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Multiple choice reciprocal equations theory of equations maths

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

  1. $\{-8, 1\}$
  2. $\{8, 1\}$
  3. $\{1, -1\}$
  4. $\{2, -2}\$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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

Multiple choice reciprocal equations theory of equations maths

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

  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
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

Multiple choice reciprocal equations theory of equations maths

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

  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
Reveal answer Fill a bubble to check yourself
D Correct answer
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.

Multiple choice reciprocal equations theory of equations maths

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

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

Multiple choice reciprocal equations theory of equations maths

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

  1. Quadratic

  2. Linear

  3. Reciprocal

  4. None
Reveal answer Fill a bubble to check yourself
D Correct answer
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.

Multiple choice reciprocal equations theory of equations maths

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

  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. $2\ell n(1 + x)$
Reveal answer Fill a bubble to check yourself
A Correct answer
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)$

Multiple choice reciprocal equations theory of equations maths

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

  1. $a=d,b=c$
  2. $a=c,b=d$
  3. $a=-d,b=-c$
  4. $a=-c,b=-d$
Reveal answer Fill a bubble to check yourself
A Correct answer
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$

Multiple choice reciprocal equations theory of equations maths

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

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

Given f(x) = 1 + integral(0 to x) t^2 * f(t) dt. Differentiating both sides gives f'(x) = x^2 * f(x). This is a separable differential equation: df/f = x^2 dx. Integrating gives ln|f(x)| = x^3/3 + C. Since f(0) = 1, we find C = 0, so f(x) = exp(x^3/3). We want to solve exp(x^3/3) = x^2 + 1. At x=0, both sides are 1. For x > 0, exp(x^3/3) grows much faster than x^2 + 1. For x < 0, exp(x^3/3) < 1 while x^2 + 1 > 1. Thus, only x=0 is a solution.

Multiple choice reciprocal equations theory of equations maths

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

  1. $\dfrac x 2$ + $\dfrac yb$ = $1$
  2. $ax + by = 1$
  3. $ax + by = ab$
  4. $ax = by = 1$
Reveal answer Fill a bubble to check yourself
B Correct answer
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= $\left(y-\cfrac {1}{b}\right)=\left(\cfrac {\cfrac {1}{b}-0}{0-\cfrac {1}{a}}\right)$
$\Rightarrow \left(\cfrac {1}{a}\right)\left(y-\cfrac {1}{b}\right)=\cfrac {1}{b}x$
$\Rightarrow \cfrac {-y}{a}+\cfrac {1}{ab}= \cfrac {x}{b}$
$\Rightarrow \cfrac {x}{b}+\cfrac {y}{a}=\cfrac {1}{ab}$
$\Rightarrow ax+by=1$

Multiple choice reciprocal equations theory of equations maths

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

  1. $a \in R$
  2. $a \in [-1, 1]$
  3. $a \in \left[0, \dfrac{1}{2}\right]$
  4. $a \in \left[-\dfrac{1}{2}, \dfrac{1}{2}\right]$
Reveal answer Fill a bubble to check yourself
D Correct answer