Revision as of 14:34, 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 so

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

2017 AIME II (Problems • Answer Key • Resources)
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Difference between revisions of "2017 AIME II Problems/Problem 7"

Revision as of 14:34, 23 March 2017

Problem

Solution

See Also

@@ Line 7: / Line 7: @@
 kx=(x+2)^2
 x^2+(4-k)x+4=0 ...(1)
-the equation has solution
+the equation has solution so
 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