Difference between revisions of "2017 AIME II Problems/Problem 7"
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For the first case, we note that this can only occur when it is a perfect square trinomal, or <math>k = 0, 8</math>. However, <math>k = 0</math> results in <math>\log(0)</math> on the LHS, which is invalid. <math>k = 8</math> yields <math>x = 2</math>, so that is one solution. | For the first case, we note that this can only occur when it is a perfect square trinomal, or <math>k = 0, 8</math>. However, <math>k = 0</math> results in <math>\log(0)</math> on the LHS, which is invalid. <math>k = 8</math> yields <math>x = 2</math>, so that is one solution. | ||
− | For the second case, we can use the quadratic formula. We have <cmath>x = \frac{k-4 \pm \sqrt{k^2-8k}}2,</cmath> so in order for there to be at least one real solution, the discriminant must be nonnegative, or <math>k < 0</math> or <math>k > 8</math>. Note that if <math>k > 8</math>, then both solutions will be positive, and therefore both valid | + | For the second case, we can use the quadratic formula. We have <cmath>x = \frac{k-4 \pm \sqrt{k^2-8k}}2,</cmath> so in order for there to be at least one real solution, the discriminant must be nonnegative, or <math>k < 0</math> or <math>k > 8</math>. |
+ | |||
+ | Note that if <math>k > 8</math>, then both solutions to <math>x</math> will be positive, and therefore both valid, which means all such <math>k</math> are unsatisfactory. | ||
+ | |||
We now wish to show that if <math>k < 0</math>, then there is exactly one solution that works. Note that whenever <math>k < 0</math>, both "solutions" in <math>x</math> are negative. One of the solutions to the equation is <math>x = \frac{k-4 + \sqrt{k^2-8k}}2</math>. We wish to prove that <math>x + 2 > 0</math>, or <math>x > -2</math> (therefore the RHS in the original equation will be defined). Substituting, we have <math>\frac{k-4 + \sqrt{k^2-8k}}2 > -2</math>, or <math>\sqrt{k^2 - 8k} > -k</math>. Since both sides are positive, we can square both sides (if <math>k < 0</math>, then <math>-k > 0</math>) to get <math>k^2-8k > k^2</math>, or <math>8k < 0 \implies k < 0</math>, which was our original assumption, so this solution satisfies the original equation. The other case is when <math>x = \frac{k-4 - \sqrt{k^2-8k}}2</math>, which we wish to show is less that <math>-2</math>, or <math>\frac{k-4 - \sqrt{k^2-8k}}2 < -2 \implies k < \sqrt{k^2-8k}</math>. However, since the square root is defined to be positive, then this is always true, which implies that whenever <math>k < 0</math>, there is exactly one real solution that satisfies the original equation. Combining this with <math>k \in [-500, 500]</math>, we find that the answer is <math>500 + 1 = \boxed{501}</math>. | We now wish to show that if <math>k < 0</math>, then there is exactly one solution that works. Note that whenever <math>k < 0</math>, both "solutions" in <math>x</math> are negative. One of the solutions to the equation is <math>x = \frac{k-4 + \sqrt{k^2-8k}}2</math>. We wish to prove that <math>x + 2 > 0</math>, or <math>x > -2</math> (therefore the RHS in the original equation will be defined). Substituting, we have <math>\frac{k-4 + \sqrt{k^2-8k}}2 > -2</math>, or <math>\sqrt{k^2 - 8k} > -k</math>. Since both sides are positive, we can square both sides (if <math>k < 0</math>, then <math>-k > 0</math>) to get <math>k^2-8k > k^2</math>, or <math>8k < 0 \implies k < 0</math>, which was our original assumption, so this solution satisfies the original equation. The other case is when <math>x = \frac{k-4 - \sqrt{k^2-8k}}2</math>, which we wish to show is less that <math>-2</math>, or <math>\frac{k-4 - \sqrt{k^2-8k}}2 < -2 \implies k < \sqrt{k^2-8k}</math>. However, since the square root is defined to be positive, then this is always true, which implies that whenever <math>k < 0</math>, there is exactly one real solution that satisfies the original equation. Combining this with <math>k \in [-500, 500]</math>, we find that the answer is <math>500 + 1 = \boxed{501}</math>. | ||
+ | |||
+ | ===Note=== | ||
+ | The key to this solution is understanding that <math>\log(x+2)</math> has a domain of <math>(-2, \infty),</math> so in the second case, when there are two possible solutions of <math>x</math> to <math>k<0,</math> we notice that only the interval of the greater solution <math>(0,-2)</math> works, which means that for any <math>k<0</math> will have exactly one solution. | ||
+ | |||
+ | ~mathboy282 | ||
+ | |||
+ | ===Reworded Solution 2=== | ||
+ | |||
+ | Immediately we notice <math>k</math> is non-zero, in fact we must have <math>kx, (x+2) > 0</math> for our sole solution <math>x = x_0</math>. Simplifying the logarithmic equation we get <math>kx = (x+2)^2 \rightarrow 0 = x^2 + (4-k)x + 4 \rightarrow x = \frac{k-4 \pm \sqrt{k^2 - 8k}}{2}</math>. Then <math>k \leq 0</math> or <math>k \geq 8</math>. When <math>k = 8</math> we have exactly one real solution (easily verifiable). Notice when <math>k > 8</math>, both solutions to <math>x</math> are positive, and so all such <math>k</math> are not satisfactory. When <math>k < 0</math> it can be shown that the greater solution to <math>x</math> is in the interval <math>(0,-2)</math> and the lesser solution is in the interval <math>(-2,-\infty)</math> which is satisfactory. Then <math>k = 8,-1,-2,...,-500 \rightarrow \boxed{501}</math> satisfactory integer values of <math>k</math>. | ||
+ | |||
+ | ~FRIDAY | ||
=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}} |
Latest revision as of 23:03, 5 January 2024
Problem
Find the number of integer values of in the closed interval
for which the equation
has exactly one real solution.
Solution 1
Note the equation
is valid for
and
.
. The equation
is derived by taking away the outside logs from the previous equation. Because
is always non-negative,
must also be non-negative; therefore this takes care of the
condition as long as
, i.e.
cannot be
. Now, we graph both
(the green graph) and
(the red graph for
) for
. It is easy to see that all negative values of
make the equation
have only one solution. However, there is also one positive value of
that makes the equation only have one solution, as shown by the steepest line in the diagram. We can show that the slope of this line is a positive integer by setting the discriminant of the equation
to be
and solving for
. Therefore, there are
negative solutions and
positive solution, for a total of
.
Solution 2
We use an algebraic approach. Since , then
(the converse isn't necessarily true!), or
. Our original equation has exactly one solution if and only if there is only one solution to the above equation, or one of the solutions is extraneous; it involves the computation of the log of a nonpositive number.
For the first case, we note that this can only occur when it is a perfect square trinomal, or . However,
results in
on the LHS, which is invalid.
yields
, so that is one solution.
For the second case, we can use the quadratic formula. We have so in order for there to be at least one real solution, the discriminant must be nonnegative, or
or
.
Note that if , then both solutions to
will be positive, and therefore both valid, which means all such
are unsatisfactory.
We now wish to show that if , then there is exactly one solution that works. Note that whenever
, both "solutions" in
are negative. One of the solutions to the equation is
. We wish to prove that
, or
(therefore the RHS in the original equation will be defined). Substituting, we have
, or
. Since both sides are positive, we can square both sides (if
, then
) to get
, or
, which was our original assumption, so this solution satisfies the original equation. The other case is when
, which we wish to show is less that
, or
. However, since the square root is defined to be positive, then this is always true, which implies that whenever
, there is exactly one real solution that satisfies the original equation. Combining this with
, we find that the answer is
.
Note
The key to this solution is understanding that has a domain of
so in the second case, when there are two possible solutions of
to
we notice that only the interval of the greater solution
works, which means that for any
will have exactly one solution.
~mathboy282
Reworded Solution 2
Immediately we notice is non-zero, in fact we must have
for our sole solution
. Simplifying the logarithmic equation we get
. Then
or
. When
we have exactly one real solution (easily verifiable). Notice when
, both solutions to
are positive, and so all such
are not satisfactory. When
it can be shown that the greater solution to
is in the interval
and the lesser solution is in the interval
which is satisfactory. Then
satisfactory integer values of
.
~FRIDAY
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.