Tag: inequalities in triangles

Questions Related to inequalities in triangles

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

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

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

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

  3. 1

  4. None of these

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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: $|z _1\pm z _2| \leq |z _1|+|z _2|$.

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

  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.

Show answer
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

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

  1. $|z _0| > 1$

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

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

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

Show answer
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

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

  1. $\displaystyle \sqrt 5 + 1$

  2. 2

  3. $\displaystyle 2 + \sqrt 2$

  4. $\displaystyle \sqrt 3 + 1$

Show answer
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

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:

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.

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

  1. $2$

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

  3. $1$

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

Show answer
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

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

  1. $n$

  2. $2n$

  3. $n(n+1)$

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

Show answer
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

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

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

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

  3. $n|z|^n$

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

Show answer
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 $|\lambda _{1}z _{1}+\lambda _{2}z _{2}+--+\lambda _{n}\mathrm{z} _{1}|?$

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

  1. $=1$

  2. $<1$

  3. $>1$

  4. $=n$

Show answer
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

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

  1. $12$

  2. $15$

  3. $13$

  4. None of these

Show answer
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.