2018 センター試験速報 数2b - toshin.com · - 1 -...
TRANSCRIPT
1
8 B
1
1
1 1
1
1 1 2
= 180
1=180
144=180144 = 4
5
2312
=2312180= 345
2 sin+ 5 - 2 cos+ 30= 1 2 x =+
5 = x -
5
2 sin x - 2 cos {x - 5 + 30 } = 1
2 sin x - 2 cosx - 6 = 1
2 cosx - 6 = 2cos x cos 6 + sin x sin 6 = 2 32 cos x + 12 sin x = 3 cos x + sin x
3601
1 1
180
1
2
2 sin x - ( 3 cos x + sin x) = 1
sin x - 3 cos x = 1
sin x - 3 cos x = 2sin x 12 - cos x 32 = 2sin x cos 3 - cos x sin 3 = 2 sinx - 3
sinx - 3 = 12
2
+5+
5 +
5
7
10 x 6
5
7
10-
3 x -
3 6
5-
3
1130
x -3 13
15
x
x -3
=56x = 7
6
+5
=76
= 76
-5
=2930
y
x
y =
1
1- 1
- 1
O
12
12
1130
1315
3
2
x log3 x xc 3 x 0c 03
log3 xlog3 x log3 xc 3
X 0 plog3 Xp = p log3 X
X 0Y 0log3
XY
= log3 X - log3 Y
(log3 x)(log3 x) 3 log3xc
(log3 x)2 3 (log3 x - log3 c)
t = log3 x
t2 3 (t - log3 c)
t 2 - 3t + 3 log3 c 0
c = 3 9 = 913 = (32)
13 = 3
23
t2 - 3t + 3 log3 323 0
t2 - 3t + 3 23
log3 3 0
log3 3 = 1
t2 - 3t + 2 0(t - 1) (t - 2) 0
t 1t 2
x ( 0)
log3 x 12 log3 x
3 log3 x 3132 3 log3 x
x 0
0 x 3x 9
4
x x 0
xtt = log3 x
t 2
x 0 c
t
f (t) = t2 - 3t + 3 log3 c
f (t) 0 c
f (t) =t - 32 2 - 94 + 3 log3 c f (t) f 32 = 3 log3 c - 94 c 3 log3 c -
94 0log3 c
34
3 log3 c 334 = (33)
14c 4 27
x
t
t = log3 x
1
31O
5
2
1
f (x) = px2 + qx + r
C A (1,1) ly = 2x - 1
2
f'(x) = 2px + qf'(1) = 2
2p + q = 2
q = - 2p + 2
C A (1,1)f (1) = 1
p + q + r = 1
p + (- 2p + 2) + r = 1r = p - 1
f (x) = px2 + (- 2p + 2) x + p - 1
S 1
S =v
1{px2 + (- 2p + 2)x + p - 1 - (2x - 1)} dx
=v
1(px2 - 2px + p) dx
=p3
x3 - px2 + pxv
1
= p3 v3 - pv2 + pv- p3 - p + p =
p3
v3 - pv2 + pv -p3
=p3
(v3 - 3v2 + 3v - 1)
T 1
2v - 1 0
T =1 + (2v - 1)
2(v - 1)
= v (v - 1)
= v 2 - v
U = S - T
=p3
(v3 - 3v2 + 3v - 1) - (v2 - v)
1
1 v
l
O
y
y = f(x)
x
1
1 vx
1
2v - 1
2
6
=p3
(v3 - 3v2 + 3v - 1) - v2 + v = g (v)
g'(v) =p3
(3v2 - 6v + 3) - 2v + 1 = p (v2 - 2v + 1) - 2v + 1
g (v) v = 2g'(2) = 0
p (22 - 22 + 1) - 22 + 1 = 0p = 3
g (v) = (v3 - 3v2 + 3v - 1) - v2 + v
= v3 - 4v2 + 4v - 1
g'(v) = 3v2 - 8v + 4
= (3v - 2) (v - 2)
g (v)
v 23 2
g'(v) + 0 - 0 +
g ( v)
v = 2
p = 3
U = 0g (v) = 0 v
v3 - 4v2 + 4v - 1 = 0
(v - 1) (v2 - 3v + 1) = 0
v = 13 5
2 v0 1
v0 =3 + 5
2
g (v) (v 1)
5 1 v0 = 3 + 52 3 + 12 = 2
U = g (v) 1 v v0 3
p = 3v 1 U = g (v)
g (2) = 23 - 422 + 42 - 1 = - 1
v (1) 2 v0 g'(v) - 0 + +
g ( v) (0) 0
v
g'(v)
12
+
23
-
7
2
F (x) f (x)
F'(x) = f (x) 7
W
W =t
1{0 - f (x)} dx
= -t
1f (x) dx
= - F (x)t
1
= - (F (t) - F (1))
= - F (t) + F (1) 4
W
h
h = (t2 + 1)2 - (t2 - 1)2
= (t4 + 2t2 + 1) - (t4 - 2t2 + 1)
= 4t2
= 2t t 1
W =12(2t2 - 2)2t
= 2t3 - 2t
- F (t) + F (1) = 2t3 - 2t
tF'(t) = f (t)
- f (t) = 6t2 - 2f (t) = - 6t 2 + 2
apqp q
p
q(x - a)2dx = 1
3(x - a)3
p
q
=13
(p - a)3 -13
(q - a)3
y = (x - a)2 x - a
x
W
t1
y =f(x)
t2 - 1
t2 + 1
2t2 - 2
h
y = x2 y = (x - a)2y
xq -a
-a
p -a pqaO
8
p
q(x - a)2dx
p - a
q - a x2dx
p - a
q - a x2dx = 1
3x3
p - a
q - a
= 13
(p - a)3 - 13
(q - a)3
S =v
1p (x2 - 2x + 1) dx
=v
1p (x - 1)2dx
=p3
(x - 1)3v
1
=p3
(v - 1)3
g (v) = p3
(v - 1)3 - v2 + v
g'(v) = p (v - 1)2 - 2v + 1 g'(2) = p12 - 22 + 1 = 0 p = 3
g (v) = (v - 1)3 - v (v - 1) = (v - 1) {(v - 1)2 - v } = (v - 1) (v2 - 3v + 1)
g (v) = 0 v = 1 v2 - 3v + 1 = 0
y = g (v) v
v = 1 3 52
1 v v0 g (v) 0
y
v1
O 3 + 523 - 5
2
y =g(v)
9
3
{an} d
a4 = a1 + 3d = 30
a1 + a2 + a3 + a4 + a5 + a6 + a7 + a8
= a1 + (a1 + d) + (a1 + 2d) ++ (a1 + 7d)
= 8a1 + (1 + 2 ++ 7) d
= 8a1 + 28d
8a1 + 28d = 288
a1 = - 6d = 12
{an}
- 6 12
an = - 6 + 12 (n - 1) = 12n - 18
Sn =a1 + an
2n
=- 6 + (12n - 18)
2n
= (6n - 12)n
= 6n2 - 12n
{bn} r
b2 = b1 r = 36
b1 + b2 + b3 = b1 + b1 r + b1 r2
= b1 (1 + r + r2) = 156
r
1 + r + r2=
36156
=3
13
3 (1 + r + r2) = 13r
3r2 - 10r + 3 = 0
(3r - 1) (r - 3) = 0
r 1r = 3
10
b1 =1336 = 12
{bn}
12 3
bn = 123n - 1
Tn =n
k = 1123k - 1 = 12 3
n - 13 - 1
= 6 (3n - 1)
cn =n
k = 1(n - k + 1) (ak - bk)
cn + 1 =n + 1
k = 1{(n + 1) - k + 1} (ak - bk)
dn = cn + 1 - cn
=n + 1
k = 1{(n + 1) - k + 1} (ak - bk) -
n
k = 1(n - k + 1) (ak - bk)
= {(n + 1) - (n + 1) + 1} (an + 1 - bn + 1)
+n
k = 1{(n + 1) - k + 1} (ak - bk) -
n
k = 1(n - k + 1) (ak - bk)
= an + 1 - bn + 1 +n
k = 1{(n + 1) - k + 1 - (n - k + 1)} (ak - bk)
= an + 1 - bn + 1 +n
k = 1(ak - bk)
dn =n
k = 1ak + an + 1 - (
n
k = 1bk + bn + 1)
=n + 1
k = 1ak -
n + 1
k = 1bk
= Sn + 1 - Tn + 1 5
dn = 6 (n + 1)2 - 12 (n + 1) - 6(3n + 1 - 1)
= 6n2 - 63n + 1
= 6n2 - 23n + 2
cn + 1 - cn = dn
11
cn + 1 - cn = 6n2 - 23n + 2
c1 = a1 - b1 = - 6 - 12 = - 18
n 2
cn = c1 +n - 1
k = 1(6k2 - 23k + 2)
= - 18 + 6 (n - 1) n {2 (n - 1) + 1}6
- 2 33 (3n - 1 - 1)
3 - 1
= 2n3 - 3n2 + n + 9 - 3n + 2
n = 1
n 1
cn = 2n3 - 3n2 + n + 9 - 3n + 2
12
4
AB = FB - FA
= q - p 2
| AB |2 = | q - p |2
= ( q - p )( q - p )
= | q |2 - 2pq + | p |2
= | p |2 - 2pq + | q |2
D AB 13
FD =3 FA + 1 FB
3 + 1
=34
p +14
q
FD = srFE = tp
sr =34
p +14
q4sr = 3p + q
q = - 3p + 4sr
E BC a(1 - a)
FE = (1 - a) FB + a FCFE = (1 - a) q + ar
FE = tp
tp = (1 - a) q + ar(1 - a) q = tp - ar
0 a 11 - a0
q =t
1 - ap -
a
1 - ar
DE ABBCFABC
FA = pFC = r0
- 3 =
t1 - a
s =- a
4 (1 - a)
4s =- a
1 - at = - 3 (1 - a)
BAD
C
E
F
1 - a
a
qp
r
13
| p | = 1
| AB |2 = 1 - 2pq + | q |2
BE = FE - FB = tp - q = - 3 (1 - a) p - q
| BE |2 = {- 3 (1 - a)p - q}{- 3 (1 - a) p - q}
= 9 (1 - a)2 | p |2 + 6 (1 - a)pq + | q |2
| p | = 1
| BE |2 = 9 (1 - a)2 + 6 (1 - a)pq + | q |2
| AB | = | BE || AB |2 = | BE |2
1 - 2pq + | q |2 = 9 (1 - a)2 + 6 (1 - a)pq + | q |2
{6 (a - 1) - 2}pq = 9a2 - 18a + 8
2 (3a - 4) pq = (3a - 4) (3a - 2)
0 a 13a - 40
pq =3a - 2
2