Tag: maths

Questions Related to maths

The solution set of the equation 
$x^{2/3} + x^{1/3} = 2 $ is

reciprocal equations theory of equations maths

  1. ${-8, 1}$

  2. ${8, 1}$

  3. ${1, -1}$

  4. ${2, -2}\$

Show answer
Correct Option: A
Explanation:

Let ${ x }^{ \cfrac { 1 }{ 3 }  }=t\ { t }^{ 2 }+t-2=0\ { t }^{ 2 }+2t-t-2=0\ t(t+2)-(t+2)=0\ (t-1)(t+2)=0\ t=1\ x^{ \cfrac { 1 }{ 3 }  }=1\ x=1\ t=-2\ x^{ \cfrac { 1 }{ 3 }  }=-2\ x=-8\ x=\left( -8,1 \right)$

Identify which of the following are reciprocal equations of 1st type.

reciprocal equations theory of equations maths

  1. $2x^4+5x^3+2x^2+5x-2=0$

  2. $2x^4-5x^3+2x^2-5x+2=0$

  3. $2x^4-5x^3+2x^2+5x-2=0$

  4. None of the above

Show answer
Correct Option: B
Explanation:

Reciprocal equation is the equation which have even numbers of roots and if one root is $x$ then the other root will be $\dfrac{1}{x}$ and the multiplication of all roots will be one.

Now $1st$ type = where cofficients  $a=e$ in 4th order equation  $ax^4 +bx^3 +cx^{2}  + dx+e = 0 $

In option [A]  $a =2$  and $e = -2$  not $1st$ type

In option [B]  $a =2$  and $e = 2$  this is a $1st$ type reciprocal equation 

In option [C]  $a =2$  and $e = -2$  not a $1st$ type reciprocal equation.
Hence, B is correct.

Identify if the following equation is a reciprocal equation by rearranging.

reciprocal equations theory of equations maths

  1. $2(x^4+1)+89x^2= 56x(x^2+1)$

  2. $2(x^4+1)+89x^2= 56x(x+1)$

  3. $2(x^4+1)+89x^2= 56x^2(x+1)$

  4. None of these

Show answer
Correct Option: D
Explanation:

To be reciprocal equation, the multiplication of the roots $(\dfrac{e}{a})$ should be $1$ 

(A)
After rearranging the equation 
$\Rightarrow $   $2x^4 -56x^3 +89x^2 -56x +2 = 0 $
$\Rightarrow $  $\dfrac{e}{a} = \dfrac{2}{2}  = 1 $
So, multiplication of roots is $1$  
Thus, it is an reciprocal equation 

(B)
$2x^4 -33x^2 -56x +2 = 0 $

$\Rightarrow $  $\dfrac{e}{a} = \dfrac{2}{2}  = 1 $

So, multiplication of roots is $1$  
Thus, it is an reciprocal equation 


(C)

$\Rightarrow $   $2x^4 -56x^3 +33x^2 +2 = 0 $

$\Rightarrow $  $\dfrac{e}{a} = \dfrac{2}{2}  = 1 $

So, multiplication of roots is $1$  
Thus, it is an reciprocal equation.

Hence, the answer is option D.

$2x^4-3x^3+7x^2+3x-2=0$ is not a reciprocal equation, because

reciprocal equations theory of equations maths

  1. The coefficients from beginning to end and vice versa are not the same.

  2. All the coefficients of terms are not same

  3. The coefficients from beginning to end and vice versa are same.

  4. None of these

Show answer
Correct Option: A
Explanation:

Here, the coefficients are not palindromic because the first and and last coefficients are opposite in sign , same is the case for second last and second coefficient.

The Equation $5x^4-3x^3+7x^2-4x+2=0$ is of the type

reciprocal equations theory of equations maths

  1. Quadratic

  2. Linear

  3. Reciprocal

  4. None

Show answer
Correct Option: D
Explanation:

The highest power of $x $ in this equation is $4$, so this is a $4th$ order equation.

Thus, it is neither linear nor quadratic.
Now to be reciprocal equation the multiplication of roots should be $1$  and in the given equation 
Multiplication of roots is $\dfrac{e}{a}=\dfrac{2}{5}$
So this not a reciprocal equation 
Hence, option D is correct.

The inverse of the function $f(x) = \frac{{{e^x} - {e^{ - x}}}}{{{e^x} + {e^{ - x}}}}$ is

reciprocal equations theory of equations maths

  1. $\dfrac{1}{2}\ell n\dfrac{{1 + x}}{{1 - x}}$

  2. $\dfrac{1}{2}\ell n\dfrac{{2 + x}}{{2 - x}}$

  3. $\dfrac{1}{2}\ell n\dfrac{{1 - x}}{{1 + x}}$

  4. $2\ell n(1 + x)$

Show answer
Correct Option: A
Explanation:

$y=\cfrac { { e }^{ x }-{ e }^{ -x } }{ { e }^{ x }+{ e }^{ -x } }$

$y=\cfrac { \cfrac { { e }^{ 2x }-1 }{ { e }^{ x } }  }{ \cfrac { { e }^{ 2x }+1 }{ { e }^{ x } }  }$
 $y=\cfrac { { e }^{ 2x }-1 }{ { e }^{ 2x }+1 }$
 for universe replace;$x$ with $y$
$\cfrac { x }{ 1 } =\cfrac { { e }^{ 2y }-1 }{ { e }^{ 2y }+1 }$
 using componendo divodendo
 $\cfrac { x+1 }{ x-1 } =\cfrac { { e }^{ 2y }-1+{ e }^{ 2y }+1 }{ { e }^{ 2y }-1-{ e }^{ 2y }-1\quad  }$
 $\cfrac { x+1 }{ x-1 } =-\cfrac { 2{ e }^{ 2y } }{ 2 }$
 $\cfrac { 1+x }{ 1-x } ={ e }^{ 2y }$
$\left(\cfrac { 1+x }{ 1-x } \right)=2y$
$ y=\cfrac { 1 }{ 2 } ln\left(\cfrac { 1+x }{ 1-x } \right)$

If $ax^{3}+bx^{2}+cx+d=0$ is a reciprocal equation of the first type, then 

reciprocal equations theory of equations maths

  1. $a=d,b=c$

  2. $a=c,b=d$

  3. $a=-d,b=-c$

  4. $a=-c,b=-d$

Show answer
Correct Option: A
Explanation:

If $ax^3+bx^2+cx+d$ is a reciprocal equation of the first type,

We know that

$a _{r}=a _{n-r}$ where $a _{n}$ are the coefficient of the equation $f(x)$

So, as $f(x)=ax^3+bx^2+cx+d$

$a=d$ and $ b=c$

If $f(x)=1+\displaystyle \int^{x} _{0}t^{2}f(t)dt$, then the number of solution of $f(x)=x^{2}+1$ is

reciprocal equations theory of equations maths

  1. $1$

  2. $2$

  3. $3$

  4. $4$

Show answer
Correct Option: A

The equation of the line, reciprocal of whose intercepts on the axes are $a$ and $b$ given by

reciprocal equations theory of equations maths

  1. $\dfrac x 2$ + $\dfrac yb$ = $1$

  2. $ax + by = 1$

  3. $ax + by = ab$

  4. $ax = by = 1$

Show answer
Correct Option: B
Explanation:

Let A & B be the part of intersection of line with X & Y axis respectively.

$\Rightarrow A= \left(\cfrac {1}{a},0\right)$
$\Rightarrow B= \left(0,\cfrac {1}{b}\right)$
$\therefore$ Equation of line= $\left(y-\cfrac {1}{b}\right)=\left(\cfrac {\cfrac {1}{b}-0}{0-\cfrac {1}{a}}\right)$
$\Rightarrow \left(\cfrac {1}{a}\right)\left(y-\cfrac {1}{b}\right)=\cfrac {1}{b}x$
$\Rightarrow \cfrac {-y}{a}+\cfrac {1}{ab}= \cfrac {x}{b}$
$\Rightarrow \cfrac {x}{b}+\cfrac {y}{a}=\cfrac {1}{ab}$
$\Rightarrow ax+by=1$

The equation $\sin^{-1}x-3\sin^{-1}a=0$ has real solutions for x if?

reciprocal equations theory of equations maths

  1. $a \in R$

  2. $a \in [-1, 1]$

  3. $a \in \left[0, \dfrac{1}{2}\right]$

  4. $a \in \left[-\dfrac{1}{2}, \dfrac{1}{2}\right]$

Show answer
Correct Option: A