2015年4月CAP微积分、线性代数考试试题及参考答案-发布版

(0) −
1 6
1 + x f (0) +
1 (f (0) x2
− 1) + ◦(x) →
1 3,
7
kf (0) = 1, f (0) = 0, f (0) = 1. ){
3 3 :^arcsin x = x + 1 6 x + ◦(x ) 1 3 Ñxf (x) = x + 2 x + ◦(x3 ),u´f (x) +
= 1, F (x) = ±1,dëY5kF (x) =
1½öF (x) = −1. ) { n:df (x)3[a, b]þ š0,Œ x4 ¦ f (x)3[a, b]þ ð ½ ð K,Ä K • 3x1 = x2 ¦ • 3x3 =
f (x1 )f (x2 ) < 0,u ´ • 3x0 ∈ (x1 , x2 )¦
g (x) h(x) Q g (x) ,h (x) = g (x+T ) h(x+T ) ,g (x)h(x + T )
= h(x)g (x +
5 T ).g (x)|g (x + T ).h(x)|h(x + T ),•Ä'ux T ) = kh(x),g (x + 2T ) = k 2 g (x),k 2 gê´ƒÓ ,kg (x + T ) = kg (x), h(x +
= L.
= L.
f (x), g (x)3[a, b]þ ë Y,… é u ? ¿
x ∈ [a, b], f (x) = 0, g 2 (x) = f 2 (x).
y²:é¤k x ∈ [a, b]½ög (x) = −f (x)½ög (x) = f (x).
xf (x)−arcsin x x3
1 2 x + ◦(x2 ), f (0) = 1, f (0) = 0, f (0) = 1. 1+ 2
5. f (x) = pi (x + kT )i
p(x) q (x) , f (x
+ T ) = f (x),Ù¥p(x) = pi xi pi
pi xi , q (x) =
qi xi ´ngõ‘ª z‡‘ )(kT )i Xê• = 0,Ï
:f (x + T ) = f (x),d Û
½ n,f (x)k à • ‡ " :,f (x) =
g (x) h(x) , f
(x) =
g (x)h (x)−g (x)h(x) ,g (x)h h2 (x)
(x)−g (x)h(x)kÕ‡":,g (x)h (x)−g (x)h(x) = 0,f (x) =
3n (n2 −3n+2)h3
x x x 1
<
4 n
→ 0.
1 ax +bx +cx ) 3
(3)limx→0+ ( a +b3 +c ) x = elimx→0+ x ln( √ ln a+ln b+ln c 3 e = 3 abc; ){ :|^þŠØ ª,|^(1 + x) x
1 (4)limx→0 ( x − 1 limx→0 x+2 1 ex −1 )
1 (4)limx→0 ( x −
1 ex −1 ).
2.(15©)y²Stolz½n:
kê yn ÷v:
(a)yn+1 > yn (n = 1, 2, · · · , ); (b)limn→+∞ yn = +∞; (c)limn→+∞ Klimn→+∞ 3.(15©)
xn+1 −xn yn+1 −yn xn yn
f (x0 ) = 0,g ñ.e ¡ ‡
g (x3 ) = f (x3 ), g (x4 ) = f (x4 ),Kg (x3 )g (x4 ) = −f (x3 )f (x4 ) < 0,u ´ •
4
3
‡È©)‰
3x5 ∈ (x3 , x4 )¦ g (x5 ) = 0,¤±f 2 (x5 ) = g 2 (x5 ) = 0, f (x5 ) = 0,gñ. ) { o:k yg (x), f (x)ð ˜ ˜K, (Ø. 4.
1
= elimx→0+
ax ln a+bx ln b+cx ln c ax +bx +cx
=
4•. = limx→0
ex −1 xex +ex −1)
= limx→0
ex −x−1 x(ex −1)
= limx→0
ex xex +2ex
=
=
1 2. ex −x−1 x(ex −1)
1 2 x +◦(x3 ) 2 x(x+◦(x2 )) 1 +◦(x) 2 1+◦(x)
3. e ( Ø Ø ¤ á,Œ f (a) = g (a) kS = {x ∈ (a, b]|f (x) = −g (x)} = ∅,K • 3c = inf S . ea < c ≤ b,? ‰x ∈ [a, c), f (x) = g (x). • Ä ë Y ¼ êF (x) = f (x) − g (x),kF (x) = 0, x ∈ [a, c) F (c) = f (c) − g (c) = 2f (c) = 0,†ëY5gñ. ec = a,•3 , 0 < 5gñ. ){ :•ÄëY¼êF (x) =
2015c‡È©! ‚5“ê•Á
1
1.(40©)¦e 4• (1)limn→+∞ ( 1 2 + (2)limn→+∞ (3)limx→0+ (
n2 an (a 3 1 2·3
‡È©
+ ··· +
1 n(n+1) );
> 1); ) x ,Ù¥a > 0, b > 0, c > 0´~ê;
1
ax +bx +cx
4 2 i=1 ai = c2 1 ≤ 1.
Ý
4 2 i=1 bi
T ¦ A = ST . =
4 2 i=1 ci , 4 i=1 ai bi
5.(15©) ai , bi , ci (i = 1, 2, 3, 4)•¢ê,… 0=
4 i=1 bi ci
1=
=
=
4 i=1 ci ai .
y²:a2 1
x2 3 x3 = 0 + x3 x 0 4 x5 0 2. 2 −1 Dn = 2 0 ··· 0 −1 2 −1 ··· 0 0 2 ··· 0 ··· ··· ··· ··· 0 0 0 ··· −1
){ :^TaylorÐmlimx→0 2.
= limx→0
= limx→0
=1 2.
∀ > 0, ∃N > 0, n > N k
xn+1 −xn yn+1 −yn
− L < . dyn+1 − yn > 0= k(L − (L − n> 2 ,=
xn+1 −xN yn+1 −yN
)(yn+1 − yn ) < xn+1 − xn < (L + )(yn+1 − yn ),én = N, N + 1, · · · ƒ \ )(yn+1 − yN ) < xn+1 − xN < (L + )(yn+1 − yN ), u ´
g (x) h(x) ,Ù¥g (x), h(x)
gê©O´m, n,Uìm, nŒ ?Øf (x)3xª ¼êŠ•~þ f •3,g
4••3,
Uì±Ï¼ê
5Ÿok¦
g (nT ) = k, h (nT ) = k ,F (x) = g (x) − kf (x)‰•õ‘ªkÕ
‡Š,kF (x) = 0, g (x) = kf (x). ){Ê:“êõ‘ª{,f (x) =
¼ ê,T ´ ˜ ‡ ¢ê. Ï d é ? ¿ qi ê ®•gñ. ){ 0gñ. ){n:éf (x) = uáž ñ. ){o:f (x) =
g (x) g (0) h(x) , h(0)
êk k0 = p(x + kT )q (x) − p(x)q (x + kT ) = qi (x + kx)i , Ï " õ ‘ ª ¼ ê
= k ,k = 1,†õ‘ª5Ÿgñ.
4
1.
‚5“ê)‰
éO2 Ý ?11C† 1 1 1 1 1 1 1 1 1 1 1 1 3 2 1 1 −3 c 0 −1 −2 −2 −6 c − 3 A= −→ 0 1 2 2 6 3 0 1 2 2 6 3 5 4 3 3 −1 d 0 −1 −2 −2 −6 d − 5 1 0 −1 −1 −5 −2 0 0 0 0 0 c −→ Ïdc = 0, d = 2žk).k)ž)• 0 1 2 2 6 3 0 0 0 0 0 d−2 x1 −2 1 1 5
0−
(kT )i q
p0 qi
(kT )i
=
(pi q0 − p0 qi f (x) =
p(x) q (x)
(pi q0 − p0 qi )xi kᇊ, •kz‡pi q0 − p0 qi = 0. duf (x)½Â =