Tag: complex numbers and linear inequations

Quizzes (4) Questions (127)

Questions Related to complex numbers and linear inequations

Multiple choice maths solving equations numerically finding roots by iteration fundamental theorem of algebra complex numbers and linear inequations

If $\quad y={ log } _{ x }({ log } _{ e }x)({ log } _{ e }x)\quad then\quad \dfrac { dy }{ dx } \quad equals$ to 

  1. $\dfrac { 1 }{ x{ log } _{ x }{ log } _{ x }x } $
  2. $\dfrac { 1 }{ x{ log } _{ e }x } $

  3. 0

  4. none of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Given,

$y=\log _x\left(\log _e\left(x\right)\right)\left(\log _e\left(x\right)\right)$

$\dfrac{dy}{dx}=\dfrac{d}{dx}\left(\log _x\left(\log _e\left(x\right)\right)\log _e\left(x\right)\right)$

$\left(f\cdot g\right)'=f\:'\cdot g+f\cdot g'\rightarrow f=\log _x\left(\log _e\left(x\right)\right),\:g=\log _e\left(x\right)$

$=\dfrac{d}{dx}\left(\log _x\left(\log _e\left(x\right)\right)\right)\log _e\left(x\right)+\dfrac{d}{dx}\left(\log _e\left(x\right)\right)\log _x\left(\log _e\left(x\right)\right)$

$=\dfrac{1-\log _e \left(\log \left(x\right)\right)}{x\log ^2\left(x\right)}\log _e\left(x\right)+\dfrac{1}{x}\log _x\left(\log _e\left(x\right)\right)$

$=\dfrac{1}{x\log _e \left(x\right)}$
Multiple choice maths solving equations numerically finding roots by iteration fundamental theorem of algebra complex numbers and linear inequations

The number of ordered pairs of integers (x,y) satisfying the equation
${ x }^{ 2 }+6x+{ y }^{ 2 }=4$ is

  1. 2

  2. 4

  3. 6

  4. 8
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Given,

$x^2+6x+y^2=4$ 

$\Rightarrow x^2+6x+9+y^2=4+9$

$\Rightarrow x^2+3^2+2(3x)+y^=13$

$\Rightarrow (x+3)^2+y^2=13$

sum of two squares is $13$

$\therefore $ when $(x+3)^2=9,x=0,-6$ and $y^2=2,-2$ $\Rightarrow 4$ ordered pairs.

$\therefore $ when $(x+3)^2=4,x=-1,-5$ and $y^2=3,-3$ $\Rightarrow 4$ ordered pairs.

A total of $8$ pairs.
Multiple choice maths solving equations numerically finding roots by iteration fundamental theorem of algebra complex numbers and linear inequations

The solution set of the system of equations $\log _{ 3 }{ x } +\log _{ 3 }{ y } =2+\log _{ 3 }{ 2 } \quad and\quad \log _{ 27 }{ (x+y) } =\dfrac { 2 }{ 3 } $ is :

  1. {6,3}

  2. {3,6}

  3. {6,12}

  4. {12,6}
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Given,

$\log _3\left(x\right)+\log _3\left(y\right)=2+\log _3\left(2\right)$

$\log _3\left(x\right)=2+\log _3\left(2\right)-\log _3\left(y\right)$

$x=3^{2+\log _3\left(2\right)-\log _3\left(y\right)}$

$=3^{\log _3\left(2\right)}\cdot \:3^{-\log _3\left(y\right)}\cdot \:3^2$

$=2\cdot \:3^{-\log _3\left(y\right)}\cdot \:3^2$

$=2y^{-1}\cdot \:3^2$

$\Rightarrow x=\dfrac{18}{y}$.......(1)

Now,

$\log _{27}\left(x+y\right)=\dfrac{2}{3}$

from (1)

$\log _{27}\left(\dfrac{18}{y}+y\right)=\dfrac{2}{3}$

$\dfrac{18}{y}+y=27^{\frac{2}{3}}$

$\Rightarrow 18+y^2=9y$

$y^2-9y+18=0$

$(y-3)(y-6)=0$

$\therefore y=3,6$

$x=\dfrac{18}{y}=\dfrac{18}{3}=6$

$x=\dfrac{18}{y}=\dfrac{18}{6}=3$

$(x,y)=\left \{ 6,3 \right \}or\left \{ 3,6 \right \}$
Multiple choice maths solving equations numerically finding roots by iteration fundamental theorem of algebra complex numbers and linear inequations

Number of real solutions of the equation $\sqrt { \log _{ 10 }{ (-x) }  } =\log _{ 10 }{ \sqrt { { x }^{ 2 } }  } $ is :

  1. zero

  2. exactly 1

  3. exactly 2

  4. 5
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Given,

$\sqrt{\log _{10}(-x)}=\log _{10}(\sqrt{x^2})$

$\sqrt{\log _{10}(-x)}=\log _{10}x$

$\log _{10}\left(-x\right)=\left(\log _{10}\left(x\right)\right)^2$

$\log _{10}\left(-1\right)+\log _{10}\left(x\right)=\left(\log _{10}\left(x\right)\right)^2$

let $\log _{10}\left(x\right)=u$

$\log _{10}\left(-1\right)+u=\left(u\right)^2........\log _{10}(-1) \ is \ not \ defined$

$u=\mathrm{Undefined}$

$\:\log _{10}\left(x\right)=\mathrm{Undefined}:\quad x=10^{\mathrm{Undefined}}$

$x=10^{\mathrm{Undefined}}\space\mathrm{False}$

No solution for $\:x\in \mathbb{R}$
Multiple choice maths solving equations numerically finding roots by iteration fundamental theorem of algebra complex numbers and linear inequations

Number of solutions satisfying, $\sqrt { 5-{ log } _{ 2 }x } =3-{ log } _{ 2 }x$ are :

  1. 1

  2. 2

  3. 3

  4. 4
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\sqrt { 5-{ \log } _{ 2 }x } =3-{ \log } _{ 2 }x$                  ---- ( 1 )


Let $\log _2x=y$                     ----- ( 2 )


$\Rightarrow$  $\sqrt{5-y}=3-y$

Squaring both sides, 

$\Rightarrow$  $5-y=9-6y+y^2$

$\Rightarrow$  $y^2-5y+4=0$

$\Rightarrow$  $y^2-4y-y+4=0$

$\Rightarrow$  $y(y-4)-1(y-4)=0$

$\Rightarrow$  $(y-4)(y-1)=0$

$\Rightarrow$  $y=4$ and $y=1$

Substituting $y=1$ in ( 1 ) we get,

$\Rightarrow$  $\log _2x=1$

We know, $\log _ba=x\Rightarrow a=b^x$

$\therefore$  $x=2^1=2$

Substituting $y=4$ in ( 1 ) we get,

$\Rightarrow$  $\log _2x=4$

We know, $\log _ba=x\Rightarrow a=b^x$

$\therefore$  $x=2^4=16$

Substituting $\log _2x=1$ in ( 1 ) we get,

$\sqrt{5-1}=3-1$

$\Rightarrow$  $\sqrt{4}=2$

$\therefore$  $2=2$

Substituting $\log _2x=4$ in ( 1 ) we get,

$\sqrt{5-4}=3-4$

$\Rightarrow$  $\sqrt{1}=-1$

$\therefore$  $1=-1$

Hence, we can see only $\log _2x=1$ satisfying.

$\therefore$  $\sqrt { 5-{ \log } _{ 2 }x } =3-{ \log } _{ 2 }x$ has only $1$ solution.

Multiple choice maths solving equations numerically finding roots by iteration fundamental theorem of algebra complex numbers and linear inequations

Number of ordered pair(s) of (x,y) satisfying the system of equations, $\log _2 xy = 5$ and $\log _{\frac{1}{2}} \frac{x}{y} = 1$ is:

  1. one

  2. two

  3. three

  4. four
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Given,

$\log _2\left(xy\right)=5,\:\log _{\frac{1}{2}}\left(\frac{x}{y}\right)=1$

$\log _2\left(xy\right)=5$

$xy=2^5=32$

$x=\dfrac{32}{y}$

$\log _{\frac{1}{2}}\left(\dfrac{x}{y}\right)=1$

$\log _{\frac{1}{2}}\left(\dfrac{\frac{32}{y}}{y}\right)=1$

$\dfrac{\frac{32}{y}}{y}=\left(\dfrac{1}{2}\right)^1$

$\dfrac{32}{y}=\dfrac{1}{2}y$

$y^2=64$

$\Rightarrow y=\pm 8$

$x=\dfrac{32}{\pm 8}=\pm 4$

$(4,8),(-4,-8)$ Therefore $2$ ordered pairs
Multiple choice maths solving equations numerically finding roots by iteration fundamental theorem of algebra complex numbers and linear inequations

Find multiplicity of the polynomial
$f(x) = (x-1)^2(2x+5)^3(x^2+1)^2(x+\pi^2)^4$

  1. $4$
  2. $2$
  3. $8$
  4. $1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The roots of the given function are $1,-\frac{5}{2} ,-i,i,-\pi^{2}$

The multiplicity of$x=1$ is $2$
The multiplicity of $x=-\frac{5}{2}$ is $3$
The multiplicity of $x=i $ is $2$
The multiplicity of $x=-i$ is $2$
The multiplicity of $x=-\pi^{2}$ is $4$
Therefore the multiplicity of polynomial is $4$
Therefor option $A$ is correct

Multiple choice maths solving equations numerically finding roots by iteration fundamental theorem of algebra complex numbers and linear inequations

The multiplicity of the root $x=1$ for the function $f(x) = x^2(x+1)^3(x-2)^2(x-1)$ is

  1. $1$
  2. $2$
  3. $3$
  4. $4$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

After factorising $f(x)$, $(x-1)$ is appearing only once 

Multiplicity of the root is the number of times a number is zero of a polynomial.
Here $x=1$ is the zero of $f(x)$ only once as it is appearing one.
So, the multiplicity of the root $x=1$x=1 for the function is one.
Hence, option A is correct.

Multiple choice maths solving equations numerically finding roots by iteration fundamental theorem of algebra complex numbers and linear inequations

List the multiplicities of the zeroes of the polynomial $P(x)=x^2-14x+49$

  1. $x=5$ is a zero of multiplicity $3$.
  2. $x=7$ is a zero of multiplicity $2$.
  3. $x=6$ is a zero of multiplicity $1$.

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given, $x^2-14x+49$

$\Rightarrow x^2-7x-7x+49$
$\Rightarrow x(x-7)-7(x-7)$
$\Rightarrow (x-7)^2$
So, $x=7$ is a zero of multiplicity $2$.