Tag: maths

Questions Related to maths

$z _0$ 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.

The value of ${ cosec }^{ 2 }{ 51 }^{ 0 }-{ cot }^{ 2 }{ 51 }^{ 0 }$ is 

using trigonometric tables trigonometric ratios of some specific angles trigonometric identities trigonometry maths

  1. 1

  2. -1

  3. 0

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

Show answer
Correct Option: A
Explanation:

We know,

$1 + \cot^2\theta = cosec^2 \theta$

$\therefore cosec^2\theta - \cot^2 \theta = 1$

Thus, $cosec^2  51^o - \cot^251^o = 1$

Hence, option a is correct.

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.

using trigonometric tables trigonometric ratios of some specific angles trigonometric identities trigonometry maths

  1. The given statement is true 

  2. The given statement is false

  3. Incomplete information

  4. None of the above.

Show answer
Correct Option: A
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.