Tag: descartes rule

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Multiple choice maths algebraic functions, equations and inequalities descartes rule sign of quadratic expression sum and product of the roots of a polynomial equation

Given $P(x) = {x^4} + a{x^3} + b{x^2} + cx + d$ such that $x=0$ is the only real root of $P(x) = 0$. If $P(-1) < P(1) $,then in the interval $[-1,1]$

  1. $P(-1)$ is the minimum and $P(1)$ is the maximum of P
  2. $P(-1)$ is not the minimum but $P(1)$ is the maximum of P
  3. $P(-1)$ is the minimum and $P(1)$ is not the maximum of P
  4. neither $P(-1)$ is the minimum nor $P(1)$ is the maximum of P
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Given:$P\left(x\right)={x}^{4}+a{x}^{3}+b{x}^{2}+cx+d$

${P}^{\prime}\left(x\right)=4{x}^{3}+3a{x}^{2}+2bx+c$

Since, $x=0$ is a solution for ${P}^{\prime}\left(x\right)=0$

$\Rightarrow\,c=0$

So, $P\left(x\right)={x}^{4}+a{x}^{3}+b{x}^{2}+d$   

Also we have $P\left(−1\right)<P\left(1\right)$

$\Rightarrow\,1-a+b+d<1+a+b+d$

$\Rightarrow\,A>0$

Since ${P}^{\prime}\left(x\right)=0,$ only when $x=0$

and $P\left(x\right)$ is differentiable in $\left(−1,1\right)$, we should have the maximum and minimum at the points

$x=−1,0$ and $1$ only.

Also, we have $P\left(−1\right)<P\left(1\right)$

So,Maximum of $P\left(x\right)=Max\left\{P\left(0\right),P\left(1\right)\right\}$ and
Minimum of $P\left(x\right)=Min\left\{P\left(−1\right),P\left(0\right)\right\}$

In the interval $\left[0,1\right]$

${P}^{\prime}{\left(x\right)}=4{x}^{3}+3a{x}^{2}+2bx=x\left(4{x}^{2}+3ax+2b\right)$

Since ${P}^{\prime}{\left(x\right)}$ has only one root $x=0$, then $4{x}^{2}+3ax+2b=0$ has no real roots.

So,${\left(3a\right)}^{2}-32b<0$

$\Rightarrow\,\dfrac{3{a}^{2}}{32}>b$

So,$b>0$

Thus, we have $a>0$ and $b>0$

So,${P}^{\prime}{\left(x\right)}=4{x}^{3}+3a{x}^{2}+2bx>0,$ for $x\in\left(0,1\right)$

Hence, $P\left(x\right)$ is increasing in $\left[0,1\right]$ and $P\left(x\right)$ is decreasing in $\left[−1,0\right]$

Therefore, Maximum of $P\left(x\right)=P\left(1\right)$ and Minimum $P\left(x\right)$ does not occur at $x=−1$ respectively.
Multiple choice maths algebraic functions, equations and inequalities descartes rule sign of quadratic expression sum and product of the roots of a polynomial equation

If $o<\alpha<\beta<\gamma<\dfrac {\pi}{2}$, then the equation $\dfrac {1}{x-\sin \alpha}+\dfrac {1}{x-\sin\beta}+\dfrac {1}{x-\sin \gamma}=0$ has

  1. Imaginary roots

  2. Real and equal roots

  3. Real and unequal roots

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

$\begin{array}{l} \frac { 1 }{ { x-\sin  \alpha  } } +\frac { 1 }{ { x-\sin  \beta  } } +\frac { 1 }{ { x-\sin  \gamma  } } =0 \ 0<\alpha <\beta <\gamma <\frac { \pi  }{ 2 }  \ let\, \alpha ={ 30^{ 0 } },\beta ={ 45^{ 0 } },\gamma ={ 60^{ 0 } } \ \Rightarrow \frac { 1 }{ { x-\frac { 1 }{ 2 }  } } =\frac { 1 }{ { x-2\sqrt { 2 }  } } =\frac { 1 }{ { x-\sqrt { \frac { 3 }{ 2 }  }  } } =0 \end{array}$

hence roots are real and unequal

Multiple choice maths algebraic functions, equations and inequalities descartes rule sign of quadratic expression sum and product of the roots of a polynomial equation

The polynomial $\displaystyle (ax^{2}+bx+c)(ax^{2}-dx-c),ac\neq 0,$ has

  1. four real zeros

  2. at least two real zeros

  3. at most two real zeros

  4. no real zeros
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For 
$ax^{2}+bx+c=0$
$b^{2}-4ac\geq 0$ for real roots ...(i)
and for 
$ax^{2}-dx-c=0$
$d^{2}+4ac\geq 0$ for real roots ...(ii)
Now, 
$ac\neq 0$
Hence, 
Case I
If $ac>0$
Hence, 
$ax^{2}-dx-x=0$ has positive roots.
Case II
If $ac<0$
Then,
$ax^{2}+bx+c=0$
has Real roots.
Hence, the above polynomial has atleast two real roots.

Multiple choice maths algebraic functions, equations and inequalities descartes rule sign of quadratic expression sum and product of the roots of a polynomial equation

If $\alpha$ and $\beta$ are the zeros of polynomial $x^{2}-ax+b$, then the value of $\alpha^{2}\left(\dfrac {\alpha^{2}}{\beta}-\beta\right)+\beta^{2}\left(\dfrac {\beta^{2}}{\alpha}-\alpha\right)$ is

  1. $\dfrac {a(a^{2}-4b)(a^{2}-b)}{b}$
  2. $\dfrac {b(a^{2}-4b)(a^{2}-b)}{a}$
  3. $\dfrac {b^{2}(a^{2}-4b)(a^{2}-b)}{a}$

  4. None

Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
We have Sum of the roots$=\alpha+\beta=a$
Product of the roots$=\alpha\beta=b$

${\alpha}^{2}\left(\dfrac{{\alpha}^{2}}{\beta}-\beta\right)+{\beta}^{2}\left(\dfrac{{\beta}^{2}}{\alpha}-\alpha\right)$
$=\dfrac{{\alpha}^{2}}{\beta}\left({\alpha}^{2}-{\beta}^{2}\right)+\dfrac{{\beta}^{2}}{\alpha}\left({\beta}^{2}-{\alpha}^{2}\right)$
$=\dfrac{{\alpha}^{2}}{\beta}\left({\alpha}^{2}-{\beta}^{2}\right)-\dfrac{{\beta}^{2}}{\alpha}\left({\alpha}^{2}-{\beta}^{2}\right)$
$=\left({\alpha}^{2}-{\beta}^{2}\right)\left(\dfrac{{\alpha}^{2}}{\beta}-\dfrac{{\beta}^{2}}{\alpha}\right)$
$=\dfrac{\left({\alpha}^{2}-{\beta}^{2}\right)}{\alpha\beta}\left({\alpha}^{3}-{\beta}^{3}\right)$
$=\dfrac{\left(\alpha-\beta\right)\left(\alpha+\beta\right)}{\alpha\beta}\left(\alpha-\beta\right)\left({\alpha}^{2}+{\beta}^{2}+\alpha\beta\right)$
$=\dfrac{{\left(\alpha-\beta\right)}^{2}\left(\alpha+\beta\right)}{\alpha\beta}\left({\alpha}^{2}+{\beta}^{2}+\alpha\beta\right)$

We know that ${\alpha}^{2}+{\beta}^{2}={\left(\alpha+\beta\right)}^{2}-2\alpha\beta$ and 
${\left(\alpha-\beta\right)}^{2}={\left(\alpha+\beta\right)}^{2}-4\alpha\beta$

Using $\alpha+\beta=a$ and $\alpha\beta=b$ we have
${\left(\alpha-\beta\right)}^{2}={\left(\alpha-\beta\right)}^{2}={a}^{2}-4b$
And ${\alpha}^{2}+{\beta}^{2}+\alpha\beta={\left(\alpha+\beta\right)}^{2}-2\alpha\beta+\alpha\beta$
$={\left(\alpha+\beta\right)}^{2}-\alpha\beta={a}^{2}-b$
$=\dfrac{a\left({a}^{2}-4b\right)\left({a}^{2}-b\right)}{b}$



Multiple choice maths algebraic functions, equations and inequalities descartes rule sign of quadratic expression sum and product of the roots of a polynomial equation

The value of $'a'$ for which the equation ${ x }^{ 3 }+ax+1=0$ and ${ x }^{ 4 }+a{ x }^{ 2 }+1=0$, have a common root is

  1. $a=2$
  2. $a=-2$
  3. $a=0$

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

Consider the following equation
$x^{4}+ax^{2}+1=0$
$x^{3}+ax+1=0$
Subtracting equation (ii) from (i), we get 
$x^{4}-x^{3}+a(x^{2}-x)=0$
$x^{3}(x-1)+ax(x-1)=0$
$(x-1)(x^{3}+ax)=0$
$x(x-1)(x^{2}+a)=0$
Hence, we get $x=0$ $x=1$ and $x^{2}=-a$
Now out of the above two, $x=0$ is not a root of the following two equations.
We do not know the nature of '$a$'. 

Hence, we cannot determine that $x^{2}=-a$ will have real or imaginary roots.
Hence, we get $x=1$ as a common root for the above two equations.
Now for both the equations to have $x=1$ as a common root, 
$f(1)=0$
$1+a+1=0$
$a=-2$
Similarly substituting in the second equation, we get $a=-2$.
Hence, the required value of $a$ is $-2$.

Multiple choice maths algebraic functions, equations and inequalities descartes rule sign of quadratic expression sum and product of the roots of a polynomial equation

Coordinates of a point P are $(a, b)$ where $a$ is a root of the equation 

$x^{2}+x-42=0$ 
and $b$ is an integral root of the equation
$x^{2}+ax+a^{2}-37=0$. 
The coordinates of P can be

  1. $(6, 4)$
  2. $(-7, 4)$
  3. $(-7, 3)$
  4. $(6, -3)$
Reveal answer Fill a bubble to check yourself
B,C Correct answer
Explanation

$a^{2}+a-42=0\Rightarrow a=-7$ or $a=6.$
Since b is a root of $x^{2}+ax+a^{2}-37=0$
For $a=-7,$ we have $x^{2}-7x+49-37=0$
$\Rightarrow x^{2}-7x+12=0\Rightarrow x=4, 3 , so, b=4$ or $3.$
So the coordinates of P can be $(-7, 4)$ or $(-7, 3)$, 

For $a=6,$ we have $x^{2}+6x-1=0$ which does not give an integral value, so $a\neq 6.$

Multiple choice maths algebraic functions, equations and inequalities descartes rule sign of quadratic expression sum and product of the roots of a polynomial equation

lf the difference of the roots of the equation $x^{2}-bx+c=0$ is equal to the differecne of the roots of the equation ${x}^{2}-{c}x+b=0$ and $b\neq c$, then $b+c=$

  1. $ 0$
  2. $2$
  3. $4$
  4. $-4$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let $(\alpha, \beta )$ and $ (\gamma, \delta )$ be the roots of the first equation and second equation respectively. 

Then, for the first equation
$ \alpha +\beta =b$ and $ \alpha \beta =c$
Now $ (\alpha +\beta )^{ 2 }={ b }^{ 2 }$
$ \Rightarrow (\alpha -\beta )^{ 2 }+4\alpha \beta ={ b }^{ 2 }$
$ \Rightarrow (\alpha -\beta )^{ 2 }={ b }^{ 2 }-4\alpha \beta $
$ \Rightarrow |\alpha -\beta |=\sqrt { { b }^{ 2 }-4c } $
Similarly for the second equation
$ |\gamma -\delta |=\sqrt { { c }^{ 2 }-4b } $
As per the given condition,
$ \sqrt { { b }^{ 2 }-4c } =\sqrt { { c }^{ 2 }-4b } $
$\Rightarrow { b }^{ 2 }-4c={ c }^{ 2 }-4b$
$\Rightarrow { b }^{ 2 }-{ c }^{ 2 }=-4(b-c)$
$ \Rightarrow (b+c)(b-c)=-4(b-c)$
Therefore, option D is correct.

Multiple choice maths algebraic functions, equations and inequalities descartes rule sign of quadratic expression sum and product of the roots of a polynomial equation

Let $\displaystyle a _{1}, a _{2},a _{3},a _{4},a _{5} \, \varepsilon \, R$ denote a rearrangement of equation $\displaystyle p _{1}x^{5}+p _{2}x^{3}+p _{3}x^{2}+p _{4}x+p _{5}=0$ then, equation  $\displaystyle a _{1}x^{4}+a _{2}x^{3}+a _{3}x^{2}+a _{4}x +a _{5}=0$ has 

  1. at least two real roots

  2. all four real roots

  3. only imaginary roots

  4. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$a _{1}x^{4}+a _{2}x^{3}+a _{3}x^{2}+a _{4}x+a _{5}=0$
for $ x=1 $
$a _{1}+a _{2}+a _{3}+a _{4}+a _{5}=0$
for the given set of equation,
sum of  $p _{1}+p _{2}+p _{3}+p _{4}+p _{5}=0$ 
So, $a _{1}+a _{2}+a _{3}+a _{4}+a _{5}\epsilon R $  for any set of arrangement
Hence at least two real roots.
Multiple choice maths algebraic functions, equations and inequalities descartes rule sign of quadratic expression sum and product of the roots of a polynomial equation

The sum of the solutions of the equation $64(81^{x})-84(144^{x})+27(256^{x})=0$  is:

  1. $1$
  2. $\dfrac{3}{2}$
  3. $\dfrac{5}{2}$

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

Given, $64(81^{ x })-84(144^{ x })+27(256^{ x })=0$

$ \Rightarrow 64\left( \left( \cfrac { 9 }{ 16 }  \right) ^{ x } \right) ^{ 2 }-84\left( \cfrac { 9 }{ 16 }  \right) ^{ x }+27=0$
Let $\left( \cfrac { 9 }{ 16 }  \right) ^{ x }=t$
$64t^{ 2 }-84t+27=0$
Solving this, we get
$\Rightarrow t=\cfrac { 3 }{ 4 } ,t=\cfrac { 9 }{ 16 } \Rightarrow x=\cfrac { 1 }{ 2 } ,x=1$
Hence, sum of roots is 
$\cfrac { 1 }{ 2 } +1=\cfrac { 3 }{ 2 } $
Hence, option 'B' is correct.

Multiple choice maths algebraic functions, equations and inequalities descartes rule sign of quadratic expression sum and product of the roots of a polynomial equation

If the sum of two roots of the equation $x^{4}+px^{3}+qx^{2}+rx+8=0$ is equal to the sum of the other two, then $p^{3}+8r=$

  1. $p^2 - 4pq$
  2. $2pq$
  3. $p^2 - pq$
  4. $4pq$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let the roots of the equation be $a,b,c,d$. 
As per the question,
$a+b = c+d$ 
From the theory of polynomials, 
$a+b+c+d = -p$
$ \Rightarrow a+b=c+d= \displaystyle \frac{-p}{2} $

Also,
$ab+ac+ad+bd+bc+cd  = q $
$ \Rightarrow (a+b)(c+d) +ab+cd  =q $
$ \Rightarrow ab+cd = q - \displaystyle \frac{p^2}{4} $

Also, 
$abc+abd+bcd+adc = -r $
$ \Rightarrow ab(c+d) +cd(a+b) = -r $
$ \Rightarrow \displaystyle \frac {-p}{2} (ab+cd) = -r $
$ \Rightarrow \displaystyle \frac {-p}{2} ( q - \displaystyle \frac{p^2}{4} ) = -r $
$ \Rightarrow -4pq + p^3 = -8r $
$ \Rightarrow p^3 + 8r = 4pq $