Tag: maths

Questions Related to maths

Let  $z _ { 1 } , z _ { 2 }$  and  $z _ { 3 }$  represent the vertices  $A, B$  and  $C$  of the triangle  $A B C$  in the argand that  $\left| z _ { 1 } \right| = \left| z _ { 2 } \right| = \left| z _ { 3 } \right| = 5,$  then  $z _ { 1 } \sin 2 A + z _ { 2 } \sin 2 B + z _ { 3 } \sin 2 C = 0.$

maths complex numbers and linear inequations argand plane and polar representation geometric representation of a complex number complex numbers and quadratic equations

  1. True

  2. False

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Correct Option: A

If $\sin \frac {6\pi}5+i(1+\cos \frac {6\pi }5)$ then

maths complex numbers and linear inequations argand plane and polar representation geometric representation of a complex number complex numbers and quadratic equations

  1. $|Z|=-2\cos \frac {3\pi}5$

  2. $Arg(Z)=\frac {\pi}5$

  3. $Arg(Z)=\frac {9\pi }{10}$

  4. none of these

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Correct Option: C

If Arg $(z + i)\, -$ Arg $(z - i)$ $= \dfrac{\pi}{2}$, then $z$ lies on a ..........

maths complex numbers and linear inequations argand plane and polar representation geometric representation of a complex number complex numbers and quadratic equations

  1. Circle

  2. Line

  3. Coordinate axes

  4. None of these

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Correct Option: A
Explanation:
Putting z = x+iy,

${tan}^{-1}\dfrac{y+1}{x}$  -  ${tan}^{-1}\dfrac{y-1}{x}$ = $\pi$/2

$\Rightarrow$ 1 + ($\dfrac{y+1}{x})$($\dfrac{y-1}{x}$) = 0

$\Rightarrow$ $x^2 +  y^2$ = 1

It is a circle of center coinciding with origin and radius 1 units.

Hence, option A is correct.

If $\overline { z } $ lies in the third quadrant then $z$ lies in the

maths complex numbers and linear inequations argand plane and polar representation geometric representation of a complex number complex numbers and quadratic equations

  1. First quadrant

  2. Second quadrant

  3. Third quadrant

  4. Fourth quadrant

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Correct Option: B
Explanation:

Fact:  $\overline z$ is image of $z$ in $x$-axis

So if $\overline z$ lies in third quadrant then $z$ will lie in second quadrant 

Let $z _1$ and $z _2$ are two complex numbers such that $(1-i)z _1=2z _2$ and $arg(z _1z _2)=\dfrac{\pi}{2}$ then $arg(z _2)$ is equals to:

maths complex numbers and linear inequations argand plane and polar representation geometric representation of a complex number complex numbers and quadratic equations

  1. $\dfrac{3 \pi}{8}$

  2. $\dfrac{\pi}{8}$

  3. $\dfrac{5 \pi}{8}$

  4. $\dfrac{-7 \pi}{8}$

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Correct Option: B
Explanation:

$(1-i)z _1=2z _2$

$\cfrac{z _2}{z _1}=\cfrac{1}{2}-\cfrac{i}{2}$
Let $z _1=r _1e^{i\theta _1}\ and\ z _=r _1r^{\theta _2}$
$arg(\cfrac{z _2}{z _1})=\tan^{-1}\cfrac{-1/2}{1/2}=-\pi/4$
$\theta _2-\theta _1=-\pi/4$  and $arg(z _1z _2)=\pi/2$ (given)
$\implies \theta _2-\theta _1=-\pi/4$  and $\theta _2
+\theta _1=\pi/2$
$\implies 2\theta _2=\pi/4,\theta _2=\pi/8$
$\implies arg(z _2)=\pi/8$

The complex number $\dfrac{1 + 2i}{1 - i}$ lies in which quadrant of the complex plane.

maths complex numbers and linear inequations argand plane and polar representation geometric representation of a complex number complex numbers and quadratic equations

  1. First

  2. Second

  3. Third

  4. Fourth

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Correct Option: B
Explanation:
$\dfrac{1+2i}{1-i}$
$\Rightarrow \dfrac{1+2i}{1-i}\times \dfrac{1+i}{1+i}$
$=\dfrac{1+i+2i-2i^2}{1-i^2}=\dfrac{1+3i-2}{2}$
$=\dfrac{-1+3i}{2}$
$\therefore$ It lies in $2^{nd}$ Quadrant.

If $arg(z) < 0$, then $arg(-z)-arg(z)=$

maths complex numbers and linear inequations argand plane and polar representation geometric representation of a complex number complex numbers and quadratic equations

  1. $\pi$

  2. $-\pi$

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

  4. $-\dfrac{\pi}{2}$

Show answer
Correct Option: A
Explanation:

Let $Z=re^{i\theta _1}$

$-Z=-re^{i\theta _1}$
$\implies -a\cos\theta _1-ib\sin\theta _1$
$\implies -a\cos(\pi+\theta _1)-ib\sin(\pi+\theta _1)$
$\implies re^{i(\pi+\theta _1)}$
$arg(-Z)-arg(Z)$
$\implies \pi+\theta _1-\theta _1\ \implies \pi$

Which of the given alternatives represent a point in Argand plane, equidistant from roots of the equation $(z+1)^4= 16z^4$?

maths complex numbers and linear inequations argand plane and polar representation geometric representation of a complex number complex numbers and quadratic equations

  1. $(0,0)$

  2. $\left(-\dfrac{1}{3},0\right)$

  3. $\left(\dfrac{1}{3},0\right)$

  4. $\left(0,\dfrac{2}{\sqrt5}\right)$

Show answer
Correct Option: C
Explanation:

Consider the given equation $(z+1)^4=16z^4$
$ \Rightarrow (z+1)^4=16z^4$

$ \Rightarrow z+1=(2^4z^4)^{\frac{1}{4}}$

$ \Rightarrow |z+1|=2|z|$

We know that $z=x+iy$

Therefore $|x+iy+1|=2|x+iy|$

$ \Rightarrow \sqrt{(x+1)^2+y^2}=2\sqrt{x^2+y^2}$

$ \Rightarrow (x+1)^2+y^2=4(x^2+y^2)$

$ \Rightarrow x^2+2x+1+y^2=4x^2+4y^2$

$ \Rightarrow 3x^2+3y^2-2x-1=0$

Divide throughout by 3 we get,
$ \Rightarrow x^2+y^2-\dfrac{2}{3}x-\dfrac{1}{3}=0$, which represents a circle.

We know that for the circle equation of the form $x^2+y^2+2gx+2hy+c=0$ the center of the circle is given by $(-g,-h)$

We have $3x^2+3y^2-2x-1=0$ where $g=-\dfrac{1}{3}, h=0$.

Hence the center is $(\dfrac{1}{3},0)$ which is equidistant from the root of the equation.

$\sin ^{ 8 }{ \theta  } -\cos ^{ 8 }{ \theta  } -\left( \sin ^{ 2 }{ \theta  } -\cos ^{ 2 }{ \theta  }  \right) \left( 1-\sin ^{ 2 }{ \theta  }  \right) $=0

maths trigonometric ratios of acute angles compound angles, multiple angles, sub multiple angles and transformation formulae trigonometric identities trigonometrical ratio and identities

  1. True

  2. False

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Correct Option: B
Explanation:

$\sin^4 \theta+\cos^4 \theta=(\sin^2 \theta+\cos^2 \theta)^2-2\sin^2 \theta\cos^2 \theta=1-2\sin^2 \theta\cos^2 \theta$

$\sin^4 \theta-\cos^4 \theta=(\sin^2 \theta-\cos^2 \theta)(\sin^2 \theta+\cos^2 \theta)=\sin^2 \theta-\cos^2 \theta$
$\sin^8\theta-\cos^8\theta-(\sin^2 \theta-\cos^2 \theta)(1-\sin^2 \theta)=(\sin^4 \theta+\cos^4 \theta)(\sin^4 \theta-\cos^4 \theta)-(\sin^2 \theta-\cos^2 \theta)(1-\sin^2 \theta)$
                                                                                 $=(1-2\sin^2 \theta\cos^2 \theta)(\sin^2 \theta-\cos^2 \theta)-(\sin ^2 \theta-\cos^2\theta)(1-\sin^2 \theta)$
                                                                                 $=(\sin^2 \theta-\cos^2 \theta)(\sin^2 \theta)(1-2\cos^2 \theta)$
So the given relation is $\text{False}$

If $tan x + cot x = 2$, then $sin^{2n}x+cos^{2n}x=$

maths compound angles, multiple angles, sub multiple angles and transformation formulae trigonometric ratios of acute angles trigonometric identities trigonometrical ratio and identities

  1. $\dfrac{1}{2}$

  2. $2^n$

  3. $\dfrac{1}{2^n}$

  4. $\dfrac{1}{2^{n-1}}$

Show answer
Correct Option: D
Explanation:

Given $\tan x+\cot x=2$

$\implies \tan x+\dfrac{1}{\tan x}=2$
$\implies \tan^2 x-2\tan x+1=0$
$\implies (\tan x-1)^2=0$
$\implies \tan x=1\implies x=\dfrac{\pi}{4}$
$\sin^{2 n} x+\cos^{2 n} x=\bigg(\dfrac{1}{\sqrt{2}}\bigg)^{2 n}+\bigg(\dfrac{1}{\sqrt{2}}\bigg)^{2 n}=\dfrac{1}{2^n}+\dfrac{1}{2^n}=\dfrac{2}{2^n}=\dfrac{1}{2^{n-1}}$