380. Functions. Range. Bijection and inverse.
by Virgil Nicula, Jul 17, 2013, 2:16 AM
PP1. The equation 
has real roots, where 
. Find the range of these roots.
Proof.
is a root of 
, i.e. 
.
![$a\in\left[\frac 23,2\right]$](//latex.artofproblemsolving.com/a/5/c/a5cf36b63fdaffe75faabd712d47d48e046a4b09.png)
.
![$\boxed{\ r\in\left[-\frac 73,-1\right]\ }$](//latex.artofproblemsolving.com/e/c/d/ecdb16e4bdac6da654d42d091aa88aa16cff88bd.png)
.
PP2. Determine the real values of the parameter
so that inequality 
hasn't real solutions.
Proof. Observe that
, where 
. Prove easily that the range of 
is ![$ \Im (f)=\left[1-\frac 2{\sqrt 3} ,1+\frac 2{\sqrt 3}\right]$](//latex.artofproblemsolving.com/0/f/3/0f3d3de361a065265779921b4e68a55fa782d93d.png)
.
Therefore, the inequality hasn't real solutions 
the inequality hasn't real solutions 
, i.e. 
.
PP3. (College Entrance Examination (China Liaoning 2013). Suppose that![$(1+x)e^{-2x}\ge ax+\frac{x^3}{2}+1+2x\cos x\ ,\ \forall\ x\in[0,1]$](//latex.artofproblemsolving.com/5/a/f/5af36ad09211624c9ac1fdeb727782c33cd96430.png)
. Find the range of real number 
.
Proof. Suppose w.l.o.g. that![$x\in (0,1]$](//latex.artofproblemsolving.com/8/8/5/885ef0336d15614e1f1fcac9cf1aa48751c547e1.png)
. Thus , ![$(1 +x)e^{-2x} \ge ax + \frac{1}{2}x^3 + 1 + 2x\cos x\ \ (\forall )\ x \in [0,1]\iff$](//latex.artofproblemsolving.com/7/4/f/74fd88f3558191713ed10840e2dd9b1689e48fbd.png)
,
where
Thus, ![$\boxed{f'(x) = \frac{1}{x^2}\cdot [g(x)+h(x)]}$](//latex.artofproblemsolving.com/8/6/e/86e8b2eeb505ed75185638a2fea820c511798d3d.png)
, where 
. Obviously 
and 
![$\boxed{g(x)>0\ ,\ (\forall )\ x\in (0,1]}$](//latex.artofproblemsolving.com/0/0/8/00827349903c43be23f3d247b996b09d348fa061.png)
. I"ll show that 
, which isn't so obvious.
Indeed,
.Thus,
and
![$\boxed{h(x)>0\ ,\ (\forall )\ x\in (0,1]}$](//latex.artofproblemsolving.com/f/3/b/f3be625ca4bad9038c4c6a587e930e76088c3df3.png)
. Hence, ![$f'(x)>0\ ,\ (\forall )x\in (0,1]$](//latex.artofproblemsolving.com/b/3/9/b39560ad8782a34494bee97719ae1a267880731d.png)
. Observe that 
.
Since is increasing obtain that 
, i.e. 
.
PP4. Prove that the function
, where 
is invertibly.
Proof. I"ll show
one and only one 
so that 
. Let the quation with parameter 
, i.e. 
![$x-2\left(x-[x]\right)=y\iff$](//latex.artofproblemsolving.com/d/8/4/d84c048c99e65aeec07655bb7b3ec023b72e7604.png)
![$\boxed{x=2[x]-y}\ (*)$](//latex.artofproblemsolving.com/c/4/0/c4038062ef6cf31d27f97328b3e02943481c0e9b.png)
and ![$[x]\le x<[x]+1\iff$](//latex.artofproblemsolving.com/2/2/d/22d486a61653217676c0f5ca583ebf67aebb6c04.png)
![$[x]\le 2[x]-y<[x]+1$](//latex.artofproblemsolving.com/3/5/e/35e6e37bb3a9fd7a30499eaf5ebde4c3dc964753.png)
![$\boxed{y\le [x]<y+1}\ (1)$](//latex.artofproblemsolving.com/2/6/8/2688a550488792d3d7860cc454636d7bb9891e83.png)
. Since ![$[x]\in\mathbb Z$](//latex.artofproblemsolving.com/d/4/e/d4e95e26205ea50f00870591a76e5a605d3c454e.png)
obtain ![$[x]=\left\{\begin{array}{ccc}
[y+1] & \mathrm{if} & y\not\in\mathbb Z\\\\
y & \mathrm{if} & y\in\mathbb Z\end{array}\right\|$](//latex.artofproblemsolving.com/c/9/5/c953a331c5e8fcafb6031f35941c0a6da5e60b60.png)
and
![$x\stackrel{(*)}{=}\left\{\begin{array}{ccc}
2[y+1]-y & \mathrm{if} & y\not\in\mathbb Z\\\\
y & \mathrm{if} & y\in\mathbb Z\end{array}\right\|$](//latex.artofproblemsolving.com/5/b/0/5b0a94f2bd5931c0ae93d67fd3a6186f176ebc24.png)
. Hence is a bijectively and its inverse is 
.
PP5. Let
be a complex number such that 
and 
. Ascertain the minimum value of 
.
Proof. Denote
, where 
and 
. Thus, 
, 
and 
. Hence 
.
Remark.
by AM-GM inequality 
.
An easy extension. Let be a complex number such that 
and ,
where
and 
. Ascertain the maximum of 
and prove that 
.
Remark. Suppose
, where 
and 
. Observe that if denote 
, then 
![$a^2\rho^4+ab\left[\left(z+\overline z\right)^2-2\rho^2\right]+b^2=$](//latex.artofproblemsolving.com/c/f/c/cfc4d2bfc1189136ba5f4198dfd4179d05df436d.png)
![$\left(a\rho^2-\frac {2ab+1}{2a}\right)^2+\left[\sqrt {ab}\cdot\left(z+\overline z\right)-\frac {c}{2\sqrt {ab}}\right]^2=$](//latex.artofproblemsolving.com/d/c/9/dc91fd0d662fc34509a6782d5c557204f5495960.png)
.
Some examples. Let be a complex number with 
and 
. Ascertain the maximum for 
and for 
, where :
(CCLVII)
has real roots, where 
. Find the range of these roots.Proof.
is a root of 
, i.e. 
.
![$a\in\left[\frac 23,2\right]$](http://latex.artofproblemsolving.com/a/5/c/a5cf36b63fdaffe75faabd712d47d48e046a4b09.png)
.
![$\boxed{\ r\in\left[-\frac 73,-1\right]\ }$](http://latex.artofproblemsolving.com/e/c/d/ecdb16e4bdac6da654d42d091aa88aa16cff88bd.png)
.PP2. Determine the real values of the parameter
so that inequality 
hasn't real solutions. Proof. Observe that
, where 
. Prove easily that the range of 
is ![$ \Im (f)=\left[1-\frac 2{\sqrt 3} ,1+\frac 2{\sqrt 3}\right]$](http://latex.artofproblemsolving.com/0/f/3/0f3d3de361a065265779921b4e68a55fa782d93d.png)
.Therefore, the inequality
hasn't real solutions 
the inequality hasn't real solutions 
, i.e. 
.PP3. (College Entrance Examination (China Liaoning 2013). Suppose that
![$(1+x)e^{-2x}\ge ax+\frac{x^3}{2}+1+2x\cos x\ ,\ \forall\ x\in[0,1]$](http://latex.artofproblemsolving.com/5/a/f/5af36ad09211624c9ac1fdeb727782c33cd96430.png)
. Find the range of real number 
.Proof. Suppose w.l.o.g. that
![$x\in (0,1]$](http://latex.artofproblemsolving.com/8/8/5/885ef0336d15614e1f1fcac9cf1aa48751c547e1.png)
. Thus , ![$(1 +x)e^{-2x} \ge ax + \frac{1}{2}x^3 + 1 + 2x\cos x\ \ (\forall )\ x \in [0,1]\iff$](http://latex.artofproblemsolving.com/7/4/f/74fd88f3558191713ed10840e2dd9b1689e48fbd.png)
,where
Thus, ![$\boxed{f'(x) = \frac{1}{x^2}\cdot [g(x)+h(x)]}$](http://latex.artofproblemsolving.com/8/6/e/86e8b2eeb505ed75185638a2fea820c511798d3d.png)
, where 
. Obviously 
and 
![$\boxed{g(x)>0\ ,\ (\forall )\ x\in (0,1]}$](http://latex.artofproblemsolving.com/0/0/8/00827349903c43be23f3d247b996b09d348fa061.png)
. I"ll show that 
, which isn't so obvious.Indeed,
.Thus, and
![$\boxed{h(x)>0\ ,\ (\forall )\ x\in (0,1]}$](http://latex.artofproblemsolving.com/f/3/b/f3be625ca4bad9038c4c6a587e930e76088c3df3.png)
. Hence, ![$f'(x)>0\ ,\ (\forall )x\in (0,1]$](http://latex.artofproblemsolving.com/b/3/9/b39560ad8782a34494bee97719ae1a267880731d.png)
. Observe that 
.Since
is increasing obtain that 
, i.e. 
.PP4. Prove that the function
, where 
is invertibly.Proof. I"ll show
one and only one 
so that 
. Let the quation with parameter 
, i.e. 
![$x-2\left(x-[x]\right)=y\iff$](http://latex.artofproblemsolving.com/d/8/4/d84c048c99e65aeec07655bb7b3ec023b72e7604.png)
![$\boxed{x=2[x]-y}\ (*)$](http://latex.artofproblemsolving.com/c/4/0/c4038062ef6cf31d27f97328b3e02943481c0e9b.png)
and ![$[x]\le x<[x]+1\iff$](http://latex.artofproblemsolving.com/2/2/d/22d486a61653217676c0f5ca583ebf67aebb6c04.png)
![$[x]\le 2[x]-y<[x]+1$](http://latex.artofproblemsolving.com/3/5/e/35e6e37bb3a9fd7a30499eaf5ebde4c3dc964753.png)
![$\boxed{y\le [x]<y+1}\ (1)$](http://latex.artofproblemsolving.com/2/6/8/2688a550488792d3d7860cc454636d7bb9891e83.png)
. Since ![$[x]\in\mathbb Z$](http://latex.artofproblemsolving.com/d/4/e/d4e95e26205ea50f00870591a76e5a605d3c454e.png)
obtain ![$[x]=\left\{\begin{array}{ccc}
[y+1] & \mathrm{if} & y\not\in\mathbb Z\\\\
y & \mathrm{if} & y\in\mathbb Z\end{array}\right\|$](http://latex.artofproblemsolving.com/c/9/5/c953a331c5e8fcafb6031f35941c0a6da5e60b60.png)
and![$x\stackrel{(*)}{=}\left\{\begin{array}{ccc}
2[y+1]-y & \mathrm{if} & y\not\in\mathbb Z\\\\
y & \mathrm{if} & y\in\mathbb Z\end{array}\right\|$](http://latex.artofproblemsolving.com/5/b/0/5b0a94f2bd5931c0ae93d67fd3a6186f176ebc24.png)
. Hence is a bijectively and its inverse is 
.PP5. Let
be a complex number such that 
and 
. Ascertain the minimum value of 
.Proof. Denote
, where 
and 
. Thus, 
, 
and 
. Hence 
.Remark.
by AM-GM inequality 
.An easy extension. Let
be a complex number such that 
and ,where
and 
. Ascertain the maximum of 
and prove that 
.Remark. Suppose
, where 
and 
. Observe that if denote 
, then 
![$a^2\rho^4+ab\left[\left(z+\overline z\right)^2-2\rho^2\right]+b^2=$](http://latex.artofproblemsolving.com/c/f/c/cfc4d2bfc1189136ba5f4198dfd4179d05df436d.png)
![$\left(a\rho^2-\frac {2ab+1}{2a}\right)^2+\left[\sqrt {ab}\cdot\left(z+\overline z\right)-\frac {c}{2\sqrt {ab}}\right]^2=$](http://latex.artofproblemsolving.com/d/c/9/dc91fd0d662fc34509a6782d5c557204f5495960.png)
.Some examples. Let
be a complex number with 
and 
. Ascertain the maximum for 
and for 
, where :
(CCLVII)This post has been edited 29 times. Last edited by Virgil Nicula, Nov 26, 2015, 12:05 PM
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