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1960 IMO Problems/Problem 2: Difference between revisions

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Which gives <math>a<\frac{7}{2}</math> and hence <math>-\frac{1}{2} \le 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 LHS indeterminate.


So, answer: <math>-\frac{1}{2} \le 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>.


==See Also==
==See Also==

Revision as of 10:41, 19 June 2017

Problem

For what values of the variable $x$ does the following inequality hold:

\[\dfrac{4x^2}{(1 - \sqrt {2x + 1})^2} < 2x + 9 \ ?\]

Solution

Set $x = -\frac{1}{2} + \frac{a^2}{2}$, where $a\ge0$. $\frac{4\left(-\frac{1}{2}+\frac{a^2}{2}\right)^2}{\left(1-\sqrt{1+2\left(-\frac{1}{2}+\frac{a^2}{2}\right)}\right)^2}<2\left(-\frac{1}{2}+\frac{a^2}{2}\right)+9$

After simplifying, we get $(a+1)^2<a^2+8$

So $a^2+2a+1<a^2+8$

Which gives $a<\frac{7}{2}$ and hence $-\frac{1}{2} \le x<\frac{45}{8}$.

But $x=0$ makes the LHS indeterminate.

So, answer: $-\frac{1}{2} \le x<\frac{45}{8}$, except $x=0$.

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
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