Difference between revisions of "1960 IMO Problems/Problem 2"
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==Solution== | ==Solution== | ||
Set <math>x = -\frac{1}{2} + \frac{a^2}{2}</math>, where <math>a\ge0</math>. | Set <math>x = -\frac{1}{2} + \frac{a^2}{2}</math>, where <math>a\ge0</math>. | ||
| − | <math>\frac{4(-\frac{1}{2}+\frac{a^2}{2})^2}{(1-\sqrt{1+2(-\frac{1}{2}+\frac{a^2}{2})})^2}< | + | <math>\frac{4\left(-\frac{1}{2}+\frac{a^2}{2}\right)^2}{\left(1-\sqrt{1+2\left(-\frac{1}{2}+\frac{a^2}{2}\right)}\right)^2}<2\left(-\frac{1}{2}+\frac{a^2}{2}\right)+9</math> |
After simplifying, we get | After simplifying, we get | ||
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<math>a^2+2a+1<a^2+8</math> | <math>a^2+2a+1<a^2+8</math> | ||
| − | Which gives <math>a<\frac{7}{2}</math> and hence <math>x<\frac{45}{8}</math>. | + | Which gives <math>a<\frac{7}{2}</math> and hence <math>-\frac{1}{2} \le x<\frac{45}{8}</math>. |
But <math>x=0</math> makes the RHS indeterminate. | But <math>x=0</math> makes the RHS indeterminate. | ||
| − | So, answer: <math>x<\frac{45}{8}</math>, except <math>x=0</math>. | + | So, answer: <math>-\frac{1}{2} \le x<\frac{45}{8}</math>, except <math>x=0</math>. |
Revision as of 16:12, 28 June 2008
Problem
For what values of the variable 
does the following inequality hold:
![\[\dfrac{4x^2}{(1 - \sqrt {2x + 1})^2} < 2x + 9 \ ?\]](http://latex.artofproblemsolving.com/c/f/8/cf8aa03d2cfb5932885cc52f52c98f7e1bd5df0b.png)
Solution
Set 
, where 
.
After simplifying, we get
So
Which gives 
and hence 
.
But 
makes the RHS indeterminate.
So, answer: , except .
See Also
| 1960 IMO (Problems) • Resources | ||
| Preceded by Problem 1 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 3 |
| All IMO Problems and Solutions | ||
