Tag: inequalities in triangle

Questions Related to inequalities in triangle

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

maths congruency of triangles triangle inequality inequalities in triangle inequalities in triangles

  1. $4$

  2. $5$

  3. $9$

  4. $6$

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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 

maths congruency of triangles triangle inequality inequalities in triangle inequalities in triangles

  1. 2

  2. 1

  3. 4

  4. 3

Show answer
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

maths congruency of triangles triangle inequality inequalities in triangle inequalities in triangles

  1. less than $2$

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

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

  4. None of these

Show answer
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

maths congruency of triangles triangle inequality inequalities in triangle inequalities in triangles

  1. $m$

  2. $n$

  3. $m+n$

  4. None of these

Show answer
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 $\displaystyle \left | z+\frac{2}{z} \right |=2$

maths congruency of triangles triangle inequality inequalities in triangle inequalities in triangles

  1. $1-\sqrt{3}$

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

  3. $1+\sqrt{3}$

  4. $\sqrt{3}$

Show answer
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

maths congruency of triangles triangle inequality inequalities in triangle inequalities in triangles

  1. 5

  2. 6

  3. 4

  4. 3

Show answer
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

maths congruency of triangles triangle inequality inequalities in triangle inequalities in triangles

  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

Show answer
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

maths congruency of triangles triangle inequality inequalities in triangle inequalities in triangles

  1. $\sqrt{3} -1$

  2. $\sqrt{3} +1$

  3. $\sqrt{3} $

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

Show answer
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

maths congruency of triangles triangle inequality inequalities in triangle inequalities in triangles

  1. 4, 2

  2. 3, 4

  3. 4, 6

  4. 2, 4

Show answer
Correct Option: D


$|\mathrm{z} _{1}-\mathrm{z} _{2}|=$

maths congruency of triangles triangle inequality inequalities in triangle inequalities in triangles

  1. $\geq||z _{1}|-|z _{2}||$

  2. $\leq|z _{1}|-|z _{2}|$

  3. $=|\mathrm{z} _{1}|-|\mathrm{z} _{2}|$

  4. $\geq|\mathrm{z} _{1}|-|\mathrm{z} _{2}|$

Show answer
Correct Option: A
Explanation:

Let $ argz _1=\theta _1  \quad argz _2=\theta _2$
we know that
$|z _{1}-z _{2}|^{2}=|z _{1}|^{2}+|z _{2}|^{2}-2|z _{1}||z _{2}|cos(\theta _{1}-\theta _{2})$


now, $+1\geq cos(\theta _{1}-\theta _{2})\geq -1$


$-2|z _{1}||z _{2}|cos(\theta _{1}-\theta _{2})\geq -2|z _{1}||z _{2}|$


$\therefore |z _{1}|^{2}+|z _{2}|^{2}-2|z _{1}||z _{2}|cos(\theta _{1}-\theta _{2})\geq |z _{1}|^{2}+|z _{2}|^{2}-2|z _{1}||z _{2}|$


$\therefore |z _{1}-z _{2}|^{2}\geq (|z _{1}|-|z _{2}|)^{2}\Rightarrow |z _{1}-z _{2}|\geq ||z _{1}|-|z _{2}||$