Difference between revisions of "1973 Canadian MO Problems/Problem 1"
Airplanes1 (talk | contribs) (Created page with "==Problem== <math>\text{(i)}</math> Solve the simultaneous inequalities, <math>x<\frac{1}{4x}</math> and <math>x<0</math>; i.e. find a single inequality equivalent to the two si...") |
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==Solution== | ==Solution== | ||
− | <math>\text{(i)}</math> <center><math>x<\frac{1}{4x} \Rightarrow 4x^2 | + | <math>\text{(i)}</math> <center><math>x<\frac{1}{4x} \Rightarrow 4x^2 > 1 \Rightarrow x^2> \frac{1}{4}</math>. |
− | Since from the second inequality <math>x<0</math>, our solution is | + | Since from the second inequality <math>x<0</math>, our solution is <math>x<-1/2</math>. |
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&=& \boxed{\frac{1}{2}} \end{matrix}</math> | &=& \boxed{\frac{1}{2}} \end{matrix}</math> | ||
</center> | </center> | ||
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==See also== | ==See also== |
Latest revision as of 15:14, 21 July 2012
Problem
Solve the simultaneous inequalities,
and
; i.e. find a single inequality equivalent to the two simultaneous inequalities.
What is the greatest integer that satisfies both inequalities
and
.
Give a rational number between
and
.
Express
as a product of two integers neither of which is an integral multiple of
.
Without the use of logarithm tables evaluate
.
Solution
![$x<\frac{1}{4x} \Rightarrow 4x^2 > 1 \Rightarrow x^2> \frac{1}{4}$](http://latex.artofproblemsolving.com/4/9/b/49b32524592c34161cab5fd7909a01e6f65080f4.png)
Since from the second inequality , our solution is
.
![$\text{(ii)}$](http://latex.artofproblemsolving.com/0/a/c/0ac3c778e6a0a8e6b77ba3855ca818e759d27013.png)
![$4x+13 < 0$](http://latex.artofproblemsolving.com/7/2/5/725d470599c101edeffb95a4f5c3aad8024bf808.png)
![$4x+13 < 0 \Rightarrow x < -\frac{13}{4} \approx -3$](http://latex.artofproblemsolving.com/3/2/4/324dd3aeb4fa91425feed899c861b14dbc611dab.png)
![$x^{2}+3x > 16$](http://latex.artofproblemsolving.com/0/9/b/09b211e7637a4e74ce29e5aa82c7ca080b9d00ad.png)
![$x^{2}+3x > 16 \Rightarrow x^{2}+3x - 16 > 0 \Rightarrow x > \frac{-3 + \sqrt{71}}{2} \approx 2 \text{ or } x < \frac{-3-\sqrt{71}}{2}\approx -5$](http://latex.artofproblemsolving.com/1/9/7/19788ea1133f5623218c1328f38b304d29be09d9.png)
With these two inequalities, we see that the greatest integer satisfying the requirements is .
. Thus, a rational number in between
and
is
Thus,
![$\text{(v)}$](http://latex.artofproblemsolving.com/9/6/c/96ca28500f069481646cdd51f5823b8a110ddf42.png)
See also
1973 Canadian MO (Problems) | ||
Preceded by 1973 Canadian MO Problems |
1 • 2 • 3 • 4 • 5 | Followed by Problem 2 |