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

$z _0$ 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. $|z _0| > \dfrac {3}{2}$
Reveal answer Fill a bubble to check yourself
A,B Correct answer
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

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

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

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

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

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. $\displaystyle \sqrt 3 + 1$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $\displaystyle \sqrt 5 + 1$.

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

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 } $
Reveal answer Fill a bubble to check yourself
B Correct answer
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 } $.

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

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. $\dfrac{8}{3}$
Reveal answer Fill a bubble to check yourself
D Correct answer
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
$|z|\geq 3$

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

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

$\Rightarrow |z+\dfrac{1}{z}|\leq \dfrac{8}{3}$
Multiple choice maths congruency of triangles triangle inequality inequalities in triangle inequalities in triangles

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. $\displaystyle \frac{n(n+1)}{2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

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

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. $n|z|^n + 1$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $[|z|^{n}-\dfrac{1}{|z|}]$.

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

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. $=n$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 )

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

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

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
Reveal answer Fill a bubble to check yourself
B Correct answer
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.
Multiple choice using trigonometric tables trigonometric ratios of some specific angles trigonometric identities trigonometry maths

Choose the correct option for the following statement.

The line of sight is the line drawn from the eye of an observer to the point in the object viewed by the observer.

  1. The given statement is true

  2. The given statement is false

  3. Incomplete information

  4. None of the above.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

By definition,

The line of sight is the line drawn from the eye of an observer to the point in the object viewed by the observer.
Therefore, the given statement is true.