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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 11 At timet= 0, a pointPstarts from the origin and moves along a straight line. For a real numberk, the velocityv(t) ofPat timet(witht≥0) is given by v(t) =t 2 −kt+ 4. From the statements in⟨Box⟩, which ones are true? (Select all that apply.) [4 points] ⟨Box⟩ (a) Ifk= 0, then the position ofPat timet= 1 is 13 3 . (b) Ifk= 3, then after starting, the direction of motion ofPchanges exactly once. (c) Ifk= 5, then the distance traveled byPfromt= 0 tot= 2 is 3. ①(a) ②(a), (b) ③(a), (c) ④(b), (c) ⑤(a), (b), (c) 4 더보기

The problem asks which statements are true.

(a) For 𝑘

0 k=0, 𝑣 ( 𝑡 )

𝑡 2 + 4 v(t)=t 2 +4. Position at 𝑡

1 t=1 is ∫ 0 1 ( 𝑡 2 + 4 )   𝑑 𝑡

1 3 + 4

13 3 ∫ 0 1 ​

(t 2 +4)dt= 3 1 ​

+4= 3 13 ​

. True.

(b) For 𝑘

3 k=3, 𝑣 ( 𝑡 )

𝑡 2 − 3 𝑡 + 4 v(t)=t 2 −3t+4. Discriminant

9 − 16

− 7 < 0 =9−16=−7<0, so 𝑣 ( 𝑡 )

0 v(t)>0 for all 𝑡 t. Velocity never changes sign, so direction never changes. False.

(c) For 𝑘

5 k=5, 𝑣 ( 𝑡 )

𝑡 2 − 5 𝑡 + 4

( 𝑡 − 1 ) ( 𝑡 − 4 ) v(t)=t 2 −5t+4=(t−1)(t−4). On [ 0 , 2 ] [0,2], 𝑣 ( 𝑡 ) v(t) changes sign at 𝑡

1 t=1.

∫ 0 1 𝑣 ( 𝑡 )   𝑑 𝑡

11 6 ∫ 0 1 ​

v(t)dt= 6 11 ​

, ∫ 1 2 − 𝑣 ( 𝑡 )   𝑑 𝑡

7 6 ∫ 1 2 ​

−v(t)dt= 6 7 ​

. Total distance

11 6 + 7 6

3

6 11 ​

6 7 ​

=3. True.

True statements: (a) and (c).

\boxed{③}

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