1-2+16[\because i^3=-i,i^2=-1]\\=-17+24i$

Let 'z' be a complex number satisfying

$|z-2-i|\le 5,$ Then |z-14-6i| lies in

0%
{8,18}
0%
{2,8}
0%
{0,2}
0%
{3,7}

Explanation

Here,

$|z-2-i|\leq 5$ ………..

$(1)$

Now,

$|z-14-6i|=|z-2-i-12-5i|=|(z-2-i)-(12+5i)|$

$\leq |z-2-i|+|12+5i|$

$\leq 5+|12+5i|$

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

$=5+\sqrt{144+25}$

$=5+\sqrt{169}$

$=5+13$

$=18$ .

$\therefore |z-14-6i|\leq 18$

and

$|z-14-6i|=|z-2-i-12-5i|$

$=|-(12+5i)+(z-2-i)|$

$\geq ||-(12+5i)|-|z-2-i||$

$\geq |12+5i|-5$

$[\because |z-2-i|\leq 5$

$\Rightarrow -|z-2-i|\geq -5]$

$=13-5$

$=8$

$\therefore |z-14-6i|\geq 8$

$\therefore |z-14-6i|$ lies in

$\{8, 18\}$ .

1188470_1315915_ans_eed83e919159412d92c24e2d369fda96.JPG

$w = \dfrac { z } { z - \dfrac { 1 } { 3 } i }$ and

$| w | = 1$ then

$z$ lies on

0%
a circle
0%
an ellipse
0%
a parabola
0%
a straight line

Explanation

$\begin{array}{l} Given \\ \omega =\frac { { 3z } }{ { 3z-i } } \, \, \, \, \, \, \therefore \left| \omega \right| =\frac { { 3\left| z \right| } }{ { \left| { 3z-i } \right| } } \\ \Rightarrow \left| { 3z-i } \right| =3\left| z \right| \\ \Rightarrow \left| { 3\left( x \right) +i\left( { 3y-1 } \right) } \right| =\left| { 3\left( { x+iy } \right) } \right| \, \, \, \, \, \left( { z=x+iy } \right) \\ \Rightarrow { \left( { 3x } \right) ^{ 2 } }+{ \left( { 3y-1 } \right) ^{ 2 } }=9\left( { { x^{ 2 } }+{ y^{ 2 } } } \right) \\ \Rightarrow 6y-1=0 \\ Which\, is\, straight\, line \\ Hence,\, option\, D\, is\, the\, correct\, answer. \end{array}$

The real part of

$\left[ 1 + \cos \left( \dfrac { \pi } { 5 } \right) + i \sin \left( \dfrac { \pi } { 5 } \right) \right] ^ { - 1 }$ is

0%
$1$
0%
$\dfrac { 1 } { 2 }$
0%
$\dfrac { 1 } { 2 } \cos \left( \dfrac { \pi } { 10 } \right)$
0%
$\dfrac { 1 } { 2 } \cos \left( \dfrac { \pi } { 5 } \right)$

Explanation

We have,

$\begin{array}{l} \dfrac { 1 }{ { 1+\cos \left( { \dfrac { \pi }{ 5 } } \right) +i\sin \left( { \dfrac { \pi }{ 5 } } \right) } } \\ =\dfrac { 1 }{ { 2{ { \cos }^{ 2 } }\dfrac { \pi }{ { 10 } } +i2\sin \left( { \dfrac { \pi }{ { 10 } } } \right) \cos \left( { \dfrac { \pi }{ { 10 } } } \right) } } \\ =\dfrac { 1 }{ { 2\cos \dfrac { \pi }{ { 10 } } \left[ { \cos \dfrac { \pi }{ { 10 } } +i\sin \dfrac { \pi }{ { 10 } } } \right] } } \\ =\dfrac { { \cos \dfrac { \pi }{ { 10 } } -i\sin \dfrac { \pi }{ { 10 } } } }{ { 2\cos \dfrac { \pi }{ { 10 } } } } \\ Thus,\, { { Re } }al\, part=\dfrac { 1 }{ 2 } \\ Hence,\, the\, option\, B\, is\, the\, required\, answer. \end{array}$

Hence, option

$B$ is the correct answer.

$\dfrac{1-2i}{2+i}+\dfrac{4-i}{3+2i}=$

0%
$\dfrac{24}{13}+\dfrac{11}{13}i$
0%
$\dfrac{24}{13}-\dfrac{11}{13}i$
0%
$\dfrac{10}{13}+\dfrac{24}{13}i$
0%
$\dfrac{10}{13}-\dfrac{24}{13}i$

Explanation

$\frac{1-2i}{2+1} +\frac{4-1}{(3+2i)}$

$\frac{(1-2i)(3+2i)+(4-i)(2+i)}{(2+i)(3+2i)}$

$\frac{3+2i-6i+4+8+4i-2i+1}{6+4i+3i-2}$

$\frac{3-4i+12+2i+1}{6+7i-2}$

$\frac{-2i+16}{7i+4}$

$\frac{16-2i}{4+7i}\times \frac{(4-7i)}{4-7i}$

$\frac{64-112i-8i-14}{16+49}$

$\frac{50-120i}{65}$

$= \frac{10-24i}{13}$

$= \frac{10}{13}-\frac{24i}{13}$

1183970_1348432_ans_517c0d7705bc4d2394b4a57a6d1c8368.jpg

The principle amplitude of

$(\sin 40^{o}+i \cos 40^{o})^{5}$ is

0%
$70^{o}$
0%
$-1100^{o}$
0%
$70^{110}$
0%
$70^{-70}$

Explanation

$40°=2\cfrac { \pi }{ 9 } ,50°=\cfrac { 5\pi }{ 18 }$

So,

$\left( \sin { 40° } +i\cos { 40° } \right) =\cos { 50° } +i\sin { 50° }$

Let

$T={ \left( \sin { 40° } +i\cos { 40° } \right) }^{ 5 }={ \left( \cos { 50° } +i\sin { 50° } \right) }^{ 5 }$

$T={ \left( { e }^{ i50° } \right) }^{ 5 }=\left( { e }^{ i\cfrac { 5\pi }{ 18 } \times 5 } \right) ={ e }^{ i\cfrac { 25\pi }{ 18 } }$

So,

$25\cfrac { \pi }{ 18 } =250°$

Thus, principal argument

$=-180°+250°=70°$

$|z-3i|<\sqrt{5}$ , then

$|i(z+1)+1|<2\sqrt{5}$ .

0%
True
0%
False

$z_1=1+i,z_2=1-i$ find

$z_1z_2$

0%
$z_1+z_2$
0%
$z_1-z_2$
0%
$z_1/z_2$
0%
None.

Explanation

$z_1=1+i$

$z_2=1-i$

$z_1z_2=1-(-1)=2$

$=1+1$

$=1-i+1+i$

$z_1z_2=z_1+z_2$

The value of the sum

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

$i=\sqrt { -1 }$ , equals

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

Explanation

As we know that

$i+i^2+i^3+i^4=0$

$\displaystyle\sum ^{13}_{n=1} (i^n+i^{n+1})=\sum ^{13}_{n=1}(1+i)(i^n)=(1+i)\sum ^{13}_{n=1} i^{n}=(1+i)(0+0+i^{13})=(1+i)i=i+i^2=i-1$

$z_1$ and

$z_2$ are two non-zero complex numbers such that

$z_1=2+4i\\z_2=5-6i$ , then

$z_2-z_1$ equals

0%
$3-10i$
0%
$3+10i$
0%
$7-2i$
0%
$10-24i$

Explanation

$z_1=2+4i\\\\z_2=5-6i\\\\z_2-z_1=5-6i-2-4i=3-10i$

The imaginary part of

$t ; t \in R$ is

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

How many prime numbers are there in the following series.

$1,2,7,9,13,15,21,23,27,29$

0%
7
0%
6
0%
5
0%
4

The real value of

$'\theta '$ , for which the expression

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

0%
$2n\pi +\frac { 3\pi }{ 2 } ,n\in I$
0%
$2n\pi -\frac { 3\pi }{ 2 } ,n\in I$
0%
$2n\pi \pm \frac { \pi }{ 2 } ,n\in I$
0%
$2n\pi +\frac { \pi }{ 4 } ,n\in I$

The greatest and least value of

$\left | z \right |$ if

$z$ satisfies

$\left | z - 5 + 5i \right |$

$\leq 5$ are

0%
$10$ , $5\sqrt{2}$
0%
$5\sqrt{2}$ , $5$
0%
$10$ , $0$
0%
$5 + 5\sqrt{2}$ , $5\sqrt{2} - 5$

Explanation

$\left | z - 5 + 5i \right |$

$\leq 5$ represents equation of circle, where centre of circle is (5,-5)

Maximum value of

$z=c+\sqrt{a^2+b^2}$

Maximum value of

$z=c-\sqrt{a^2-b^2}$

Here,

$a=5, b=-5$ and

$c=5$

Maximum value of

$|z|=|5+\sqrt{25+25}|$

$=|5+\sqrt{50}|$

$=5+5\sqrt{2}$

Minimum value of

$|z|=|5-\sqrt{50}|$

$=|5-5\sqrt{2}|$

$=5\sqrt{2}-5$

let

$z=\left| \begin{matrix} 1 & 1+2i & -5i \\ 1-2i & -3 & 5+3i \\ 5i & 5-3i & 7 \end{matrix} \right|$ then

0%
z is purely real
0%
z is purely imaginary
0%
$\left( z-\bar { z } \right)$ i is real and imaginary both
0%
$\left( z+\bar { z } \right) =0$

Explanation

From given, we have,

$z=\begin{vmatrix}1&1+2i&-5i\\ 1-2i&-3&5+3i\\ 5i&5-3i&7\end{vmatrix}$

$=1 \times \begin{vmatrix}-3&5+3i\\ 5-3i&7\end{vmatrix}-(1+2i) \times \begin{vmatrix}1-2i&5+3i\\ 5i&7\end{vmatrix}-5i \times \begin{vmatrix}1-2i&-3\\ 5i&5-3i\end{vmatrix}$

$=1\cdot \left(-55\right)-\left(1+2i\right)\left(22-39i\right)-5i\left(-1+2i\right)$

$=-145$

Hence

$z$ is purely real

Which of the following is a prime number

0%
391
0%
899
0%
621
0%
199

Given

$z _ { 1 } + 3 z _ { 2 } - 4 z _ { 3 } = 0$ then

$z _ { 1 } , z _ { 2 } , z _ { 3 }$ are

0%
collinear
0%
can form sides of equilateral $\Delta$
0%
lie on circle
0%
none of these

The imaginary roots of the equation

${ ({ x }^{ 2 }+2) }^{ 2 }+8{ x }^{ 2 }=6x({ x }^{ 2 }+2)$ are ____________.

0%
$1+i$
0%
$2\pm i$
0%
$-1\pm i$
0%
$none of these$

Explanation

${\left({x}^{2}+2\right)}^{2}+8{x}^{2}=6x\left({x}^{2}+2\right)$

${\left({x}^{2}+2\right)}^{2}-6x\left({x}^{2}+2\right)+8{x}^{2}=0$

Let

$t={x}^{2}+2$

$\Rightarrow\,{t}^{2}-6xt+8{x}^{2}=0$

$\Rightarrow\,{t}^{2}-2t\times 3x+9{x}^{2}-{x}^{2}=0$

$\Rightarrow\,{\left(t-3x\right)}^{2}-{x}^{2}=0$

$\Rightarrow\,{\left({x}^{2}+2-3x\right)}^{2}-{x}^{2}=0$ where

$t={x}^{2}+2$

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

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

$\Rightarrow\,\left({x}^{2}-2x+2\right)\left({x}^{2}-4x+2\right)=0$

$\Rightarrow\,{x}^{2}-2x+2=0,\,or\,{x}^{2}-4x+2=0$

$\Rightarrow\,x=\dfrac{2\pm\sqrt{4-4\times\,1\times 2}}{2},\,or\,x=\dfrac{-4\pm\sqrt{16-4\times 1\times 2}}{2}$

$\Rightarrow\,x=\dfrac{2\pm\sqrt{4-8}}{2},\,or\,x=\dfrac{-4\pm\sqrt{16-8}}{2}$

$\Rightarrow\,x=\dfrac{2\pm\sqrt{-4}}{2},\,or\,x=\dfrac{-4\pm\sqrt{8}}{2}$

$\Rightarrow\,x=\dfrac{2\pm 2i}{2},\,or\,x=\dfrac{-4\pm\,2\sqrt{2}}{2}$

$\Rightarrow\,x=1\pm i,\,or\,x=-2\pm\,\sqrt{2}$

Hence the roots are

$\left\{1+i,\,1-i,\,-2+\sqrt{2},\,-2-\sqrt{2}\right\}$

Imaginary roots are

$\left\{1+i,\,1-i\right\}$

If z be a complex number satisfying

$z^{4}+z^{3}+2z^{2}+z+1=0$ then

$\left|z\right|=$

0%
$\dfrac{1}{2}$
0%
$\dfrac{3}{4}$
0%
$1$
0%
none of these

Explanation

$z^{4}+z^{3}+2z^{2}+z+1=0$

$\Rightarrow z^{4}+z^{2}+z^{3}+z+z^{2}+1=0$ by rearranging

$\Rightarrow z^{2}(z^{2}+1)+z(z^{2}+1)+1(z^{2}+1)=0$

$\Rightarrow z^{2}(z^{2}+1)+(z^{2}+1)(z+1)=0$

$\Rightarrow (z^{2}+1)(z^{2}+z+1)=0$

$\Rightarrow z^{2}+1=0, z^{2}+2+1=0$

$\Rightarrow z=\pm i, z=\frac{-1\pm i\sqrt{3}}{2}$

$\therefore z=i, -i, \frac{-1+i\sqrt{3}}{2},\frac{-1-i\sqrt{3}}{2}$

$\therefore |z|=+1$

1240698_1507366_ans_b4f400c823594905ac18096a8656a951.PNG

CBSE Questions for Class 11 Commerce Applied Mathematics Number Theory Quiz 9 - MCQExams.com

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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

CBSE Questions for Class 11 Commerce Applied Mathematics Number Theory Quiz 9 - MCQExams.com

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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers

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