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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 30 Letf(x) be an increasing continuous function on the entire real line, and letf −1(x) be its inverse function. The functionf −1(x) satisfies the following conditions: (A) For|x| ≤1, 4 � f −1(x)
2 =x 2(x2 −5) 2. (B) For|x|>1,

f −1(x)

=e |x|−1 + 1. For a real numberm, letg(m) denote the number of intersection points between the line of slopempassing through (1,0) and the curvey=f(x). Suppose thatg(m) is discontinuous atm=aandm=bwitha < b. Evaluate g(a)×  lim m→a+ g(m)  +g(b)× lnb b 2 . (Here, you may use lim x→∞ lnx x = 0.) [4 points] Mathematics (Geometry) 더보기

Looking at this problem, I need to find 𝑓 ( 𝑥 ) f(x) from its inverse, determine where 𝑔 ( 𝑚 ) g(m) is discontinuous, and evaluate the expression.

Step 1: Find 𝑓 ( 𝑥 ) f(x)

For ∣ 𝑥 ∣ ≤ 1 ∣x∣≤1: 𝑓 − 1 ( 𝑥 )

2 𝑥 2 ( 𝑥 2 − 5 ) 2 f −1 (x)=2x 2 (x 2 −5) 2 . Let 𝑦

𝑓 − 1 ( 𝑥 ) y=f −1 (x), so 𝑦

2 𝑥 2 ( 𝑥 2 − 5 ) 2 y=2x 2 (x 2 −5) 2 . Setting 𝑡

𝑥 2 t=x 2 :

𝑦

2 𝑡 ( 𝑡 − 5 ) 2

2 𝑡 3 − 20 𝑡 2 + 50 𝑡 y=2t(t−5) 2 =2t 3 −20t 2 +50t

For ∣ 𝑥 ∣

1 ∣x∣>1: 𝑓 − 1 ( 𝑥 )

𝑒 ∣ 𝑥 ∣ − 1 + 1 f −1 (x)=e ∣x∣−1 +1. Since 𝑓 f is increasing, 𝑓 − 1 f −1 is increasing, so for 𝑥

1 x>1: 𝑓 − 1 ( 𝑥 )

𝑒 𝑥 − 1 + 1 f −1 (x)=e x−1 +1, giving 𝑓 ( 𝑥 )

1 + ln ⁡ 𝑥 f(x)=1+lnx for 𝑥

2 x>2.

Step 2: Determine 𝑓 ( 𝑥 ) f(x) on [ − 1 , 1 ] [−1,1]

Since 𝑓 f is continuous and increasing, 𝑓 − 1 f −1 must be continuous at 𝑥

± 1 x=±1. At 𝑥

1 x=1: 𝑓 − 1 ( 1 )

2 ( 1 ) ( 1 − 5 ) 2

32 f −1 (1)=2(1)(1−5) 2 =32. At 𝑥

− 1 x=−1: 𝑓 − 1 ( − 1 )

32 f −1 (−1)=32.

For ∣ 𝑥 ∣

1 ∣x∣>1: 𝑓 − 1 ( 1 + )

𝑒 0 + 1

2 f −1 (1 + )=e 0 +1=2.

There's a discontinuity at 𝑥

± 1 x=±1 unless the formula is different. Reconsidering: if 𝑓 − 1 ( 𝑥 )

𝑥 2 ( 𝑥 2 − 5 ) 2 8 f −1 (x)= 8 x 2 (x 2 −5) 2 ​

for ∣ 𝑥 ∣ ≤ 1 ∣x∣≤1, then 𝑓 − 1 ( ± 1 )

16 8

2 f −1 (±1)= 8 16 ​

=2, matching the right piece.

With 𝑦

𝑡 ( 𝑡 − 5 ) 2 8 y= 8 t(t−5) 2 ​

where 𝑡

𝑥 2 ∈ [ 0 , 1 ] t=x 2 ∈[0,1], and 𝑓 − 1 f −1 maps [ − 1 , 1 ] [−1,1] to [ 0 , 2 ] [0,2], we get 𝑓 f maps [ 0 , 2 ] [0,2] to [ − 1 , 1 ] [−1,1].

Step 3: Find 𝑎 a and 𝑏 b

The line 𝑦

𝑚 ( 𝑥 − 1 ) y=m(x−1) passes through ( 1 , 0 ) (1,0). For 𝑚

0 m>0, the line intersects 𝑓 f on [ 0 , 2 ] [0,2] and possibly on ( 2 , ∞ ) (2,∞) where 𝑓 ( 𝑥 )

1 + ln ⁡ 𝑥 f(x)=1+lnx.

Discontinuity at 𝑏 b: When 𝑚

1 m=1, the line 𝑦

𝑥 − 1 y=x−1 is tangent to 𝑓 ( 𝑥 )

1 + ln ⁡ 𝑥 f(x)=1+lnx at 𝑥

1 x=1 (but 𝑥

1 x=1 is not in the domain 𝑥

2 x>2). Actually, for 𝑥

2 x>2, 𝑓 ′ ( 𝑥 )

1 𝑥 f ′ (x)= x 1 ​

, and 𝑓 ′ ( 𝑥 )

𝑚 f ′ (x)=m gives 𝑥

1 𝑚 x= m 1 ​

. For tangency, 𝑚 ( 𝑥 − 1 )

1 + ln ⁡ 𝑥 m(x−1)=1+lnx with 𝑥

1 𝑚 x= m 1 ​

:

𝑚 ( 1 𝑚 − 1 )

1 + ln ⁡ 1 𝑚    ⟹    1 − 𝑚

1 − ln ⁡ 𝑚    ⟹    𝑚

ln ⁡ 𝑚 m( m 1 ​

−1)=1+ln m 1 ​

⟹1−m=1−lnm⟹m=lnm This has no solution for 𝑚

0 m>0 since ln ⁡ 𝑚 < 𝑚 lnm<m.

Discontinuity at 𝑎 a: When 𝑚

0 m=0, the line 𝑦

0 y=0 passes through ( 1 , 0 ) (1,0). This line intersects 𝑓 f at 𝑥

1 x=1 (where 𝑓 ( 1 )

0 f(1)=0) and potentially other points.

Step 4: Count intersections

For 𝑚

0 m=0: line 𝑦

0 y=0 intersects 𝑓 ( 𝑥 )

0 f(x)=0 at 𝑥

1 x=1 (since 𝑓 − 1 ( 0 )

0 f −1 (0)=0). On [ 0 , 2 ] [0,2], 𝑓 − 1 ( 𝑥 )

𝑥 2 ( 𝑥 2 − 5 ) 2 8 f −1 (x)= 8 x 2 (x 2 −5) 2 ​

has 𝑓 − 1 ( 0 )

0 f −1 (0)=0, so 𝑓 ( 0 )

0 f(0)=0 as well. Wait, 𝑓 − 1 ( 0 )

0 f −1 (0)=0 means 𝑓 ( 0 )

0 f(0)=0. So the line 𝑦

0 y=0 intersects at 𝑥

0 x=0 and 𝑥

1 x=1.

For 𝑚 m slightly positive: the line 𝑦

𝑚 ( 𝑥 − 1 ) y=m(x−1) is below 𝑦

0 y=0 for 𝑥 < 1 x<1 and above for 𝑥

1 x>1. On [ 0 , 1 ] [0,1], 𝑓 ( 𝑥 ) ≥ 0 f(x)≥0 (since 𝑓 − 1 ( 𝑥 ) ≥ 0 f −1 (x)≥0 for 𝑥 ∈ [ 0 , 1 ] x∈[0,1]), so the line 𝑦

𝑚 ( 𝑥 − 1 ) < 0 y=m(x−1)<0 doesn't intersect 𝑓 f on [ 0 , 1 ] [0,1]. On [ 1 , 2 ] [1,2], 𝑓 ( 𝑥 ) ≥ 0 f(x)≥0 and the line 𝑦

𝑚 ( 𝑥 − 1 ) ≥ 0 y=m(x−1)≥0, so they intersect once. On ( 2 , ∞ ) (2,∞), 𝑓 ( 𝑥 )

1 + ln ⁡ 𝑥 f(x)=1+lnx and the line may intersect once.

So 𝑔 ( 0 )

2 g(0)=2 (intersections at 𝑥

0 x=0 and 𝑥

1 x=1), and lim ⁡ 𝑚 → 0 + 𝑔 ( 𝑚 )

1 lim m→0 + ​

g(m)=1 (one intersection on [ 1 , 2 ] [1,2]).

For 𝑚

1 m=1: the line 𝑦

𝑥 − 1 y=x−1 intersects 𝑓 ( 𝑥 )

1 + ln ⁡ 𝑥 f(x)=1+lnx on ( 2 , ∞ ) (2,∞). At 𝑥

2 x=2: 𝑓 ( 2 )

1 + ln ⁡ 2 ≈ 1.69 f(2)=1+ln2≈1.69, line gives 1

1. At 𝑥 = 𝑒 x=e: 𝑓 ( 𝑒 ) = 2 f(e)=2, line gives 𝑒 − 1 ≈ 1.72 e−1≈1.72. They intersect once on ( 2 , ∞ ) (2,∞).

For 𝑚 m slightly less than 1: still one intersection on ( 2 , ∞ ) (2,∞). For 𝑚 m slightly more than 1: the line 𝑦

𝑚 ( 𝑥 − 1 ) y=m(x−1) grows faster, may lose the intersection on ( 2 , ∞ ) (2,∞).

So 𝑏

1 b=1, 𝑔 ( 𝑏 )

1 g(b)=1, lim ⁡ 𝑚 → 1 − 𝑔 ( 𝑚 )

1 lim m→1 − ​

g(m)=1.

Step 5: Evaluate

With 𝑎

0 a=0, 𝑏

1 b=1:

𝑔 ( 𝑎 ) × lim ⁡ 𝑚 → 𝑎 + 𝑔 ( 𝑚 ) + 𝑔 ( 𝑏 ) × ( ln ⁡ 𝑏 𝑏 ) 2

2 × 1 + 1 × ( ln ⁡ 1 1 ) 2

2 + 0

2 g(a)×lim m→a + ​

g(m)+g(b)×( b lnb ​

) 2 =2×1+1×( 1 ln1 ​

) 2 =2+0=2

\boxed{2}

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