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Triangle inequality - class-VIII

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The points $\left( 0,\dfrac { 8 }{ 3 }  \right),(1,3)$ and $(82,30)$ are the vertices of:

  1. an equilateral triangle

  2. an isosceles triangle

  3. a right angled triangle

  4. none of these


Correct Option: A
Explanation:

According to the problem :

$AB^2=(0-1)^2+(\dfrac{8}{3}-3)^2$
$=1+\dfrac{1}{9}=\dfrac{10}{9}=1.11$

Similarly,
$BC^2=(82-1)^2+(30-3)^2=7290$
and
$AC^2=(82-0)^2+(30-\dfrac{8}{3})^2=7471.11$

Therefore,
$AB^2+BC^2<AC^2$

Hence the answer is acute-angled triangle.

If $\left| z+4 \right| \le 3$, then the maximum value of $\left| z+4 \right| $ is

  1. $3$

  2. $10$

  3. $6$

  4. $0$


Correct Option: A
Explanation:

If |Z+4| <= 3

then maximum value of |Z+4| = 3

The triangle inequality theorem states that 

  1. The sum of the lengths of the $2$ sides of a triangle is equal than the third side of the triangle

  2. The sum of the lengths of the $2$ sides of a triangle is less than the third side of the triangle

  3. The sum of the lengths of the $2$ sides of a triangle is more than the third side of the triangle

  4. None of these


Correct Option: C
Explanation:

The triangle inequality theorem states that the sum of the lengths of the $2$ sides of a triangle is greater than the third side of the triangle.

State the following statement is True or False
It is possible to have a triangle of sides $3,4,8$

  1. True

  2. False


Correct Option: B
Explanation:

It is not possible to have a triangle of sides $3,4,8$

Since, sum of the $2$ sides ($3$ and $4$) is not greater than the third side that is $8$.
$3+4<8$
According to triangle inequality theorem, it is not possible to construct such triangle.

State the following statement is True or False
The triangle inequality theorem states that the sum of the lengths of the $2$ sides of a triangle is equal than the third side of the triangle

  1. True

  2. False


Correct Option: B
Explanation:

The triangle inequality theorem states that the sum of the lengths of the $2$ sides of a triangle is greater than the third side of the triangle.

State whether the following statement is True or False.
It is possible to have a triangle of sides $8,10,14$.

  1. True

  2. False


Correct Option: A
Explanation:

It is possible to have a triangle of sides $8,10,14$

Since, sum of any $2$ sides of the triangle is greater than the $3^{rd}$ side.
$8+14>10\ 10+14>8\ 10+8>14$

A triangle cannot be drawn with the following three sides:

  1. $2m, 3m, 4m$

  2. $3m, 4m, 8m$

  3. $4m, 6m, 9m$

  4. $5m, 7m, 10m$


Correct Option: B
Explanation:

A triangle with three sides a,b and c will be possible when:

$a+b>c$
$ b+c>a$
$ a+c>b$
$ Here,a=2,b=3,c=4$
$ 2+3>4$
$ 3+4>2$
$ 2+4>3$
$ \therefore A)is\quad possible.$
$ Here,a=3,b=4,c=8$
$ 3+4=7$
$7<8$
$ \therefore a+b>c\quad is\quad not\quad satisfied.$
$ C)&amp; D)\quad will\quad also\quad be\quad possible.$
$ \therefore B)Correct\quad answer.$

The complex number z having least positive argument which satisfies the condition $|z - 25i| \le 15$   is:

  1. $25i$

  2. $12+5i$

  3. $16+12i$

  4. $12+16i$


Correct Option: D
Explanation:
Solution:

$|z-25 i| \leq 15$

Let $z= r(cos \theta + i \, sin \theta)$

$\theta$ must be minimum

$| r \, cos \theta +i ( r\, sin \theta-25)|\leq 15$

$|\sqrt{r^2cos^2\theta+r^2 sin^2 \theta+ 625- 50 r \, sin \theta} \,|\leq 15$

square both side

$r^2 (cos^2 \theta+ sin^2 \theta)+625 - 50 r \, sin \theta \leq 225$

$r^250 r \, sin \theta \leq - 400$

$f(r)=\dfrac {400+r^2}{50 \, r}\leq sin \theta $

Find maximum value of $f(r)=\dfrac {400+r^2}{50\, r}$

$f'(r)=\dfrac {100 r^2- 50(400+r^2)}{2500 r^2}=0$

$50 r^2- 50 \times 400=0$

$r= 20$

$f(r=20)=\dfrac {800}{1000}\leq sin \theta $

$\dfrac {4}{5}\leq  sin \theta $

Least value of $sin \theta $ is $4/5$

$ tan \, \theta = 4/3 \,\,\,\,\,\,\,\,\,\, cos \theta  = 3/5$

$z= 20(3/5+4/5 \,i)$

$z= 12+16\, i$

D is correct.

If $|z^2-3|=3|z|$, then the maximum value of |z| is

  1. $1$

  2. $\displaystyle \frac {3+\sqrt {21}}{2}$

  3. $\displaystyle \frac {\sqrt {21}-3}{2}$

  4. $None\ of\ these$


Correct Option: B
Explanation:

By the law of inequality, 
$|{ z }^{ 2 }-3|\ge { |z| }^{ 2 }-3$
$ \Longrightarrow 3|z|\ge { |z| }^{ 2 }-3\ \Longrightarrow { |z| }^{ 2 }-3|z|-3\le 0\ \Longrightarrow 0\le |z|\le \displaystyle\frac { 3+\sqrt { 21 }  }{ 2 } $
Hence the maximum value of $|z|=\displaystyle\frac { 3+\sqrt { 21 }  }{ 2 } $

If $z$ is a complex number satisfying the equation $\left| z+i \right| +\left| z-i \right| =8$, on the complex plane then maximum value of $\left| z \right| $ is

  1. $2$

  2. $4$

  3. $6$

  4. $8$


Correct Option: B

$\begin{array} { l } { \text { If } z _ { 1 } \text { and } z _ { 2 } \text { are complex numbers, then } \left| z _ { 1 } + z _ { 2 } \right| ^ { 2 } = \left| z _ { 1 } \right| ^ { 2 } + \left| z _ { 2 } \right| ^ { 2 } \text { if and only if } z _ { 1 } \overline { z } _ { 2 } \text { is } } \ { \text { purely imaginary. } } \end{array}$

  1. True

  2. False


Correct Option: A

If $|z| < \sqrt 2-1$, then $|z^2+2z cos\alpha|$ is

  1. Less than 1

  2. $\sqrt 2+1$

  3. $\sqrt 2-1$

  4. None of these


Correct Option: A
Explanation:

$|z^2+2z cs \alpha| \leq |z^2|+|2z cos \alpha|$
$=|z^2|+|2z^2| |cos\alpha|$
$\leq |z|^2 + 2 |z| $
$ < (\sqrt 2-1)^2 + 2(\sqrt 2-1) =1$

If $P$ and $Q$ are represented by complex numbers $z _{1}$ and $z _{2}$ such that $\left| \dfrac { 1 }{ { z } _{ 1 } } +\dfrac { 1 }{ { z } _{ 2 } }  \right| =\left| \dfrac { 1 }{ { z } _{ 1 } } -\dfrac { 1 }{ { z } _{ 2 } }  \right| $ then the circumference of $\triangleOPQ(O is origin)$ is

  1. $\dfrac{{ z } _{ 1 } -{ z } _{ 2 } }{2}$

  2. $\dfrac{{ z } _{ 1 } +{ z } _{ 2 } }{2}$

  3. $\dfrac{{ z } _{ 1 } +{ z } _{ 2 } }{3}$

  4. ${ z } _{ 1 } +{ z } _{ 2 } $


Correct Option: A

The sum of all sides of a quadrilateral is lessthan the sum of its diagonals.

  1. True

  2. False


Correct Option: B

If $\left| {z - 1} \right| + \left| {z + 3} \right| \le 8$ then the range of values of $\left| {z - 4} \right|$

  1. $[1,\,7]$

  2. $[1,\,8]$

  3. $[1,\,9]$

  4. $[2,\,5]$


Correct Option: A
Explanation:

$|z-1|+|z+3|\le 8$

Using the triangle inequality 
$|z _1\pm z _2|\le |z _1|+|z _2|$
We have $|z-1|+|z+3|\le 8$
$\implies |z-1+z+3|\le 8$
$\implies |z+1|\le 4$
Using triangle inequality again 
$|z|+1\le 4\implies |z|\le 3$
So, the maximum value of $|z _1+ _2|$ is $|z _1|+|z _2|$
And  the minimum value of $|z _1+ _2|$ is $|z _1|-|z _2|$
Hence the maximum value of $|z-4|$ is $|z|+|4|=3+4=7$
 the minimum value of $|z-4|$ is $|z|-|4|=3-4=-1$
Hence the range of $|z-4|$
$1\le|z-4|\le 7$

If $\left|z\right| <\sqrt{2} -1$, then $\left|z^2 + 2 z  cos  \alpha \right|$ is

  1. less than 1

  2. $\sqrt{2} + 1$

  3. $\sqrt{2} -1$

  4. none of these


Correct Option: A
Explanation:

$\left| z \right| <\sqrt { 2 } -1\ \left| { z }^{ 2 }+2z\cos { \alpha  }  \right| \le \left| { z }^{ 2 } \right|+ \left| 2z\cos { \alpha  }  \right| \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad ...\left{ \quad \because \quad \left| { z } _{ 1 }+{ z } _{ 2 } \right| \le \left| { z } _{ 1 } \right| +\left| { z } _{ 2 } \right|  \right} \ \left| { z }^{ 2 }+2z\cos { \alpha  }  \right| \le |z|(|z|+2) \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad ...\left{ \quad \because \quad \left| \cos { \alpha  }  \right| \le 1\quad  \right} \ \Rightarrow \left| { z }^{ 2 }+2z\cos { \alpha  }  \right| <(\sqrt{2}-1){ \left( \sqrt { 2 } +1 \right)  }<1\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad ...\left{ \quad \because \quad \left| z \right| <\sqrt { 2 } -1\quad  \right} \ \therefore \quad \left| { z }^{ 2 }+2z\cos { \alpha  }  \right| <1\ $
Hence, option 'A' is correct.

If z be a complex number for which $|2z  cos  \theta + z^2| = 1$, then the minimum value of |z|
 is ......................

  1. $\sqrt{3} -1$

  2. $\sqrt{3} +1$

  3. $\sqrt{2} -1$

  4. $\sqrt{2} +1$


Correct Option: C
Explanation:

$|z^{2}+2zcos\theta|$
$=|z(z+2cos\theta)|$
$=|z|.|z+2cos\theta|$
$=1$
Now 
$|z|=1$ and 
$|z+2cos\theta|=1$
Now 
$|z+2cos\theta|\leq |z|+|2cos\theta|$
Considering 
$|z+2\cos\theta|=|z|+|2cos\theta|=1$
Hence
$|z|=|2cos\theta|\pm1$
Considering $z=|2cos\theta|-1$ we get the minimum value at multiples of $\theta=45^{0}$
Hence
$z=\sqrt{2}-1$.

$sin^{-1}\left { \frac{1}{i} (z-1)\right }$ ,Where Z is non - real, can be the angle  of a triangle, if 

  1. $Re(z)=1, Im(z)=2$

  2. $Re(z)=1,-1\leq Im(z)\leq 1$

  3. $Re(z)=1,Im(z)=0$

  4. $Re(z)=1,Im(z)=-2$


Correct Option: A

Let $z$ be any point in $\displaystyle A\cap B\cap C$ and let $w$ be any point satisfying $\displaystyle \left | w-2-i \right |< 3.$ Then, $\displaystyle \left | z \right |-\left | w \right |+3$ lies between

  1. $-6$ and $3$

  2. $-3$ and $6$

  3. $-6$ and $6$

  4. $-3$ and $9$


Correct Option: D
Explanation:

$\left| w-\left( 2+i \right)  \right| <3\Rightarrow \left| w \right| -\left| 2+i \right| <3\ \Rightarrow -3+\sqrt { 5 } <\left| w \right| <3+\sqrt { 5 } $
$\Rightarrow -3-\sqrt { 5 } <-\left| w \right| <3-\sqrt { 5 } $   ...(1)
Also, $\left| z-\left( 2+i \right)  \right| =3$
$\Rightarrow -3+\sqrt { 5 } <-\left| z \right| \le 3+\sqrt { 5 } $   ...(2)
$\therefore -3<\left| z \right| -\left| w \right| +3<9$

If $z=a+ib$ where $a>0,b>0$, then

  1. $\displaystyle \left| z \right| \ge \frac { 1 }{ \sqrt { 2 }  } \left( a-b \right) $

  2. $\displaystyle \left| z \right| \ge \frac { 1 }{ \sqrt { 2 }  } \left( a+b \right) $

  3. $\displaystyle \left| z \right| < \frac { 1 }{ \sqrt { 2 }  } \left( a+b \right) $

  4. None of these


Correct Option: B
Explanation:

As ${ \left( a-b \right)  }^{ 2 }\ge 0,{ a }^{ 2 }+{ b }^{ 2 }\ge 2ab$   ...(1)

But $\left| z \right| =\sqrt { { a }^{ 2 }+{ b }^{ 2 } } ;$ si from (1), ${ \left| z \right|  }^{ 2 }\ge 2ab$
$\therefore { \left| z \right|  }^{ 2 }+{ a }^{ 2 }+{ b }^{ 2 }\ge { a }^{ 2 }+{ b }^{ 2 }+2ab\ \Rightarrow { \left| z \right|  }^{ 2 }+{ \left| z \right|  }^{ 2 }\ge { \left( a+b \right)  }^{ 2 }\Rightarrow 2{ \left| z \right|  }^{ 2 }\ge { \left( a+b \right)  }^{ 2 }$
$\Rightarrow \sqrt { 2 } \left| z \right| \ge a+b$ as $\left| z \right| $ is positive
$\displaystyle \left| z \right| \ge \frac { 1 }{ \sqrt { 2 }  } \left( a+b \right) $

The minimum value of $\displaystyle \left | z-1 \right |+\left | z \right |$for complex values of z is

  1. $2$

  2. $\displaystyle \frac{1}{2}$

  3. $0$

  4. $1$


Correct Option: D
Explanation:

$\left| w \right| =\left| \left( w-z \right) +z \right| $ 
Using Triangle Inequality.
$\left| w-z \right| +\left| z \right| \ge \left| \left( w-z \right) +z \right| =\left| w \right| $
$\Rightarrow \left| z \right| +\left| z-w \right| \ge \left| w \right| $
$\Rightarrow \left| z \right| +\left| z-1 \right| \ge 1$
Therefore, minimum value of $\left| z \right| +\left| z-1 \right| $ is 1
Hence, option 'D' is correct.

If $|z| < 4$, then $|iz+3-4i|$ is less then

  1. $4$

  2. $5$

  3. $9$

  4. $6$


Correct Option: C
Explanation:

Given, $|z| < 4$......(1).

Now 
$|iz+3-4i|$$\le |iz|+|(3-4i)|$ [ Using standard inequility of sum of two complex number]
or, $|iz+3-4i|$$\le |z|+5$ [ Since $|iz|=|z|$]
or, $|iz+3-4i|$$\lt 4+5$ [ Using (1)]
or, $|iz+3-4i|$$\lt 9$.

If $\displaystyle \left | z-\frac{2}{z} \right |=1$, then the greatest value of $\left | z \right |$ is 

  1. 2

  2. 1

  3. 4

  4. 3


Correct Option: A
Explanation:

$\left| z-\frac { 2 }{ z }  \right| =1$        ...(1)
Let $ \dfrac{2}{z}=w$

$\left| z \right| =\left| \left( z-w \right) +w \right| \le \left| z-w \right| +\left| w \right| $      ..(Triangle inequality)

$\Rightarrow \left| z \right| -\left| w \right| \le \left| z-w \right| $

$\Rightarrow \left| z \right| -\left| \frac { 2 }{ z }  \right| \le \left| z-\frac { 2 }{ z }  \right| $

$\Rightarrow \left| z \right| -\left| \frac { 2 }{ z }  \right| \le 1$         ...{ from 1 }

$\Rightarrow { \left| z \right|  }^{ 2 }-\left| z \right| -2\le 0\ \Rightarrow -1\le \left| z \right| \le 2\ \Rightarrow 0\le \left| z \right| \le 2$

Therefore,  maximum value of $\left| z \right| $ is 2
Hence, option A is correct. 

If $\displaystyle \left | z \right |< \sqrt{3}-1 $ then $\displaystyle \left | z^{2}+2z\cos\alpha  \right | $ is

  1. less than $2$

  2. $\displaystyle \sqrt{3}+1$

  3. $\displaystyle \sqrt{3}-1 $

  4. None of these


Correct Option: A
Explanation:

$\displaystyle \left | z^{2}+2z \cos \alpha  \right |\leq \left | z \right |^{2}+2\left | z \right |\left | \cos \alpha  \right |  \leq \left | z \right |^{2}+2\left | z \right |$

$|z^2+2z \cos \alpha|  < \left ( \sqrt{3}-1 \right )^{2}+2\left ( \sqrt{3}-1 \right ) = 3+1-2\sqrt{3}+2\sqrt{3}-2=2$

$\displaystyle \therefore \left | z^{2}+2z \cos  \alpha  \right |< 2$

If $|z-4+3i|\le 1$ and $m$ and $n$ are the least and greatest values of $|z|$ and $k$ is the least value of $\displaystyle \frac { { x }^{ 4 }+{ x }^{ 2 }+4 }{ x } $ on the interval $(0,\infty)$, then $k$ is equal to

  1. $m$

  2. $n$

  3. $m+n$

  4. None of these


Correct Option: B
Explanation:

We have, 

$1\ge \left| z-\left( 4-3i \right)  \right| $
$\Rightarrow 1\ge \left| z \right| -\left| 4-3i \right| \quad ,\quad \left| 4-3i \right| -\left| z \right| $
$\Rightarrow 1\ge \left| z \right| -5\quad ,\quad 5-\left| z \right| $
$\left| z \right| \le 6,\left| z \right| \ge 4\Rightarrow 4\le \left| z \right| \le 6\Rightarrow m=4,n=6$
Let $y=\displaystyle\frac { 4+{ x }^{ 2 }+{ x }^{ 4 } }{ x } ={ x }^{ 3 }+x+\displaystyle\frac { 4 }{ x } ={ x }^{ 3 }+x+\frac { 1 }{ x } +\frac { 1 }{ x } +\frac { 1 }{ x } +\frac { 1 }{ x } $
$\because x\in \left( 0,\infty  \right) $, then ${ x }^{ 3 },x,\frac { 1 }{ x } ,\frac { 1 }{ x } ,\frac { 1 }{ x } ,\frac { 1 }{ x } $ are all positive numbers whose product is 1.
Thus their sum y will be least when 
${ x }^{ 3 }=x=\displaystyle\frac { 1 }{ x } \Rightarrow x=1$
So least value of $y=6,k=6$
So $k=n$

The maximum value of $|z|$ when $z$ satisfies the condition $\displaystyle \left | z+\frac{2}{z} \right |=2$

  1. $1-\sqrt{3}$

  2. $\sqrt{3}+\sqrt{3}$

  3. $1+\sqrt{3}$

  4. $\sqrt{3}$


Correct Option: C
Explanation:

$\left| z+\dfrac { 2 }{ z }  \right| =2$


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

$\Rightarrow 2\ge \left| z \right| -\dfrac { 2 }{ \left| z \right|  } \ \Rightarrow { \left| z \right|  }^{ 2 }-2\left| z

\right| -2\le 0\ \Rightarrow \left| z \right| \le \sqrt { 3 } +1$

Ans: C

If $\displaystyle z\epsilon C \; and \; \left | z+4 \right |\leq 3$ then the greatest value of $\left | z+1 \right |$ is

  1. 5

  2. 6

  3. 4

  4. 3


Correct Option: B
Explanation:

$\left| z+4 \right| \le 3$      ...(1)

$\left| \left( z+4 \right) -3 \right| \le \left| z+4 \right| +\left| -3 \right| \ \Rightarrow \left| z+1 \right| \le \left| z+4 \right| +3$

$\Rightarrow \left| z+1 \right| \le 6$       ....{ $\because \quad \left| z+4 \right| \le 3$}

Ans: B

If $\left| z  - \displaystyle \frac{1}{z}\right| = 1$ then

  1. $|z| _{max} = \displaystyle \frac {1+\sqrt 5}{2}$

  2. $|z| _{min} = \displaystyle \frac {1+\sqrt 5}{2}$

  3. $|z| _{max} =\displaystyle \frac {-1+\sqrt 5}{2}$

  4. none of these


Correct Option: A
Explanation:

$\left| z  - \displaystyle \frac{1}{z}\right| = 1$
$|z|-\displaystyle\frac{1}{|z|}\leq |z-\displaystyle\frac{1}{z}|$
$\Rightarrow |z|-\displaystyle\frac{1}{|z|}\leq 1$
$\Rightarrow |z|^2-|z|-1\leq 0$
$\Rightarrow\displaystyle \frac {1-\sqrt 5}{2}\leq |z|\leq \frac {1+\sqrt 5}{2} $
$\therefore |z| _{max}=\displaystyle \frac {1+\sqrt 5}{2}$
Hence, option A.

The maximum value of |z| where z satisfies the condition $\displaystyle \left | z + \frac{2}{z} \right | = 2$ is

  1. $\sqrt{3} -1$

  2. $\sqrt{3} +1$

  3. $\sqrt{3} $

  4. $\sqrt{2} +\sqrt{3} $


Correct Option: B
Explanation:

Given, $\displaystyle \left | z + \frac{2}{z} \right | = 2   $

$   \Rightarrow |z| - \dfrac{2}{|z|} \leq 2      $
$ \Rightarrow |z|^2 - 2 |z| - 2 \leq 0$
$\Rightarrow |z| \leq \displaystyle \frac{2 \pm \sqrt{4 + 8}}{2} \leq 1 \pm \sqrt 3$
Hence, max. value of |z| is $1 + \sqrt 3$

If $|z| \leq 1$ then the minimum and maximum value of |z - 3| are

  1. 4, 2

  2. 3, 4

  3. 4, 6

  4. 2, 4


Correct Option: D


 lf $|\mathrm{z} _{1}|=2,\ |\mathrm{z} _{2}|=3$, then $|\mathrm{z} _{1}+\mathrm{z} _{2}+5+12\mathrm{i}|$ is less than or equal to

  1. $8$

  2. $18$

  3. $10$

  4. $5$


Correct Option: B
Explanation:

We know that $|z _{1}+z _{2}+ z _3 |\leq |z _{1}|+|z _{2}| + | z _3| $
$|z _{1}|+|z _{2}|=5$
$z _3 = 5+12 i $

$|z _3 | = 13 $
$\therefore 18\geq |z _{1}+z _{2}+5+12i|$
Hence, option B is correct

A point M is taken inside a parallelogram ABCD, then area of $\displaystyle \Delta AMD,$ $\displaystyle \Delta AMB,$ $\displaystyle \Delta AMC$ can take which of of the following values, respectively.

  1. 15, 6, 11

  2. 9, 6, 4

  3. 13, 5, 8

  4. 25, 7, 24


Correct Option: C

The complex number $z$ satisfies the condition $\left|\displaystyle {z}-\frac{25}{z}\right|=24$. Then the maximum distance from the origin to the point '$z$' in the argand plane is

  1. 20

  2. 25

  3. 30

  4. 35


Correct Option: B
Explanation:

$|z| = |z - \frac{25}{z} + \frac{25}{z}| \leq 24 + \frac{25}{|z|}$
$\therefore$ $|z|^2 - 24|z| - 25 \leq 0.$
$\therefore$ $(|z| - 25)(|z| + 1) \leq 0., \Rightarrow |z| \leq 25.$
Hence maximum distance of z from origin is 25.

If $|z+4|\leq 3$, then the maximum value of $|{z}+1|$ is

  1. $0$

  2. $4$

  3. $10$

  4. $6$


Correct Option: D

A point $'z'$ moves on the curve $|z - 4 - 3i| = 2$ in an argand plane. The maximum and minimum values of $|z|$ are

  1. $2, 1$

  2. $6, 5$

  3. $4, 3$

  4. $7, 3$


Correct Option: D
Explanation:
Let $w = 4 + 3i$. 
We can write, $|z| = |(z-w) + w|$. Hence by triangle inequality ($|z _1+z _2| \leq |z _1| + |z _2|$), we can write $|z| \leq |z-w| + |w|$. It is given in the question that, $|z-w| = 2$ and $|w| = \sqrt{4^2 + 3^2} = 5$.
Putting the values, we get $|z| \leq 7$. 
Using another result of triangle inequality ($\big||z _1| - |z _2| \big| \leq |z _1 + z _2|$), we can write $\big||z-w| - |w|\big| \leq |z - w + w|$.
Hence, we get $|z| \geq 3$. The minimum value is 3 and maximum value is 7.
Hence, (D) is the correct option

If $|{z _1}| = |{z _2}| = |{z _3}| = 1$ and ${z _1} + {z _2} + {z _3} = 0$ then the area of the triangle whose vertices are $z _1, z _2, z _3$ is

  1. $\frac{3\sqrt{3}}{4}$

  2. $\frac{\sqrt{3}}{4}$

  3. 1

  4. None of these


Correct Option: A
Explanation:

given  $|Z _1|=|Z _2|=|Z _3|=1$     and     $Z _1+Z _2+Z _3=0$
$\Rightarrow |Z _1 -Z _2|=2 (Cos 30)=\sqrt{3}$
$\Rightarrow  area =\frac{\sqrt{3}}{4}a^2$    &     $ a=\sqrt{3}$
So, area $=\frac{\sqrt{3}}{4}\cdot 3 =\frac{3\sqrt{3}}{4}$

Statement 1: $|z _1-a| < a, |z _2-b| < b, |z _3-c| < c$, where a, b, c are positive real numbers, then $|z _1+z _2+z _3|$ is greater than $2|a+b+c|$.
Statement 2: $|z _1\pm z _2| \leq |z _1|+|z _2|$.

  1. Both the statements are true, and Statement 2 is the correct explanation for Statement 1.

  2. Both the statements are true, but Statement 2 is not the correct explanation for Statement 1.

  3. Statement 1 is true and Statement 2 is false.

  4. Statement 1 is false and Statement 2 is true.


Correct Option: D
Explanation:

$|z _1+z _2+z _3| = |z _1-a+z _2-b+z _3-c+(a+b+c)|$
$\leq |z _1-a|+|z _2-b|+|z _3-c|+|a+b+c|$
$\leq 2|a+b+c| $
Hence, $|z _1+z _2+z _3|$ is less than $2|a+b+c|$.
Statement 1 is false and Statement 2 is true

$z _0$ is a root of the equation $z^n cos \theta _o+z^{n-1} cos\theta _1+....+z cos\theta _{n-1}+cos\theta _n=2$, where $\theta, \epsilon R$, then

  1. $|z _0| > 1$

  2. $|z _0| > \dfrac {1}{2}$

  3. $|z _0| > \dfrac {1}{4}$

  4. $|z _0| > \dfrac {3}{2}$


Correct Option: A,B
Explanation:

$z^n cos\theta _0+z^{n-1} cos\theta _1+.....+z cos\theta _{n-1}+cos\theta _n=2$

or $2=|z _0^n cos\theta _0+z _0^{n-1} cos\theta _1+....+z _0 cos\theta _{n-1}+cos\theta _n|$

or $2\leq |z _0|^n |cos\theta _0|+|z|^{n-1}|cos\theta _1|+....+|z _0||cos\theta _{n-1}|+|cos\theta _n|$

or $2\leq |z _0|^n+|z _0|^{n-1}+|z _0|^{n-2}+.....+|z _0|+1$

which is clearly satisfied for $|z _0| \geq 1$. If $|z _0| < 1$, then

$2 < 1+|z _0|+|z _0|^2+.....+|z|^n+....\infty$

$\Rightarrow 2 < \dfrac {1}{1-|z _0|}$

$\Rightarrow |z _0| > \dfrac {1}{2}$

If $\displaystyle |Z - \frac {4}{Z}| = 2$, then the maximum value of $\displaystyle |Z|$ is equal to

  1. $\displaystyle \sqrt 5 + 1$

  2. 2

  3. $\displaystyle 2 + \sqrt 2$

  4. $\displaystyle \sqrt 3 + 1$


Correct Option: A
Explanation:

We have for any two complex numbers $\displaystyle \alpha$ and $\displaystyle \beta$
$\displaystyle ||\alpha|| \leq |\alpha - \beta|$
Now $\displaystyle ||Z|-|\frac {4}{|Z|}||\leq|Z-\frac {4}{Z}|$
$\displaystyle \Rightarrow |Z| - \frac {4}{|Z|}|\leq 2$
Set $\displaystyle |Z| = r > 0$, then $\displaystyle |r-\frac {4}{r}|\leq 2$
$\displaystyle \Rightarrow -2 \leq r - \frac {4}{r} \leq 2$
The left inequality gives
$\displaystyle r^2 + 2r - 4 \geq 0$
The corresponding roots are
$\displaystyle r = \frac {-2\pm \sqrt {20}}{2} = -1 \pm \sqrt 5$
Thus $\displaystyle r \geq \sqrt 5 - 1$ or $\displaystyle r \leq -1 - \sqrt 5$
implies that $\displaystyle r \geq \sqrt 5 - 1$ (As r > 0) ... (i)
Again consider the right inequality
$\displaystyle r - \frac {4}{r} \leq 2 \Rightarrow r^2 - 2 r - 4 \leq 0$
The corresponding roots are
$\displaystyle r = \frac {2 \pm \sqrt {20}}{2} = 1 \pm \sqrt 5$
Thus $\displaystyle 1 - \sqrt 5 \leq r \leq 1 + \sqrt 5$
But r > 0, hence $\displaystyle r \leq 1 + \sqrt 5$ .... (ii)
(i) and (ii) gives
$\displaystyle \sqrt 5 - 1 \leq r \leq \sqrt 5 + 1$
So, the greatest value is $\displaystyle \sqrt 5 + 1$.

The maximum value of $\left| z \right| $ when $z$ satisfies the condition $\displaystyle \left| z+\dfrac { 2 }{ z }  \right| =2$ is

  1. $\sqrt { 3 } -1$

  2. $\sqrt { 3 } +1$

  3. $\sqrt { 3 } $

  4. $\sqrt { 2 } +\sqrt { 3 } $


Correct Option: B
Explanation:

We have $\displaystyle \left| z \right| =\left| z+\frac { 2 }{ z } -\frac { 2 }{ z }  \right| \le \left| z+\frac { 2 }{ z }  \right| +\frac { 2 }{ \left| z \right|  } $

$\displaystyle \Rightarrow \left| z \right| \le 2+\frac { 2 }{ \left| z \right|  } \Rightarrow { \left| z \right|  }^{ 2 }\le 2\left| z \right| +2\ \Rightarrow { \left| z \right|  }^{ 2 }-2\left| z \right| +1\le 1+2\Rightarrow { \left( \left| z \right| -1 \right)  }^{ 2 }\le 3\ \Rightarrow -\sqrt { 3 } \le \left| z \right| -1\le \sqrt { 3 } \Rightarrow 1-\sqrt { 3 } \le \left| z \right| \le 1+\sqrt { 3 } $
That is , the maximum value of $\left| z \right| $ is $1+\sqrt { 3 } $.

If the complex number z satisfies the condition |z| $\geq$ 3, then the least value of $\displaystyle \left | z + \frac{1}{z} \right |$ is equal to.

  1. $2$

  2. $\dfrac{4}{3}$

  3. $1$

  4. $\dfrac{8}{3}$


Correct Option: D
Explanation:
By using triangle inequality:  $||z _1-|z _2||\le |z _1+z _2|\le |z _1|+|z _2|$

We have    $|z+\dfrac{1}{z}|\leq |z|+|\dfrac{1}{z}|$

Now Given that
$|z|\geq 3$

$\therefore |z+\dfrac{1}{z}|\leq |3|+|\dfrac{1}{3}|$

$\Rightarrow |z+\dfrac{1}{z}|\leq 3-\dfrac{1}{3}$

$\Rightarrow |z+\dfrac{1}{z}|\leq \dfrac{8}{3}$

Let $\left| { z } _{ r }-r \right| \le r$, for all $ r = 1, 2, 3, ..., n.$ Then $\left| \sum _{ r=1 }^{ n }{ { z } _{ r } }  \right| $ is less than

  1. $n$

  2. $2n$

  3. $n(n+1)$

  4. $\displaystyle \frac{n(n+1)}{2}$


Correct Option: C
Explanation:

$\left| { z } _{ 1 }-1 \right| \le 1,\quad \left| { z } _{ 2 }-2 \right| \le 2,\quad \left| { z } _{ 3 }-3 \right| \le 3...\left| { z } _{ n }-n \right| \le n$
Adding these and using triangle inequality:
$\left| { z } _{ 1 }+{ z } _{ 2 }+...{ z } _{ n }-(1+2+...n) \right| \le 1+2+...n\quad =>\quad \left| { z } _{ 1 }+{ z } _{ 2 }+...{ z } _{ n }-\left(\dfrac { n(n+1) }{ 2 } \right) \right| \le \dfrac { n(n+1) }{ 2 } $
Thus, $\left| { z } _{ 1 }+{ z } _{ 2 }+...{ z } _{ n } \right| -\left(\dfrac { n(n+1) }{ 2 } \right)\le \dfrac { n(n+1) }{ 2 } \quad =>\quad \left| { z } _{ 1 }+{ z } _{ 2 }+...{ z } _{ n } \right| \le n(n+1)$
Hence, (c) is correct.

If $Re(z)$ is a positive integer, then value of the $|1+z+...+z^n|$ cannot be less than

  1. $|z^n| - \displaystyle\frac{1}{|z|}$

  2. $|z^n| + \displaystyle\frac{1}{|z|}$

  3. $n|z|^n$

  4. $n|z|^n + 1$


Correct Option: A
Explanation:

$|1+z+z^{2}+..z^{n}|\leq 1+|z|+|z|^{2}+..|z|^{n}$
$\leq \dfrac{|z|^{n+1}-1}{|z|-1}$

$\leq\dfrac{(|z|.|z|^{n}-1)}{|z|-1}$

$\leq\dfrac{|z|}{|z|-1}.[|z|^{n}-\dfrac{1}{|z|}]$
Hence 
It cannot be less than $[|z|^{n}-\dfrac{1}{|z|}]$.

If $z _{1},\ z _{2}--,\ z _{n}$ are complex numbers such that $|z _{i}|<\mathrm{l}\mathrm{a}\mathrm{n}\mathrm{d}\lambda _{i}>0$ for $i=1,2,---n$ and $\lambda _{1}+\lambda _{2}+--+\lambda _{n}=1$ then $|\lambda _{1}z _{1}+\lambda _{2}z _{2}+--+\lambda _{n}\mathrm{z} _{1}|?$

  1. $=1$

  2. $<1$

  3. $>1$

  4. $=n$


Correct Option: B
Explanation:

Given:

$\lambda _i>0$  and  $\lambda _1+\lambda _2+...+\lambda _n=1$
$\therefore 0<\lambda _i<1$
Also,  $|z _i|<1$
$\therefore \lambda _1|z _1|+\lambda _2|z _2|+.....+\lambda _n|z _n|<1$                ......( 1 )

$\therefore |\lambda _1z _1+\lambda _2z _2+.....+\lambda _nz _n|$
$\leq |\lambda _1z _1|+|\lambda _2z _2|+.....+|\lambda _nz _n|$
$\leq \lambda _1|z _1|+\lambda _2|z _2|+.....+\lambda _n|z _n|$
$<1$

If $\left | z-i \right |\leq 2$ and $z _{0}=13+5i$, then the maximum value of $\left | iz+z _{0} \right |$ is

  1. $12$

  2. $15$

  3. $13$

  4. None of these


Correct Option: B
Explanation:
$\left | iz+z _{0} \right |=\left | iz +1 + z _{0} -1\right |$
$\left | iz+z _{0} \right |=\left | iz -i^{2} + z _{0} -1\right |$
$=|{i}({z}-{i})+13+5{i}-1|$
$\leq|{i}||{z}-{i}|+|12+5i|\leq 1\times2+13\le15$

Hence, option B.
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