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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 15 Consider the function f(x) = ( −x2 (x <0), x2 −x(x≥0), and, for a positive real numbera, define the function g(x) =    ax+a(x <−1), 0 (−1≤x <1), ax−a(x≥1). Define h(x) = Z x 0 � g(t)−f(t)
dt. Letkbe the maximum value ofafor which the functionh(x) has exactly one extremum (local maximum or local minimum). Whena=k, compute the value ofk+h(3). [4 points] ① 9 2 ② 11 2 ③ 13 2 ④ 15 2 ⑤ 17 2 6 Numerical answer 더보기

Let me work through this problem carefully.

Step 1: Write out the piecewise functions

𝑓 ( 𝑥 )

{ − 𝑥 2

( 𝑥 < 0 )

𝑥 2 − 𝑥

( 𝑥 ≥ 0 ) f(x)={ −x 2 x 2 −x ​

(x<0) (x≥0) ​

𝑔 ( 𝑥 )

{ 𝑎 ( 𝑥 + 1 )

( 𝑥 < − 1 )

0

( − 1 ≤ 𝑥 < 1 )

𝑎 ( 𝑥 − 1 )

( 𝑥 ≥ 1 ) g(x)= ⎩ ⎨ ⎧ ​

a(x+1) 0 a(x−1) ​

(x<−1) (−1≤x<1) (x≥1) ​

Step 2: Compute ℎ ′ ( 𝑥 )

𝑔 ( 𝑥 ) − 𝑓 ( 𝑥 ) h ′ (x)=g(x)−f(x)

Piecewise:

𝑥 < − 1 x<−1: ℎ ′ ( 𝑥 )

𝑎 ( 𝑥 + 1 ) + 𝑥 2

𝑥 2 + 𝑎 𝑥 + 𝑎 h ′ (x)=a(x+1)+x 2 =x 2 +ax+a − 1 ≤ 𝑥 < 0 −1≤x<0: ℎ ′ ( 𝑥 )

0 + 𝑥 2

𝑥 2 h ′ (x)=0+x 2 =x 2 0 ≤ 𝑥 < 1 0≤x<1: ℎ ′ ( 𝑥 )

0 − ( 𝑥 2 − 𝑥 )

𝑥 ( 1 − 𝑥 ) h ′ (x)=0−(x 2 −x)=x(1−x) 𝑥 ≥ 1 x≥1: ℎ ′ ( 𝑥 )

𝑎 ( 𝑥 − 1 ) − ( 𝑥 2 − 𝑥 )

− 𝑥 2 + ( 𝑎 + 1 ) 𝑥 − 𝑎

− ( 𝑥 − 1 ) ( 𝑥 − 𝑎 ) h ′ (x)=a(x−1)−(x 2 −x)=−x 2 +(a+1)x−a=−(x−1)(x−a) Step 3: Analyze zeros and sign changes On ( − 1 , 0 ) (−1,0): ℎ ′ ( 𝑥 )

𝑥 2

0 h ′ (x)=x 2

0 On ( 0 , 1 ) (0,1): ℎ ′ ( 𝑥 ) = 𝑥 ( 1 − 𝑥 )

0 h ′ (x)=x(1−x)>0 𝑥

0 x=0: zero, but no sign change (positive on both sides) 𝑥

1 x=1: zero; sign after 𝑥

1 x=1 depends on 𝑎 a

For 𝑥 < − 1 x<−1: 𝑥 2 + 𝑎 𝑥 + 𝑎 x 2 +ax+a has discriminant 𝑎 ( 𝑎 − 4 ) a(a−4).

0 < 𝑎 < 4 0<a<4: no real roots, always positive 𝑎

4 a=4: double root at 𝑥

− 2 x=−2 𝑎

4 a>4: two roots 𝑟 1 < 𝑟 2 < − 1 r 1 ​

<r 2 ​

<−1

For 𝑥 ≥ 1 x≥1: − ( 𝑥 − 1 ) ( 𝑥 − 𝑎 ) −(x−1)(x−a).

0 < 𝑎 < 1 0<a<1: negative for 𝑥

1 x>1 → sign change at 𝑥

1 x=1 (local max) 𝑎

1 a=1: − ( 𝑥 − 1 ) 2 ≤ 0 −(x−1) 2 ≤0, no sign change 1 < 𝑎 < 4 1<a<4: positive on ( 1 , 𝑎 ) (1,a), negative on ( 𝑎 , ∞ ) (a,∞) → two sign changes (local max at 𝑥

1 x=1, local min at 𝑥

𝑎 x=a) 𝑎

4 a=4: positive on ( 1 , 4 ) (1,4), negative on ( 4 , ∞ ) (4,∞) → two sign changes 𝑎

4 a>4: positive on ( 1 , 𝑎 ) (1,a), negative on ( 𝑎 , ∞ ) (a,∞) → two sign changes; plus two sign changes from the left roots 𝑟 1 , 𝑟 2 r 1 ​

,r 2 ​

→ four extrema Step 4: Find 𝑘 k

Exactly one extremum occurs when 0 < 𝑎 < 1 0<a<1. The maximum value is:

𝑘

1 k=1

Step 5: Compute ℎ ( 3 ) h(3) when 𝑎

1 a=1

ℎ ( 3 )

∫ 0 3 ( 𝑔 ( 𝑡 ) − 𝑓 ( 𝑡 ) )   𝑑 𝑡 h(3)=∫ 0 3 ​

(g(t)−f(t))dt

Break into intervals:

[ 0 , 1 ] [0,1]: 𝑔 − 𝑓

0 − ( 𝑡 2 − 𝑡 )

𝑡 − 𝑡 2 g−f=0−(t 2 −t)=t−t 2

∫ 0 1 ( 𝑡 − 𝑡 2 )   𝑑 𝑡

[ 𝑡 2 2 − 𝑡 3 3 ] 0 1

1 6 ∫ 0 1 ​

(t−t 2 )dt=[ 2 t 2 ​

− 3 t 3 ​

] 0 1 ​

= 6 1 ​

[ 1 , 3 ] [1,3]: 𝑔 − 𝑓

( 𝑡 − 1 ) − ( 𝑡 2 − 𝑡 )

− 𝑡 2 + 2 𝑡 − 1

− ( 𝑡 − 1 ) 2 g−f=(t−1)−(t 2 −t)=−t 2 +2t−1=−(t−1) 2

∫ 1 3 − ( 𝑡 − 1 ) 2   𝑑 𝑡

[ − ( 𝑡 − 1 ) 3 3 ] 1 3

− 8 3 ∫ 1 3 ​

−(t−1) 2 dt=[− 3 (t−1) 3 ​

] 1 3 ​

=− 3 8 ​

ℎ ( 3 )

1 6 − 8 3

1 6 − 16 6

− 15 6

− 5 2 h(3)= 6 1 ​

− 3 8 ​

= 6 1 ​

− 6 16 ​

=− 6 15 ​

=− 2 5 ​

Step 6: Final answer

𝑘 + ℎ ( 3 )

1 + ( − 5 2 )

− 3 2 k+h(3)=1+(− 2 5 ​

)=− 2 3 ​

\boxed{-\dfrac{3}{2}}

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