0, t°α, β5M
j: J lim x→−∞(p x2+ 3x + 2 − αx − β

 } (B) ‚25tæDj–(I) 1. t°”Ì lim

x→0

x²sin_x¹ sin x 5M

j: Jx 6= 0, †| sin¹_x| ≤ 1, †

x→0lim

x²sin¹_x

sin x ≤ | x² sin x| Ä lim

x→0

x² sin x = lim

x→

x

sin x x

= 0

1 = 0, ]limx→|_{sin x}^x2 | = 0
âHUìÜø lim

x→0|x²sin_x¹ sin x | = 0, ])lim

x→

x²sin¹_x sin x = 0

2. qα,βÑùb, cq lim

x→−∞(p

x²+ 3x + 2 − αx − β) = 0, t°α, β5M

j: J lim

x→−∞(p

x²+ 3x + 2 − αx − β) = 0, † lim

x→−∞

√

x²+ 3x + 2 − αx − β

x = 0,

¹ lim

x→−∞(−

r 1 + 3

x+ 2

x² − α −β

x) = 0, ]α = lim

x→−∞(−1 r

1 + 3 x+ 2

x² −β

x) = −1, 7

β = lim

x→−∞(p

x²+ 3x + 2 + x)

= lim

x→−∞

(x²+ 3x + 2) − x²

√x²+ 3x + 2 − x

= lim

x→−∞

2x + 3

√

x²+ 3x + 2 − x

= lim

x→−∞

3 + ²_x

−q

1 + ³_x+_x²2 − 1

= 3

−2 = −3 2

3. qf (x), g(x)Ñùª}ƒb/f (1) = 1,f⁰(1) = 0,g(0) = 0,g⁰(0) = 1, t°”

Ì lim

x→^π₂

f (sin x)g(cos x) x²−^π₂x 5M

j: Ih(x) = f (sin x)g(cos x), †h(^π₂) = f (1)g(0) = 0, /

h⁰(x) = f⁰(sin x) cos xg(cos x) + f (sin x)g⁰(cos x)(− sin x)

, ]h⁰(^π₂) = f⁰(1) · 1 · g(0) + f (1) · g⁰(0) · (−1) = 0 · 1 · 0 + 1 · 1 · (−1) = −1

Ĥ

x→lim^π₂

f (sin x)g(cos x) x²−^π₂x

= lim

x→^π₂

1

x·h(x) − h(^π₂) x − ^π₂

= 1

π 2

· h⁰(π 2) = −2

π

1

4. qƒbf (x)ù¼ûƒb, /Å—x²+ xf (x) + [f (x)]² = k, w2kÑøb
˛ øf⁰(a) = f⁰⁰(a) = 1, t°aDk5M

j: øx²+ xf (x) + [f (x)]²= køúx}, )

2x + f (x) + xf⁰(x) + 2f (x)f⁰(x) = 0, (1) yø (1) }, )

2 + 2f⁰(x) + xf⁰⁰(x) + 2[f⁰(x)]²+ 2f (x)f⁰⁰(x) = 0 (2) Jx = a}Hp (1) £ (2) , cÜ)

 f (a) + a = 0 2f (a) + a = −6

j)a = 6,f (a) = −6, ]k = a²+ af (a) + [f (a)]²= 36

5. t°Ó(y²= 2x,×õ(1, 1)|¡5õ

j: qP (x, y)ÑÓ(y²= 2x,øõ, õP DìõP0(1, 1)5×Ñd, † d² = (x − 1)²+ (y − 1)²

= (y²

2 − 1)²+ (y − 1)²

= y⁴

4 − y²+ 1 + y²− 2y + 1

= y⁴

4 − 2y + 2 If (y) = ^y₄⁴−2y, †f⁰(y) = y³−2 = (y−√³

2)(y²+√³ 2y+√³

4), ]f⁰(y)Ê(−∞,√³ 2)0 Š, Ê(√³

2, ∞)0£
Ĥ,f (y)Ê(−∞,√³

2]]Á, Ê[√³

2, ∞)]Ó, ]Êy =√³ 2

|üM, ¹(³

√ 4 2 ,√³

2)¹ÑF°

6. t„j˙x⁵− 5x + 1 = 0úõ;/cúõ;

j: qf (x) = x⁵− 5x + 1, †f⁰(x) = 5x⁴− 5 = 5(x − 1)(x + 1)(x²+ 1),

]f⁰(x)Ê(−∞, −1)£(1, ∞)Ï]Ó, Ê[−1, 1]Ï]Á
Älim x → −∞f (x) = −∞,f (−1) = 5 > 0,f (1) = −3 < 0, lim

x→∞f (x) = ∞, ]â¡;ìÜøj˙f (x) = 0Ê(−∞, −1),(−1, 1),(1, ∞)®

/ø;, Ĥ/úõ;

7. t°.ì }

Z 1

√x(1 +√ x)²dx

j: Iu = 1 +√

x, †x = (u − 1)², ]dx = 2(u − 1)du
Ĥ

Z 1

√x(1 +√ x)²dx

=

Z 2(u − 1) (u − 1)u²du

= Z 2

u²du = −2

u+ C = − 1 1 +√

x+ C 2

w2CÑøb

8. t°ƒbg(x) = 3 + Z x²

1

sec(t − 1)dtÊx = −15øŸV¡

j: g(−1) = 3 +R(−1)²

1 sec(t − 1)dt = 3 +R1

1 sec(t − 1)dt = 3 g⁰(x) = _dx^d g(x) =

d

dx(3 +Rx²

1 sec(t − 1)dt) = _dx^d2

Rx²

1 sec(t − 1)dt · ^dx_dx² = 2x sec(x²− 1)

Ĥg⁰(−1) = −2 sec 0 = −2
FJ, g(x)Êx = −1íøŸV¡Ñg(x) = 3 − 2(x + 3) = −2x − 3

9. t°ÀPÆ2WýíÌÅ

jø:(Dj¶.Éø.w…d¶ÇWt0) âÆíú˚4, Bb.^cqýW kyW
à¬Ç, ¦¬õ(x, 0)5ýÅÑ2√

1 − x², ?¹2 sin θ, w2θ[ó@íÆ- i
Ĥ, ýÅóúkxíÌÑ

1 1 − (−1)

Z 1

−1

2p

1 − x²dx = Z 1

−1

p1 − x²dx = π 2, ÄÑ|(ø_ì }ÑÀPÆÞ 5š; 7ýÅóúkθíÌÑ

1 π − 0

Z π 0

2 sin θdθ = 2 π

Z π 0

sin θdθ = −2

πcos θ|^π₀ = 4 π

10. ˛ø®Òíxê§Dw[Þ A£ª
DøšÑ0.3…í7$®Ò, %¬3}

 (š‰A0.25…, t½v®Òr¶xê, ,uÛbÖývÈ

j: qV Ñ®Òíñ ,AÑw[Þ , Yæ<ø^dV_dt = kA, w2kÑøb
q7$®Òš

Ñr, †A = 4πr²,V = ⁴₃πr³, ]^dV_dt = 4πr^{2 dr}_dt
Ĥ,^dr_dt = k, j)r = kt+C, w2CÑøb
çt = 0v,r = 0.3, ])C = 0.3
7t = 3v,r = 0.25, ] )k = −¹₃× 0.05
cqt} (,r = 0, ¹(−¹₃× 0.05)t + 0.3 = 0, †t = 18, ?

¹r¶xêuÛ18} 

3