高等学校学習指導要領解説　数学編

2 (1) a = ( 2, 2), b = (1, 2), c = (4, 4) c = l a + k b l, k (2) a = (3, 5) (1) (4, 4) = l( 2, 2) + k(1, 2), (4, 4) = ( 2l + k, 2l 2k) 2l + k = 4, 2l

ABCDEF a = AB, b = a b (1) AC (3) CD (2) AD (4) CE AF B C a A D b F E (1) AC = AB + BC = AB + AO = AB + ( AB + AF) = a + ( a + b) = 2 a + b (2) AD = 2 AO = 2( AB + AF) = 2( a + b) (3) CD = AF = b (4) CE

Part y mx + n mt + n m 1 mt n + n t m 2 t + mn 0 t m 0 n 18 y n n a 7 3 ; x α α 1 7α +t t 3 4α + 3t t x α x α y mx + n

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A (1) = 4 A( 1, 4) 1 A 4 () = tan A(0, 0) π A π

4 4.1 4.1.1 A = f() = f() = a f (a) = f() (a, f(a)) = f() (a, f(a)) f(a) = f 0 (a)( a) 4.1 (4, ) = f() = f () = 1 = f (4) = 1 4 4 (4, ) = 1 ( 4) 4 = 1 4 + 1 17 18 4 4.1 A (1) = 4 A( 1, 4) 1 A 4 () = tan

4 4 4 a b c d a b A c d A a da ad bce O E O n A n O ad bc a d n A n O 5 {a n } S n a k n a n + k S n a a n+ S n n S n n log x x {xy } x, y x + y 7 fx

4 4 5 4 I II III A B C, 5 7 I II A B,, 8, 9 I II A B O A,, Bb, b, Cc, c, c b c b b c c c OA BC P BC OP BC P AP BC n f n x xn e x! e n! n f n x f n x f n x f k x k 4 e > f n x dx k k! fx sin x cos x tan

高等学校学習指導要領

高等学校学習指導要領

1 12 ( )150 ( ( ) ) x M x 0 1 M 2 5x 2 + 4x + 3 x 2 1 M x M 2 1 M x (x + 1) 2 (1) x 2 + x + 1 M (2) 1 3 M (3) x 4 +

( )5 ( ( ) ) 4 6 7 9 M M 5 + 4 + M + M M + ( + ) () + + M () M () 4 + + M a b y = a + b a > () a b () y V a () V a b V n f() = n k= k k () < f() = log( ) t dt log () n+ (i) dt t (n + ) (ii) < t dt n+ n

t θ, τ, α, β S(, 0 P sin(θ P θ S x cos(θ SP = θ P (cos(θ, sin(θ sin(θ P t tan(θ θ 0 cos(θ tan(θ = sin(θ cos(θ ( 0t tan(θ

4 5 ( 5 3 9 4 0 5 ( 4 6 7 7 ( 0 8 3 9 ( 8 t θ, τ, α, β S(, 0 P sin(θ P θ S x cos(θ SP = θ P (cos(θ, sin(θ sin(θ P t tan(θ θ 0 cos(θ tan(θ = sin(θ cos(θ ( 0t tan(θ S θ > 0 θ < 0 ( P S(, 0 θ > 0 ( 60 θ

1 29 ( ) I II III A B (120 ) 2 5 I II III A B (120 ) 1, 6 8 I II A B (120 ) 1, 6, 7 I II A B (100 ) 1 OAB A B OA = 2 OA OB = 3 OB A B 2 :

9 ( ) 9 5 I II III A B (0 ) 5 I II III A B (0 ), 6 8 I II A B (0 ), 6, 7 I II A B (00 ) OAB A B OA = OA OB = OB A B : P OP AB Q OA = a OB = b () OP a b () OP OQ () a = 5 b = OP AB OAB PAB a f(x) = (log

高校生の就職への数学II

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8 ) ) [ ] [ ) 8 5 5 II III A B ),,, 5, 6 II III A B ) ),,, 7, 8 II III A B ) [ ]),,, 5, 7 II III A B ) [ ] ) ) 7, 8, 9 II A B 9 ) ) 5, 7, 9 II B 9 ) A, ) B 6, ) l ) P, ) l A C ) ) C l l ) π < θ < π sin

熊本県数学問題正解

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... 0 60 Q,, = QR PQ = = PR PQ = = QR PR = P 0 0 R 5 6 θ r xy r y y r, x r, y x θ x θ θ (sine) (cosine) (tangent) sin θ, cos θ, tan θ. θ sin θ = = 5 cos θ = = 4 5 tan θ = = 4 θ 5 4 sin θ = y r cos θ =

17 ( ) II III A B C(100 ) 1, 2, 6, 7 II A B (100 ) 2, 5, 6 II A B (80 ) 8 10 I II III A B C(80 ) 1 a 1 = 1 2 a n+1 = a n + 2n + 1 (n = 1,

17 ( ) 17 5 1 4 II III A B C(1 ) 1,, 6, 7 II A B (1 ), 5, 6 II A B (8 ) 8 1 I II III A B C(8 ) 1 a 1 1 a n+1 a n + n + 1 (n 1,,, ) {a n+1 n } (1) a 4 () a n OA OB AOB 6 OAB AB : 1 P OB Q OP AQ R (1) PQ

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No2 4 y =sinx (5) y = p sin(2x +3) (6) y = 1 tan(3x 2) (7) y =cos 2 (4x +5) (8) y = cos x 1+sinx 5 (1) y =sinx cos x 6 f(x) = sin(sin x) f 0 (π) (2) y

No1 1 (1) 2 f(x) =1+x + x 2 + + x n, g(x) = 1 (n +1)xn + nx n+1 (1 x) 2 x 6= 1 f 0 (x) =g(x) y = f(x)g(x) y 0 = f 0 (x)g(x)+f(x)g 0 (x) 3 (1) y = x2 x +1 x (2) y = 1 g(x) y0 = g0 (x) {g(x)} 2 (2) y = µ

18 ( ) I II III A B C(100 ) 1, 2, 3, 5 I II A B (100 ) 1, 2, 3 I II A B (80 ) 6 8 I II III A B C(80 ) 1 n (1 + x) n (1) n C 1 + n C

8 ( ) 8 5 4 I II III A B C( ),,, 5 I II A B ( ),, I II A B (8 ) 6 8 I II III A B C(8 ) n ( + x) n () n C + n C + + n C n = 7 n () 7 9 C : y = x x A(, 6) () A C () C P AP Q () () () 4 A(,, ) B(,, ) C(,,

*3 i 9 (1,) i (i,) (1,) 9 (i,) i i 2 1 ( 1, ) (1,) 18 2 i, 2 i i r 3r + 4i 1 i 1 i *4 1 i 9 i 1 1 i i 3 9 +

1 2 IT 1 *1 1 2 3 π i 1i 2i 3i πi i 2 1 *2 2 + 3 + 4i π ei 3 4 4 2 2 *1 *2 x 2 + 1 = x 2 + x + 1 = 2 3 1 2 2 2 2 *3 i 9 (1,) i (i,) (1,) 9 (i,) i i 2 1 ( 1, ) (1,) 18 2 i, 2 i 1 2 1 2 2 1 i r 3r + 4i 1

A(6, 13) B(1, 1) 65 y C 2 A(2, 1) B( 3, 2) C 66 x + 2y 1 = 0 2 A(1, 1) B(3, 0) P 67 3 A(3, 3) B(1, 2) C(4, 0) (1) ABC G (2) 3 A B C P 6

1 1 1.1 64 A6, 1) B1, 1) 65 C A, 1) B, ) C 66 + 1 = 0 A1, 1) B, 0) P 67 A, ) B1, ) C4, 0) 1) ABC G ) A B C P 64 A 1, 1) B, ) AB AB = 1) + 1) A 1, 1) 1 B, ) 1 65 66 65 C0, k) 66 1 p, p) 1 1 A B AB A 67

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arctan 1 arctan arctan arctan π = = ( ) π = 4 = π = π = π = =

arctan arctan arctan arctan 2 2000 π = 3 + 8 = 3.25 ( ) 2 8 650 π = 4 = 3.6049 9 550 π = 3 3 30 π = 3.622 264 π = 3.459 3 + 0 7 = 3.4085 < π < 3 + 7 = 3.4286 380 π = 3 + 77 250 = 3.46 5 3.45926 < π < 3.45927

1 (1) ( i ) 60 (ii) 75 (iii) 315 (2) π ( i ) (ii) π (iii) 7 12 π ( (3) r, AOB = θ 0 < θ < π ) OAB A 2 OB P ( AB ) < ( AP ) (4) 0 < θ < π 2 sin θ

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Chap10.dvi

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名古屋工業大の数学 2000 年 ~2015 年 大学入試数学動画解説サイト

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x () g(x) = f(t) dt f(x), F (x) 3x () g(x) g (x) f(x), F (x) (3) h(x) = x 3x tf(t) dt.9 = {(x, y) ; x, y, x + y } f(x, y) = xy( x y). h (x) f(x), F (x

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入試の軌跡

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1 1.1 (1) (1 + x) + (1 + y) = 0 () x + y = 0 (3) xy = x (4) x(y + 3) + y(y + 3) = 0 (5) (a + y ) = x ax a (6) x y 1 + y x 1 = 0 (7) cos x + sin x cos y = 0 (8) = tan y tan x (9) = (y 1) tan x (10) (1 +

DVIOUT-HYOU

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(1) θ a = 5(cm) θ c = 4(cm) b = 3(cm) (2) ABC A A BC AD 10cm BC B D C 99 (1) A B 10m O AOB 37 sin 37 = cos 37 = tan 37

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IMO 1 n, 21n n (x + 2x 1) + (x 2x 1) = A, x, (a) A = 2, (b) A = 1, (c) A = 2?, 3 a, b, c cos x a cos 2 x + b cos x + c = 0 cos 2x a

1 40 (1959 1999 ) (IMO) 41 (2000 ) WEB 1 1959 1 IMO 1 n, 21n + 4 13n + 3 2 (x + 2x 1) + (x 2x 1) = A, x, (a) A = 2, (b) A = 1, (c) A = 2?, 3 a, b, c cos x a cos 2 x + b cos x + c = 0 cos 2x a = 4, b =

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1 1 3 ABCD ABD AC BD E E BD 1 : 2 (1) AB = AD =, AB AD = (2) AE = AB + (3) A F AD AE 2 = AF = AB + AD AF AE = t AC = t AE AC FC = t = (4) ABD ABCD 1 1

ABCD ABD AC BD E E BD : () AB = AD =, AB AD = () AE = AB + () A F AD AE = AF = AB + AD AF AE = t AC = t AE AC FC = t = (4) ABD ABCD AB + AD AB + 7 9 AD AB + AD AB + 9 7 4 9 AD () AB sin π = AB = ABD AD

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1 26 ( ) ( ) 1 4 I II III A B C (120 ) ( ) 1, 5 7 I II III A B C (120 ) 1 (1) 0 x π 0 y π 3 sin x sin y = 3, 3 cos x + cos y = 1 (2) a b c a +

6 ( ) 6 5 ( ) 4 I II III A B C ( ) ( ), 5 7 I II III A B C ( ) () x π y π sin x sin y =, cos x + cos y = () b c + b + c = + b + = b c c () 4 5 6 n ( ) ( ) ( ) n ( ) n m n + m = 555 n OAB P k m n k PO +

さくらの個別指導 ( さくら教育研究所 ) A a 1 a 2 a 3 a n {a n } a 1 a n n n 1 n n 0 a n = 1 n 1 n n O n {a n } n a n α {a n } α {a

... A a a a 3 a n {a n } a a n n 3 n n n 0 a n = n n n O 3 4 5 6 n {a n } n a n α {a n } α {a n } α α {a n } a n n a n α a n = α n n 0 n = 0 3 4. ()..0.00 + (0.) n () 0. 0.0 0.00 ( 0.) n 0 0 c c c c c

さくらの個別指導 ( さくら教育研究所 ) A AB A B A B A AB AB AB B

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mugensho.dvi

1 1 f (t) lim t a f (t) = 0 f (t) t a 1.1 (1) lim(t 1) 2 = 0 t 1 (t 1) 2 t 1 (2) lim(t 1) 3 = 0 t 1 (t 1) 3 t 1 2 f (t), g(t) t a lim t a f (t) g(t) g(t) f (t) = o(g(t)) (t a) = 0 f (t) (t 1) 3 1.2 lim

ORIGINAL TEXT I II A B 1 4 13 21 27 44 54 64 84 98 113 126 138 146 165 175 181 188 198 213 225 234 244 261 268 273 2 281 I II A B 292 3 I II A B c 1 1 (1) x 2 + 4xy + 4y 2 x 2y 2 (2) 8x 2 + 16xy + 6y 2

() x + y + y + x dy dx = 0 () dy + xy = x dx y + x y ( 5) ( s55906) 0.7. (). 5 (). ( 6) ( s6590) 0.8 m n. 0.9 n n A. ( 6) ( s6590) f A (λ) = det(a λi)

0. A A = 4 IC () det A () A () x + y + z = x y z X Y Z = A x y z ( 5) ( s5590) 0. a + b + c b c () a a + b + c c a b a + b + c 0 a b c () a 0 c b b c 0 a c b a 0 0. A A = 7 5 4 5 0 ( 5) ( s5590) () A ()

lim lim lim lim 0 0 d lim 5. d 0 d d d d d d 0 0 lim lim 0 d

lim 5. 0 A B 5-5- A B lim 0 A B A 5. 5- 0 5-5- 0 0 lim lim 0 0 0 lim lim 0 0 d lim 5. d 0 d d d d d d 0 0 lim lim 0 d 0 0 5- 5-3 0 5-3 5-3b 5-3c lim lim d 0 0 5-3b 5-3c lim lim lim d 0 0 0 3 3 3 3 3 3

D xy D (x, y) z = f(x, y) f D (2 ) (x, y, z) f R z = 1 x 2 y 2 {(x, y); x 2 +y 2 1} x 2 +y 2 +z 2 = 1 1 z (x, y) R 2 z = x 2 y

5 5. 2 D xy D (x, y z = f(x, y f D (2 (x, y, z f R 2 5.. z = x 2 y 2 {(x, y; x 2 +y 2 } x 2 +y 2 +z 2 = z 5.2. (x, y R 2 z = x 2 y + 3 (2,,, (, 3,, 3 (,, 5.3 (. (3 ( (a, b, c A : (x, y, z P : (x, y, x

1990 IMO 1990/1/15 1:00-4:00 1 N N N 1, N 1 N 2, N 2 N 3 N 3 2 x x + 52 = 3 x x , A, B, C 3,, A B, C 2,,,, 7, A, B, C

0 9 (1990 1999 ) 10 (2000 ) 1900 1994 1995 1999 2 SAT ACT 1 1990 IMO 1990/1/15 1:00-4:00 1 N 1990 9 N N 1, N 1 N 2, N 2 N 3 N 3 2 x 2 + 25x + 52 = 3 x 2 + 25x + 80 3 2, 3 0 4 A, B, C 3,, A B, C 2,,,, 7,

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数学Ⅲ立体アプローチ.pdf

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