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

(Solution)
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==Solution==
 
==Solution==
 
<math>\boxed{501}</math>
 
<math>\boxed{501}</math>
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kx=(x+2)^2
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x^2+(4-k)x+4=0 ...(1)
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the equation has solution
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D=(4-k)^2-16=k(k-8)>=0
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so k<=0 or k>=8
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becuase k can't be zero or the original equation will be meanless.
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there are 3 cases
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1:k=8
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then x=2, which is satisified the question.
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2:k<0
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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.
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then we get 2 inequities
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-2<( k-4 + sqrt( k(k-8) ) )/2<0
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( k-4 - sqrt( k(k-8) ) )/2<-2
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notice k<0<sqrt(k(k-8)) and (4-k)^2=k(k-8)+16>k(k-8)
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we know in this case, there is always and only one solution for the orign equation.
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3:k>8
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similar to case2 we can get inequity
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( k-4 - sqrt( k(k-8) ) )/2 < 0 <( k-4 + sqrt( k(k-8) ))/2
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and there are always 2 solution for the origin equation, so this case is not satisfied.
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so we get k<0 or k=8
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because k belong to [-500,500], the answer is 501
  
 
=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 14:33, 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

$\boxed{501}$

kx=(x+2)^2 x^2+(4-k)x+4=0 ...(1) the equation has solution D=(4-k)^2-16=k(k-8)>=0 so k<=0 or k>=8 becuase k can't be zero or the original equation will be meanless. 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 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.

3:k>8 similar to case2 we can get inequity ( k-4 - sqrt( k(k-8) ) )/2 < 0 <( 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 501

See Also

2017 AIME II (ProblemsAnswer KeyResources)
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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