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Difference between revisions of "2017 AIME II Problems/Problem 7"

(→‎Solution)
(fixed \LaTeX)
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==Solution==
 
==Solution==
−
<math>\boxed{501}</math>
+
<math>kx=(x+2)^2</math>
−
 
+
<math>x^2+(4-k)x+4=0 ...(1)</math>
−
kx=(x+2)^2
 
−
x^2+(4-k)x+4=0 ...(1)
 
  
 
the equation has solution so
 
the equation has solution so
  
−
D=(4-k)^2-16=k(k-8)>=0
+
<math>D=(4-k)^2-16=k(k-8)\geq0</math>
  
−
so k<=0 or k>=8
+
so <math>k<0</math> or <math>k\geq8</math> because k can't be zero or the original equation will be meaningless.
 +
there are 3 cases
  
−
becuase k can't be zero or the original equation will be meanless.
+
1:<math>k=8</math>
−
there are 3 cases
+
then <math>x=2</math>, which is satisified the question.
  
−
1:k=8
+
2:<math>k<0</math>
−
then x=2, which is satisified the question.
+
then one solution of the equation(1) should be in <math>(-2,0)</math> and another is out of it or the origin equation will be meanless.
 +
then we get 2 inequalities
 +
<math>-2<\frac{k-4+\sqrt{k(k-8)}}{2}<0</math>
  
−
2:k<0
+
<math>\frac{k-4-\sqrt{k(k-8)}}{2}<-2</math>
−
then one solution of the equation(1) should be in (-2,0) and another is out of it or the origin equation will be meanless.
+
notice <math>k<0<\sqrt{k(k-8)}</math> and <math>(4-k)^2=k(k-8)+16>k(k-8)</math>
−
then we get 2 inequities
 
−
-2<( k-4 + sqrt( k(k-8) ) )/2<0
 
−
( k-4 - sqrt( k(k-8) ) )/2<-2
 
−
notice k<0<sqrt(k(k-8)) and (4-k)^2=k(k-8)+16>k(k-8)
 
 
we know in this case, there is always and only one solution for the orign equation.
 
we know in this case, there is always and only one solution for the orign equation.
  
−
3:k>8
+
3:<math>k>8</math>
−
similar to case2 we can get inequity
+
similar to case2 we can get inequality
−
( k-4 - sqrt( k(k-8) ) )/2 < 0 <( k-4 + sqrt( k(k-8) ))/2
+
<math>\frac{k-4-\sqrt{k(k-8)}}{2}<0<\frac{k-4+\sqrt{k(k-8)}}{2}</math>
 
and there are always 2 solution for the origin equation, so this case is not satisfied.
 
and there are always 2 solution for the origin equation, so this case is not satisfied.
  
−
so we get k<0 or k=8
+
so we get <math>k<0</math> or <math>k=8</math>
  
−
because k belong to [-500,500], the answer is 501
+
because k belong to <math>[-500,500]</math>, the answer is <math>\boxed{501}</math>
  
 
=See Also=
 
=See Also=
 
{{AIME box|year=2017|n=II|num-b=6|num-a=8}}
 
{{AIME box|year=2017|n=II|num-b=6|num-a=8}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 14:52, 23 March 2017

Problem

Find the number of integer values of $k$ in the closed interval $[-500,500]$ for which the equation $\log(kx)=2\log(x+2)$ has exactly one real solution.

Solution

$kx=(x+2)^2$ $x^2+(4-k)x+4=0 ...(1)$

the equation has solution so

$D=(4-k)^2-16=k(k-8)\geq0$

so $k<0$ or $k\geq8$ because k can't be zero or the original equation will be meaningless. there are 3 cases

1:$k=8$ then $x=2$, which is satisified the question.

2:$k<0$ then one solution of the equation(1) should be in $(-2,0)$ and another is out of it or the origin equation will be meanless. then we get 2 inequalities $-2<\frac{k-4+\sqrt{k(k-8)}}{2}<0$

$\frac{k-4-\sqrt{k(k-8)}}{2}<-2$ notice $k<0<\sqrt{k(k-8)}$ and $(4-k)^2=k(k-8)+16>k(k-8)$ we know in this case, there is always and only one solution for the orign equation.

3:$k>8$ similar to case2 we can get inequality $\frac{k-4-\sqrt{k(k-8)}}{2}<0<\frac{k-4+\sqrt{k(k-8)}}{2}$ and there are always 2 solution for the origin equation, so this case is not satisfied.

so we get $k<0$ or $k=8$

because k belong to $[-500,500]$, the answer is $\boxed{501}$

See Also

2017 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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