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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 9 For a positive real numbera, define f(x) =x 3 + 3ax2 −9a 2x+ 4. Suppose that the liney= 5 is tangent to the curvey=f(x). Determine the value of f(2). [4 points] 3 ①11 ②12 ③13 ④14 ⑤15 더보기

The line 𝑦

5 y=5 is tangent to the curve 𝑦

𝑓 ( 𝑥 ) y=f(x) where 𝑓 ( 𝑥 )

𝑥 3 + 3 𝑎 𝑥 2 − 9 𝑎 2 𝑥 + 4 f(x)=x 3 +3ax 2 −9a 2 x+4. Tangency requires 𝑓 ( 𝑥 )

5 f(x)=5 and 𝑓 ′ ( 𝑥 )

0 f ′ (x)=0 at the point of contact.

First, find the derivative:

𝑓 ′ ( 𝑥 )

3 𝑥 2 + 6 𝑎 𝑥 − 9 𝑎 2 . f ′ (x)=3x 2 +6ax−9a 2 .

Set 𝑓 ′ ( 𝑥 )

0 f ′ (x)=0:

3 𝑥 2 + 6 𝑎 𝑥 − 9 𝑎 2

0 ⟹ 𝑥 2 + 2 𝑎 𝑥 − 3 𝑎 2

3x 2 +6ax−9a 2 =0⟹x 2 +2ax−3a 2 =0.

Factor:

( 𝑥 + 3 𝑎 ) ( 𝑥 − 𝑎 )

0 ⟹ 𝑥

− 3 𝑎    or    𝑥

𝑎 . (x+3a)(x−a)=0⟹x=−3aorx=a.

Since 𝑎

0 a>0, evaluate 𝑓 ( 𝑥 ) f(x) at these critical points:

𝑓 ( 𝑎 )

𝑎 3 + 3 𝑎 ( 𝑎 2 ) − 9 𝑎 2 ( 𝑎 ) + 4

𝑎 3 + 3 𝑎 3 − 9 𝑎 3 + 4

− 5 𝑎 3 + 4. f(a)=a 3 +3a(a 2 )−9a 2 (a)+4=a 3 +3a 3 −9a 3 +4=−5a 3 +4.

Setting 𝑓 ( 𝑎 )

5 f(a)=5 gives − 5 𝑎 3 + 4

5 ⇒ 𝑎 3

− 1 5 −5a 3 +4=5⇒a 3 =− 5 1 ​

, impossible for 𝑎

0 a>0.

𝑓 ( − 3 𝑎 )

( − 3 𝑎 ) 3 + 3 𝑎 ( − 3 𝑎 ) 2 − 9 𝑎 2 ( − 3 𝑎 ) + 4

− 27 𝑎 3 + 27 𝑎 3 + 27 𝑎 3 + 4

27 𝑎 3 + 4. f(−3a)=(−3a) 3 +3a(−3a) 2 −9a 2 (−3a)+4=−27a 3 +27a 3 +27a 3 +4=27a 3 +4.

Setting 𝑓 ( − 3 𝑎 )

5 f(−3a)=5 gives 27 𝑎 3 + 4

5 ⇒ 𝑎 3

1 27 ⇒ 𝑎

1 3 27a 3 +4=5⇒a 3

27 1 ​

⇒a= 3 1 ​

.

Now compute 𝑓 ( 2 ) f(2) with 𝑎

1 3 a= 3 1 ​

:

𝑓 ( 2 )

2 3 + 3 ( 1 3 ) ( 2 2 ) − 9 ( 1 9 ) ( 2 ) + 4

8 + 4 − 2 + 4

f(2)=2 3 +3( 3 1 ​

)(2 2 )−9( 9 1 ​

)(2)+4=8+4−2+4=14.

The value 14 14 corresponds to choice ④.

\boxed{④}

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