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로그인 회원가입 You are solving a Korean CSAT mathematics problem. Reason carefully, but the final response must contain exactly one final answer inside \boxed{}.

For multiple-choice problems, put only one of ①, ②, ③, ④, ⑤ inside \boxed{}. For short-answer problems, put only the final integer inside \boxed{}. Do not use external tools.

Problem 21 Letf(x) be a cubic function whose leading coefficient is positive, and lettbe a real number. Define g(x) = ( −f(x) (x < t), f(x) (x≥t), so thatgis continuous on the entire real line and satisfies the following conditions. 8 (A) For every real numbera, the limit lim x→a+ g(x) x(x−2) exists. (B) The set of natural numbersmfor which lim x→m+ g(x) x(x−2) is negative is  g(−1),− 7 2 g(1)  . Assuming thatg(−1)̸=− 7 2 g(1), computeg(−5). [4 points] 더보기

Let me work through this carefully.

We have a cubic function 𝑓 ( 𝑥 ) f(x) with positive leading coefficient, and 𝑡 ∈ 𝑅 t∈R. Define

𝑔 ( 𝑥 )

{ − 𝑓 ( 𝑥 ) ,

𝑥 < 𝑡 ,

𝑓 ( 𝑥 ) ,

𝑥 ≥ 𝑡 . g(x)={ −f(x), f(x), ​

x<t, x≥t. ​

𝑔 g is continuous on 𝑅 R.

Step 1: Continuity at 𝑡 t. At 𝑥

𝑡 x=t, left limit is − 𝑓 ( 𝑡 ) −f(t), right limit is 𝑓 ( 𝑡 ) f(t). Continuity forces − 𝑓 ( 𝑡 )

𝑓 ( 𝑡 ) −f(t)=f(t), so 𝑓 ( 𝑡 )

0 f(t)=0. Thus 𝑡 t is a root of 𝑓 f.

Step 2: Condition (A). For every real 𝑎 a, lim ⁡ 𝑥 → 𝑎 + 𝑔 ( 𝑥 ) 𝑥 ( 𝑥 − 2 ) x→a + lim ​

x(x−2) g(x) ​

exists. The denominator vanishes at 𝑎

0 , 2 a=0,2. For the limit to be finite, the numerator must also vanish there, so 𝑔 ( 0 )

0 g(0)=0 and 𝑔 ( 2 )

0 g(2)=0. Checking the sign of 𝑡 t relative to 0 , 2 0,2 shows this forces 𝑓 ( 0 )

0 f(0)=0 and 𝑓 ( 2 )

0 f(2)=0. Hence 0 0 and 2 2 are roots of 𝑓 f.

Since 𝑓 f is cubic with positive leading coefficient, write

𝑓 ( 𝑥 )

𝑘 𝑥 ( 𝑥 − 2 ) ( 𝑥 − 𝑟 ) , 𝑘

0 ,    𝑟 ∈ 𝑅 . f(x)=kx(x−2)(x−r),k>0,r∈R.

Step 3: Condition (B). The set of natural numbers 𝑚 m for which lim ⁡ 𝑥 → 𝑚 + 𝑔 ( 𝑥 ) 𝑥 ( 𝑥 − 2 ) < 0 x→m + lim ​

x(x−2) g(x) ​

<0 is { 𝑔 ( − 1 ) , − 7 2 𝑔 ( 1 ) } {g(−1),− 2 7 ​

g(1)}, and 𝑔 ( − 1 ) ≠ − 7 2 𝑔 ( 1 ) g(−1)  =− 2 7 ​

g(1).

For 𝑚 ≠ 2 m  =2, the limit is 𝑔 ( 𝑚 ) 𝑚 ( 𝑚 − 2 ) m(m−2) g(m) ​

. For 𝑚

2 m=2, since 𝑔 ( 2 )

0 g(2)=0, the limit is lim ⁡ 𝑥 → 2 + 𝑔 ( 𝑥 ) 𝑥 ( 𝑥 − 2 ) lim x→2 + ​

x(x−2) g(x) ​

, which equals ± 𝑘 ( 2 − 𝑟 ) ±k(2−r) depending on whether 𝑡 ≤ 2 t≤2 or 𝑡

2 t>2.

Testing the three possible roots for 𝑡 t:

𝑡

0 t=0: gives 𝑔 ( 1 )

𝑓 ( 1 )

𝑘 ( 𝑟 − 1 ) g(1)=f(1)=k(r−1). Then − 7 2 𝑔 ( 1 )

7 2 𝑘 ( 1 − 𝑟 ) − 2 7 ​

g(1)= 2 7 ​

k(1−r), requiring 𝑟 < 1 r<1. But then no natural 𝑚 m gives negative limit, contradicting a two-element set.

𝑡

2 t=2: gives 𝑔 ( 1 )

− 𝑓 ( 1 )

𝑘 ( 1 − 𝑟 ) g(1)=−f(1)=k(1−r), so − 7 2 𝑔 ( 1 )

7 2 𝑘 ( 𝑟 − 1 ) − 2 7 ​

g(1)= 2 7 ​

k(r−1), requiring 𝑟

1 r>1. The limit at 𝑚

1 m=1 is 𝑘 ( 𝑟 − 1 )

0 k(r−1)>0, and for 𝑚 ≥ 3 m≥3 it is 𝑘 ( 𝑚 − 𝑟 ) k(m−r). The negative limits occur for 𝑚

2 m=2 (if 𝑟

2 r>2) and for 𝑚

3 , … , ⌊ 𝑟 ⌋ m=3,…,⌊r⌋. To have exactly two such 𝑚 m, we need 𝑟 ∈ ( 3 , 4 ) r∈(3,4). Then 𝑔 ( − 1 )

3 𝑘 ( 1 + 𝑟 ) g(−1)=3k(1+r) and − 7 2 𝑔 ( 1 )

7 2 𝑘 ( 𝑟 − 1 ) − 2 7 ​

g(1)= 2 7 ​

k(r−1). Setting these equal to 2 2 and 3 3 respectively gives 𝑘

2 3 ( 1 + 𝑟 ) k= 3(1+r) 2 ​

and 𝑟

13 5 r= 5 13 ​

, which lies in ( 3 , 4 ) (3,4). This works.

𝑡

𝑟 t=r: forces 𝑟 ∈ ( 0 , 1 ) r∈(0,1) and yields no negative limits, contradiction.

Thus 𝑡

2 t=2, 𝑟

13 5 r= 5 13 ​

, 𝑘

5 27 k= 27 5 ​

.

Step 4: Compute 𝑔 ( − 5 ) g(−5). Since − 5 < 2

𝑡 −5<2=t, we have 𝑔 ( − 5 )

− 𝑓 ( − 5 ) g(−5)=−f(−5). Now

𝑓 ( − 5 )

𝑘 ( − 5 ) ( − 7 ) ( − 5 − 13 5 )

5 27 ⋅ 35 ⋅ ( − 38 5 )

− 1330 27 . f(−5)=k(−5)(−7)(−5− 5 13 ​

)= 27 5 ​

⋅35⋅(− 5 38 ​

)=− 27 1330 ​

.

Therefore

𝑔 ( − 5 )

− 𝑓 ( − 5 )

1330 27 . g(−5)=−f(−5)= 27 1330 ​

.

\boxed{\dfrac{1330}{27}}

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