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예비 남고딩 [760922] · MS 2017 (수정됨) · 쪽지

2025-08-09 00:31:39
조회수 156

수능수학 ai 성능

게시글 주소: https://orbi.kr/00074214522

출시한지 5시간된 epsilon 모의고사의 21번을




제미나이 2.5 pro에게 풀어달라함.


 한글로 물어보니 헛소리만 늘어놓음. 영어로 "solve"라고만 하니  정답을 맞추었다.


Here is a step-by-step solution to the problem.

1. Understanding the Problem Statement

We are given a quadratic function f(x) with the following properties:

 * The leading coefficient is a, and a > 0.

 * f(0) = 0.

 * There's a positive constant k.

This means the function can be written as f(x) = ax^2 + px for some constant p.

The function f(x) must satisfy two conditions:

(Condition Ga)

\lim_{x \to b} \frac{f(x+k)}{\sqrt{\{f(x)\}^2+3} - 2\{f(x)\}^2}

The limit does not exist for exactly one real value of b, and that value is b=2.

The original Korean "실수 b의 값은 2뿐이다" is best interpreted as "the only value for the real number b is 2". This means there is only one value of b for which the limit fails to exist, and that value is b=2.

(Condition Na)

f(a) = \frac{1}{4}a^2.

We need to find the value of k + f(6).

2. Analyzing Condition (Ga)

The limit may not exist if the denominator approaches zero. Let's find the values of f(x) that make the denominator zero.

Let y = f(x). The denominator is \sqrt{y^2+3} - 2y^2.

Set the denominator to zero:

\sqrt{y^2+3} - 2y^2 = 0

\sqrt{y^2+3} = 2y^2

For this equation to hold, 2y^2 must be non-negative, which is always true. Squaring both sides:

y^2 + 3 = (2y^2)^2 = 4y^4

4y^4 - y^2 - 3 = 0

This is a quadratic equation in y^2. Let Y=y^2.

4Y^2 - Y - 3 = 0

Factoring the quadratic:

(4Y+3)(Y-1) = 0

This gives Y = -\frac{3}{4} or Y = 1. Since Y=y^2 must be non-negative, we have Y=1.

So, y^2=1, which means y = f(x) = 1 or y = f(x) = -1.

The denominator is zero when f(x)=1 or f(x)=-1. These are the candidate values for b.

The limit at x \to b fails to exist if the denominator approaches 0, but the numerator does not. The numerator is f(x+k), which approaches f(b+k).

So, the limit does not exist if (f(b)=1 or f(b)=-1) AND f(b+k) \neq 0.

The limit might exist if the form is \frac{0}{0}, i.e., (f(b)=1 or f(b)=-1) AND f(b+k)=0. In this case, the value b is "cancelled" from the set of points where the limit does not exist (assuming the limit resolves to a finite value, which is generally true when the vertex is not involved).

The roots of f(x)=ax^2+px=x(ax+p)=0 are x=0 and x=-p/a.

So, cancellation occurs for a candidate b if b+k=0 or b+k=-p/a.

3. Finding the form of f(x)

We are given that the limit fails to exist only for b=2.

This means that f(2) must be 1 or -1, and f(2+k) \neq 0.

Also, for any other value b' such that f(b') \in \{1, -1\}, it must be that f(b'+k)=0.

Let's analyze the graph of f(x) = ax^2+px. It's a parabola opening upwards (a>0) passing through the origin. Its vertex is at x_v = -p/(2a), with minimum value y_v = f(x_v) = -p^2/(4a) \le 0.

The equation f(x)=1 will always have two distinct solutions, as 1 > y_v.

The equation f(x)=-1 can have 0, 1, or 2 solutions.

The total number of candidates for b is at least 2. Since only one value (b=2) remains, at least one candidate must be cancelled.

This suggests a simple structure. Let's assume the equation f(x)=1 has solutions b_1 and b_2, and the equation f(x)=-1 has no solutions. This occurs if y_v > -1.

The set of candidates for b is \{b_1, b_2\}.

We need one to be 2 and the other to be cancelled. Let b_2=2.

So, f(2)=1.

And b_1 must be cancelled, which means f(b_1+k)=0.

The roots of f(x)=1 are symmetric about the vertex axis, so b_1+b_2 = 2x_v \implies b_1+2 = -p/a \implies b_1 = -p/a - 2.

The cancellation condition for b_1 means b_1+k=0 or b_1+k=-p/a.

 * If b_1+k=-p/a, then (-p/a-2)+k=-p/a, which gives k=2.

 * If b_1+k=0, then (-p/a-2)+k=0, which gives k=p/a+2.

Let's proceed with the simpler case, k=2.

From f(2)=1:

f(x) = ax^2+px

f(2) = a(2^2) + p(2) = 4a+2p=1.

Now, use Condition (Na): f(a) = \frac{1}{4}a^2.

f(a) = a(a^2)+p(a) = a^3+pa.

So, a^3+pa = \frac{1}{4}a^2. Since a>0, we can divide by a:

a^2+p = \frac{1}{4}a.

We have a system of two equations for a and p:

 * 4a+2p=1 \implies p = \frac{1-4a}{2}

 * a^2+p = \frac{a}{4}

Substitute (1) into (2):

a^2 + \frac{1-4a}{2} = \frac{a}{4}

Multiply by 4:

4a^2 + 2(1-4a) = a

4a^2 + 2 - 8a = a

4a^2 - 9a + 2 = 0

Factoring this quadratic equation:

(4a-1)(a-2) = 0

This gives two possible values for a: a=1/4 or a=2.

Let's check each case:

 * Case a=1/4:

   p = \frac{1-4(1/4)}{2} = \frac{1-1}{2}=0.

   So f(x) = \frac{1}{4}x^2.

   Let's verify our assumptions.

   The minimum value is y_v=0, so y_v>-1 is satisfied.

   The candidates for b are solutions to f(x)=1 \implies \frac{1}{4}x^2=1 \implies x^2=4 \implies x=\pm 2.

   The set of candidates is \{-2, 2\}.

   The value b=2 is one of them. The other is b_1=-2.

   We must check if b_1=-2 is cancelled. Cancellation occurs if f(b_1+k)=0.

   With p=0 and k=2, f(x)=0 only at x=0. The cancellation condition is b_1+k=0.

   -2+2=0. So f(-2+2)=f(0)=0.

   The candidate b=-2 is indeed cancelled.

   So the only value of b for which the limit doesn't exist is b=2. This perfectly matches Condition (Ga).

   Condition (Na) is also satisfied by construction. So this solution (a=1/4, p=0, k=2) is valid.

 * Case a=2:

   p = \frac{1-4(2)}{2} = -\frac{7}{2}.

   We must check the condition y_v > -1.

   y_v = -p^2/(4a) = -(-7/2)^2 / (4 \cdot 2) = -(49/4)/8 = -49/32.

   -49/32 = -1.53125. This is less than -1. So the condition y_v > -1 is not satisfied. This case is invalid.

Therefore, the only valid solution is a=1/4, p=0, and k=2.

The function is f(x) = \frac{1}{4}x^2.

4. Calculating the Final Answer

We need to find k+f(6).

We found k=2.

We find f(6):

f(6) = \frac{1}{4}(6^2) = \frac{36}{4} = 9.

Finally,

k + f(6) = 2 + 9 = 11.

The final answer is 11.


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