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Questions Related to maths

$x$ and $y$ are real numbers such that ${7^x} - 16y = 0\;{\text{and}}\;{4^x} - 49y = 0,$ then the value of $\left( {y - x} \right)$ is

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

  1. $\dfrac{5}{2}$

  2. $\dfrac{{19}}{5}$

  3. $\dfrac{{4115}}{{2013}}$

  4. $\dfrac{{1569}}{{784}}$

Show answer
Correct Option: D
Explanation:
$7^{x}=16y$

$4^{x}=49 y$

$\Rightarrow \dfrac{7^{x}}{4^{x}} = \dfrac{16}{49}$

$\Rightarrow \left( \dfrac{7}{4} \right)^{x} = \left( \dfrac{4}{7} \right)^{2} = \left( \dfrac{7}{4} \right)^{-2}$

$\Rightarrow x=-2$

$y= \dfrac{7^{x}}{16}$

$\Rightarrow y= \dfrac{1}{49 \times 16}$

So, $y-x = \dfrac{1}{49 \times 16}+2$

$=\dfrac{1}{784}+2$

$=\dfrac{1569}{784}$

if $3^{x}-3^{x-1}=18$, then $x^{x}$ is equal to

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

  1. $3$

  2. $8$

  3. $27$

  4. $216$

Show answer
Correct Option: C
Explanation:

$3^x-3^{x-1}=18$

$\Rightarrow 3^x-\dfrac{3^x}{3}=18$
$\Rightarrow 3^x\left(\dfrac{2}{3}\right)=18$
     $3^{x-1}.2=3^2.2$
     $x-1=2$
$\therefore x=3$
    $x^x=3^3=27.$
Hence, the answer is $27.$

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

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

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

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

  3. 0

  4. none of these

Show answer
Correct Option: B
Explanation:
Given,

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

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

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

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

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

$=\dfrac{1}{x\log _e \left(x\right)}$

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

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

  1. 2

  2. 4

  3. 6

  4. 8

Show answer
Correct Option: D
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.

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 :

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

  1. {6,3}

  2. {3,6}

  3. {6,12}

  4. {12,6}

Show answer
Correct Option: A
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)

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

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

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

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

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

$\therefore y=3,6$

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

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

$(x,y)=\left \{ 6,3 \right \}or\left \{ 3,6 \right \}$

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

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

  1. zero

  2. exactly 1

  3. exactly 2

  4. 5

Show answer
Correct Option: A
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 $\:x\in \mathbb{R}$

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

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

  1. 1

  2. 2

  3. 3

  4. 4

Show answer
Correct Option: A
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.

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:

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

  1. one

  2. two

  3. three

  4. four

Show answer
Correct Option: B
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

If $H.C.F. \left(a,b\right) = 9$ and $a  . b = 100$, then $L.C.M.\left(a,b\right) =$

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

  1. $200$

  2. $100/3$

  3. $100/9$

    1. $150$
Show answer
Correct Option: C
Explanation:

$H.C.F.(a,b)=9\ a.b=100\ L.C.M.(a,b)=\frac { a.b }{ H.C.F } \ =\quad \frac { 100 }{ 9 } $

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

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

  1. $4$

  2. $2$

  3. $8$

  4. $1$

Show answer
Correct Option: A
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