305. Position of the roots w.r.t. a given real number.
by Virgil Nicula, Jul 29, 2011, 9:39 PM
Let 
be an equation with real coefficients , 
and with the roots 
, 
. Assume is given a real 
. Denote the discriminant 
and the sum of roots
. I"ll study the positions of real roots , w.r.t. the given real number , i.e. the position of w.r.t. the interval 
. Distinquish the following nine cases :
Remark. In the particular case
we have 
(same sign). Thus, our debate becomes the study of the sign for the
roots, i.e. the study of the signs for
, 
, 
. Observe two interesting properties : 
.
Example. Let
be three real numbers. Consider the equation 
.
Prove that its roots , is really and 
. Indeed, from 
, 
and
obtain required conclusion. Similarly the equation 
has same properties.
Applications.
PP1. Ascertain
so that 
for any ![$x\in [0,1]$](//latex.artofproblemsolving.com/5/6/1/561b0f8577e618149adc3fae918f20a4a4d4c5ca.png)
.
Proof. I"ll debate w.r.t. the parameter
the sign of 
and the positions of the roots of 
w.r.t. the numbers 
and 
. Therefore,
From the upper table obtain that the sign of
is constant on the interval ![$[0,1]$](//latex.artofproblemsolving.com/e/8/6/e861e10e1c19918756b9c8b7717684593c63aeb8.png)
iff 
and
namely : for
we have 
and for 
we have 
, of course for any .
PP2. Ascertain so that the equation 
has a maximum number of zeroes in the first dial, i.e. ![$\left[0,\frac {\pi}{2}\right]$](//latex.artofproblemsolving.com/2/1/8/218957f184ff318bffc0781452c79af681ced9a0.png)
.
Proof. With the substitution
, where ![$\phi :\left[0,\frac {\pi}{2}\right]\rightarrow\left[1,\sqrt 2\right]$](//latex.artofproblemsolving.com/e/9/c/e9c55a805899de88bd42618851170c7f963b2aa1.png)
is a surjective function, the initial equation becomes
. Must to study the relative position of the real roots 
, 
of the equation 
w.r.t. the real numbers , 
.
The transfer from this debate to the study of the nature/kind of zeroes of the initial equation I"ll perform so :
. I"ll study the signs (see the table from the previous problem) for
. Therefore obtain ![$m\in \left(2\sqrt 2-1,2\right]$](//latex.artofproblemsolving.com/f/b/c/fbc59186d9642743444676aed323cf58d0b27fd2.png)
and in this case the maximum number of zeroes is two !
PP3. Ascertain so that for any 
we have 
.
Proof. Prove easily that the function
is 
strict increasing and 
. Thus, 
![$\left|\left(x+\frac 1x\right)-m\right|\le3\cdot\left[\left(x+\frac 1x\right)+1\right]\ (\forall )\ x\ge 1\iff$](//latex.artofproblemsolving.com/2/7/1/271e87fbade4124520289c24bab094d4b5a81718.png)
. Let 
and 
. Prove easily that 
for any . Therefore,
the solutions of the inequation
are ![$t\in S_m=\left\{\begin{array}{cccc}
\mathbb R & \iff & m=-1 & \left(t_1=t_2\right)\\\\
\left(-\infty,t_1\right]\cup\left[t_2,\infty\right) & \iff & m<-1 & \left(t_1<t_2\right)\\\\
\left(-\infty,t_2\right]\cup\left[t_1,\infty\right) & \iff & m>-1 & \left(t_2<t_1\right)\end{array}\right\|$](//latex.artofproblemsolving.com/6/7/a/67a83e78f6e03940c983fcf787cd8d88df5f0396.png)
.
The required values of the parameter so that 
obtain from 
, i.e. 
![$m\in\{-1\}\cup[-7,-1)\cup (-1,11]\implies$](//latex.artofproblemsolving.com/0/8/0/08055eff48e602b3907f443299f27c68a521a689.png)
![$\boxed{m\in [-7,11]}$](//latex.artofproblemsolving.com/f/1/7/f17b6c42943587f3545350231c0badaa0846b1e2.png)
.
PP4. Prove that
for any ![$x\in [-1,1]\ \iff\ h(-1)\le 4$](//latex.artofproblemsolving.com/9/0/0/900fd5d86ed14921317291806120d082394467ca.png)
.
Proof. Observe that
. Thus, 
for any ![$x\in [-1,1]\iff$](//latex.artofproblemsolving.com/1/3/3/13366138e74badbba2f38bc50f1bfa3f3a7c4726.png)
.
PP5. Ascertain so that ![$\left\{y\in\mathbb R\left|\ (\exists )\ x\in\mathbb R\ \mathrm {so\ that\ } y=\frac {x^2+mx+1}{x^2-x+1}\right|\right\}\subset [-3,2]$](//latex.artofproblemsolving.com/c/d/2/cd2b82cefa0c790a1acb74e46c17091875905058.png)
.
Proof.
be an equation with real coefficients , 
and with the roots 
, 
. Assume is given a real 
. Denote the discriminant 
and the sum of roots
. I"ll study the positions of real roots , w.r.t. the given real number , i.e. the position of w.r.t. the interval 
. Distinquish the following nine cases :
Remark. In the particular case
we have 
(same sign). Thus, our debate becomes the study of the sign for theroots, i.e. the study of the signs for
, 
, 
. Observe two interesting properties : 
.Example. Let
be three real numbers. Consider the equation 
.Prove that its roots
, is really and 
. Indeed, from 
, 
and
obtain required conclusion. Similarly the equation 
has same properties. Applications.
PP1. Ascertain
so that 
for any ![$x\in [0,1]$](http://latex.artofproblemsolving.com/5/6/1/561b0f8577e618149adc3fae918f20a4a4d4c5ca.png)
.Proof. I"ll debate w.r.t. the parameter
the sign of 
and the positions of the roots of 
w.r.t. the numbers 
and 
. Therefore,
From the upper table obtain that the sign of
is constant on the interval ![$[0,1]$](http://latex.artofproblemsolving.com/e/8/6/e861e10e1c19918756b9c8b7717684593c63aeb8.png)
iff 
andnamely : for
we have 
and for 
we have 
, of course for any .PP2. Ascertain
so that the equation 
has a maximum number of zeroes in the first dial, i.e. ![$\left[0,\frac {\pi}{2}\right]$](http://latex.artofproblemsolving.com/2/1/8/218957f184ff318bffc0781452c79af681ced9a0.png)
.Proof. With the substitution
, where ![$\phi :\left[0,\frac {\pi}{2}\right]\rightarrow\left[1,\sqrt 2\right]$](http://latex.artofproblemsolving.com/e/9/c/e9c55a805899de88bd42618851170c7f963b2aa1.png)
is a surjective function, the initial equation becomes
. Must to study the relative position of the real roots 
, 
of the equation 
w.r.t. the real numbers , 
.The transfer from this debate to the study of the nature/kind of zeroes of the initial equation I"ll perform so :
. I"ll study the signs (see the table from the previous problem) for
. Therefore obtain ![$m\in \left(2\sqrt 2-1,2\right]$](http://latex.artofproblemsolving.com/f/b/c/fbc59186d9642743444676aed323cf58d0b27fd2.png)
and in this case the maximum number of zeroes is two !PP3. Ascertain
so that for any 
we have 
.Proof. Prove easily that the function
is 
strict increasing and 
. Thus, 
![$\left|\left(x+\frac 1x\right)-m\right|\le3\cdot\left[\left(x+\frac 1x\right)+1\right]\ (\forall )\ x\ge 1\iff$](http://latex.artofproblemsolving.com/2/7/1/271e87fbade4124520289c24bab094d4b5a81718.png)
. Let 
and 
. Prove easily that 
for any . Therefore,the solutions of the inequation
are ![$t\in S_m=\left\{\begin{array}{cccc}
\mathbb R & \iff & m=-1 & \left(t_1=t_2\right)\\\\
\left(-\infty,t_1\right]\cup\left[t_2,\infty\right) & \iff & m<-1 & \left(t_1<t_2\right)\\\\
\left(-\infty,t_2\right]\cup\left[t_1,\infty\right) & \iff & m>-1 & \left(t_2<t_1\right)\end{array}\right\|$](http://latex.artofproblemsolving.com/6/7/a/67a83e78f6e03940c983fcf787cd8d88df5f0396.png)
.The required values of the parameter
so that 
obtain from 
, i.e. 
![$m\in\{-1\}\cup[-7,-1)\cup (-1,11]\implies$](http://latex.artofproblemsolving.com/0/8/0/08055eff48e602b3907f443299f27c68a521a689.png)
![$\boxed{m\in [-7,11]}$](http://latex.artofproblemsolving.com/f/1/7/f17b6c42943587f3545350231c0badaa0846b1e2.png)
.PP4. Prove that
for any ![$x\in [-1,1]\ \iff\ h(-1)\le 4$](http://latex.artofproblemsolving.com/9/0/0/900fd5d86ed14921317291806120d082394467ca.png)
. Proof. Observe that
. Thus, 
for any ![$x\in [-1,1]\iff$](http://latex.artofproblemsolving.com/1/3/3/13366138e74badbba2f38bc50f1bfa3f3a7c4726.png)
.PP5. Ascertain
so that ![$\left\{y\in\mathbb R\left|\ (\exists )\ x\in\mathbb R\ \mathrm {so\ that\ } y=\frac {x^2+mx+1}{x^2-x+1}\right|\right\}\subset [-3,2]$](http://latex.artofproblemsolving.com/c/d/2/cd2b82cefa0c790a1acb74e46c17091875905058.png)
. Proof.
This post has been edited 91 times. Last edited by Virgil Nicula, Nov 20, 2015, 7:05 PM
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