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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 13 Consider the functionf(x) =x 2 −4x−3. Letℓbe the tangent line to the curvey=f(x) at the point (1,−6). For the functiong(x) = (x 3 −2x)f(x), letmbe the tangent line to the curvey=g(x) at the point (1,6). Find the area of the region bounded by the two linesℓ,m, and they-axis. [4 points] ①21 ②28 ③35 ④42 ⑤49 더보기

The tangent line ℓ to 𝑓 ( 𝑥 )

𝑥 2 − 4 𝑥 − 3 f(x)=x 2 −4x−3 at ( 1 , − 6 ) (1,−6) has slope 𝑓 ′ ( 1 )

2 ( 1 ) − 4

− 2 f ′ (1)=2(1)−4=−2, so its equation is 𝑦

− 2 𝑥 − 4 y=−2x−4.

For 𝑔 ( 𝑥 )

( 𝑥 3 − 2 𝑥 ) 𝑓 ( 𝑥 ) g(x)=(x 3 −2x)f(x), we have 𝑔 ( 1 )

6 g(1)=6. Using the product rule,

𝑔 ′ ( 𝑥 )

( 3 𝑥 2 − 2 ) 𝑓 ( 𝑥 ) + ( 𝑥 3 − 2 𝑥 ) 𝑓 ′ ( 𝑥 ) . g ′ (x)=(3x 2 −2)f(x)+(x 3 −2x)f ′ (x).

At 𝑥

1 x=1, 𝑔 ′ ( 1 )

1 ⋅ ( − 6 ) + ( − 1 ) ⋅ ( − 2 )

− 4 g ′ (1)=1⋅(−6)+(−1)⋅(−2)=−4. Thus the tangent line 𝑚 m is 𝑦

− 4 𝑥 + 10 y=−4x+10.

The two lines intersect when − 2 𝑥 − 4

− 4 𝑥 + 10 −2x−4=−4x+10, giving 𝑥

7 x=7 and 𝑦

− 18 y=−18. Their 𝑦 y-intercepts are ( 0 , − 4 ) (0,−4) and ( 0 , 10 ) (0,10). The region bounded by ℓ, 𝑚 m, and the 𝑦 y-axis is a triangle with base 10 − ( − 4 )

14 10−(−4)=14 and height 7 7. Its area is

1 2 × 14 × 7

2 1 ​

×14×7=49.

\boxed{⑤}

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