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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 7 Find the area of the region enclosed by the curves y=x 2 + 3, y=− 1 5 x2 + 3 and the linex= 2. [3 points] ① 18 5 ② 7 2 ③ 17 5 ④ 33 10 ⑤ 16 5 더보기

The curves are 𝑦

𝑥 2 + 3 y=x 2 +3 and 𝑦

− 1 5 𝑥 2 + 3 y=− 5x 2 +3 1 ​

. To find their intersection, set them equal:

𝑥 2 + 3

− 1 5 𝑥 2 + 3 x 2 +3=− 5x 2 +3 1 ​

Multiply both sides by 5 𝑥 2 + 3 5x 2 +3 (which is always positive):

( 𝑥 2 + 3 ) ( 5 𝑥 2 + 3 )

− 1 (x 2 +3)(5x 2 +3)=−1

Expand the left side:

5 𝑥 4 + 3 𝑥 2 + 15 𝑥 2 + 9

5 𝑥 4 + 18 𝑥 2 + 9 5x 4 +3x 2 +15x 2 +9=5x 4 +18x 2 +9

So the equation becomes:

5 𝑥 4 + 18 𝑥 2 + 9

− 1 ⟹ 5 𝑥 4 + 18 𝑥 2 + 10

0 5x 4 +18x 2 +9=−1⟹5x 4 +18x 2 +10=0

Let 𝑢

𝑥 2 u=x 2 . Then 5 𝑢 2 + 18 𝑢 + 10

0 5u 2 +18u+10=0. The discriminant is 18 2 − 4 ⋅ 5 ⋅ 10

324 − 200

124

0 18 2 −4⋅5⋅10=324−200=124>0, so there are two real roots for 𝑢 u:

𝑢

− 18 ± 124 10

− 18 ± 2 31 10

− 9 ± 31 5 u= 10 −18± 124 ​

​

= 10 −18±2 31 ​

​

= 5 −9± 31 ​

​

Since 31 ≈ 5.57 31 ​

≈5.57, both roots are negative: − 9 + 5.57 5 < 0 5 −9+5.57 ​

<0 and − 9 − 5.57 5 < 0 5 −9−5.57 ​

<0. Thus there is no real 𝑥 x satisfying the equation; the curves never intersect.

The region is described as “enclosed by the curves 𝑦

𝑥 2 + 3 y=x 2 +3, 𝑦

− 1 5 𝑥 2 + 3 y=− 5x 2 +3 1 ​

, and the line 𝑥

2 x=2.” Because the two curves do not meet, the only way a finite region can be enclosed is if the left boundary is implicitly the 𝑦 y-axis ( 𝑥

0 x=0). The area is then the integral of the upper curve minus the lower curve from 𝑥

0 x=0 to 𝑥

2 x=2:

Area

∫ 0 2 [ ( 𝑥 2 + 3 ) − ( − 1 5 𝑥 2 + 3 ) ] 𝑑 𝑥

∫ 0 2 ( 𝑥 2 + 3 + 1 5 𝑥 2 + 3 ) 𝑑 𝑥 Area=∫ 0 2 ​

[(x 2 +3)−(− 5x 2 +3 1 ​

)]dx=∫ 0 2 ​

(x 2 +3+ 5x 2 +3 1 ​

)dx

Compute each part:

∫ 0 2 𝑥 2   𝑑 𝑥

[ 𝑥 3 3 ] 0 2

8 3 ∫ 0 2 ​

x 2 dx=[ 3 x 3 ​

] 0 2 ​

= 3 8 ​

∫ 0 2 3   𝑑 𝑥

[ 3 𝑥 ] 0 2

6 ∫ 0 2 ​

3dx=[3x] 0 2 ​

=6 ∫ 0 2 1 5 𝑥 2 + 3   𝑑 𝑥 ∫ 0 2 ​

5x 2 +3 1 ​

dx

For the third integral, use the formula ∫ 𝑑 𝑥 𝑎 2 + 𝑥 2

1 𝑎 arctan ⁡ 𝑥 𝑎 ∫ a 2 +x 2 dx ​

= a 1 ​

arctan a x ​

. Rewrite:

1 5 𝑥 2 + 3

1 3 ⋅ 1 5 3 𝑥 2 + 1 5x 2 +3 1 ​

= 3 1 ​

⋅ 3 5 ​

x 2 +1 1 ​

Let 𝑢

5 3   𝑥 u= 3 5 ​

​

x, so 𝑑 𝑢

5 3   𝑑 𝑥 du= 3 5 ​

​

dx. Then:

∫ 𝑑 𝑥 5 𝑥 2 + 3

1 3 ∫ 𝑑 𝑥 5 3 𝑥 2 + 1

1 3 ⋅ 3 5 ∫ 𝑑 𝑢 𝑢 2 + 1

1 15 arctan ⁡ ( 5 3   𝑥 ) ∫ 5x 2 +3 dx ​

= 3 1 ​

∫ 3 5 ​

x 2 +1 dx ​

= 3 1 ​

⋅ 5 3 ​

​

∫ u 2 +1 du ​

= 15 ​

1 ​

arctan( 3 5 ​

​

x)

Evaluate from 0 to 2:

[ 1 15 arctan ⁡ ( 5 3   𝑥 ) ] 0 2

1 15 arctan ⁡ ( 2 5 3 ) [ 15 ​

1 ​

arctan( 3 5 ​

​

x)] 0 2 ​

= 15 ​

1 ​

arctan(2 3 5 ​

​

)

This is not a rational number, so the total area would be irrational. None of the given options (all rational) match this. Therefore the region cannot be from 𝑥

0 x=0 to 𝑥

2 x=2.

Re‑examine the problem statement: “enclosed by the curves 𝑦

𝑥 2 + 3 y=x 2 +3, 𝑦

− 1 5 𝑥 2 + 3 y=− 5x 2 +3 1 ​

, and the line 𝑥

2 x=2.” The only way a finite region is enclosed is if the two curves intersect at some 𝑥 x and the line 𝑥

2 x=2 forms the right boundary. But we proved they do not intersect. Hence the problem as written has no finite enclosed region. However, if we assume a typo and that the second curve is actually 𝑦

− 1 5 𝑥 2 − 3 y=− 5x 2 −3 1 ​

(or similar), we might get intersections. But given the options, the intended answer is likely 17 5 5 17 ​

(option ③) from a standard problem where the curves are 𝑦

𝑥 2 + 3 y=x 2 +3 and 𝑦

− 1 5 𝑥 2 + 3 y=− 5x 2 +3 1 ​

and the line is 𝑥

2 x=2, but the region is bounded by the curves and the 𝑦 y-axis, and the integral of the rational part simplifies to a rational number? Let's check if ∫ 0 2 1 5 𝑥 2 + 3 𝑑 𝑥 ∫ 0 2 ​

5x 2 +3 1 ​

dx could be rational? No, it's transcendental.

Wait, maybe the first curve is 𝑦

𝑥 2 − 3 y=x 2 −3? If it were 𝑦

𝑥 2 − 3 y=x 2 −3, then intersection: 𝑥 2 − 3

− 1 5 𝑥 2 + 3 x 2 −3=− 5x 2 +3 1 ​

leads to 5 𝑥 4 − 12 𝑥 2 − 8

0 5x 4 −12x 2 −8=0, giving a positive root 𝑥 2

6 + 2 19 5 x 2

5 6+2 19 ​

​

. Then the area from that root to 2 would be messy.

Given the options are all simple fractions, the area likely comes from a polynomial integrand. Perhaps the second curve is 𝑦

− 1 5 𝑥 2 + 3 y=− 5 1 ​

x 2 +3? No.

Let's test if the curves are 𝑦

𝑥 2 + 3 y=x 2 +3 and 𝑦

− 1 5 𝑥 2 + 3 y=− 5 1 ​

x 2 +3? Then intersection: 𝑥 2 + 3

− 1 5 𝑥 2 + 3 ⇒ 𝑥 2

− 1 5 𝑥 2 ⇒ 6 5 𝑥 2

0 ⇒ 𝑥

0 x 2 +3=− 5 1 ​

x 2 +3⇒x 2 =− 5 1 ​

x 2 ⇒ 5 6 ​

x 2 =0⇒x=0. Then area from 0 to 2: ∫ 0 2 ( 𝑥 2 + 3 − ( − 1 5 𝑥 2 + 3 ) ) 𝑑 𝑥

∫ 0 2 ( 6 5 𝑥 2 ) 𝑑 𝑥

6 5 ⋅ 8 3

48 15

16 5 ∫ 0 2 ​

(x 2 +3−(− 5 1 ​

x 2 +3))dx=∫ 0 2 ​

( 5 6 ​

x 2 )dx= 5 6 ​

⋅ 3 8 ​

= 15 48 ​

= 5 16 ​

. That matches option ⑤.

But the problem says 𝑦

− 1 5 𝑥 2 + 3 y=− 5x 2 +3 1 ​

, not − 1 5 𝑥 2 + 3 − 5 1 ​

x 2 +3. Could it be a formatting issue? In many Korean CSAT problems, they write 𝑦

− 1 5 𝑥 2 + 3 y=− 5 1 ​

x 2 +3 as 𝑦

− 1 5 𝑥 2 + 3 y=− 5 1 ​

x 2 +3. The given string “y=−1/(5x^2+3)” might be a misrendering of 𝑦

− 1 5 𝑥 2 + 3 y=− 5 1 ​

x 2 +3. Indeed, if the original was 𝑦

− 1 5 𝑥 2 + 3 y=− 5 1 ​

x 2 +3, then the area is 16 5 5 16 ​

.

Let's verify: 𝑦

𝑥 2 + 3 y=x 2 +3 and 𝑦

− 1 5 𝑥 2 + 3 y=− 5 1 ​

x 2 +3. They intersect at 𝑥

0 x=0. The line 𝑥

2 x=2 is the right boundary. The region enclosed is from 𝑥

0 x=0 to 𝑥

2 x=2. The upper curve is 𝑦

𝑥 2 + 3 y=x 2 +3, the lower is 𝑦

− 1 5 𝑥 2 + 3 y=− 5 1 ​

x 2 +3. The difference is 𝑥 2 + 3 − ( − 1 5 𝑥 2 + 3 )

6 5 𝑥 2 x 2 +3−(− 5 1 ​

x 2 +3)= 5 6 ​

x 2 . Integrate from 0 to 2: ∫ 0 2 6 5 𝑥 2 𝑑 𝑥

6 5 ⋅ 8 3

48 15

16 5 ∫ 0 2 ​

5 6 ​

x 2 dx= 5 6 ​

⋅ 3 8 ​

= 15 48 ​

= 5 16 ​

. That is option ⑤.

Given the options, this is the most plausible intended problem. The problem statement likely had a formatting error, and the correct curves are 𝑦

𝑥 2 + 3 y=x 2 +3 and 𝑦

− 1 5 𝑥 2 + 3 y=− 5 1 ​

x 2 +3. The area is 16 5 5 16 ​

.

\boxed{⑤}

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