# 2013年高中函数竞赛专题详细答案及其函数第二课练习题

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1! ? g ? êf (x) = ax2 + bx + c(a = 0),? ?(a < b),… é ? ? ?êx,? kf (x) = a+b+c ax2 + bx + c ≥ 0,?“ê? ? ? b?a ):w,a > 0,…b2 ? 4ac ≤ 0,= b2 c≥ 4a ¤± b2 a + b + 2 a+b+c 4a = (2a + b) ≥ 4(b ? a) · 3a = 3 ≥ b?a b?a 4a(b ? a) 4a(b ? a) ){2:w,b ? a > 0, a+b+c ≥x>0 b?a K (x + 1)a ? (x ? 1)b + c ≥ 0 (x + 1) = (x ? 1)2 =? x = 3 ){3:5?? ? a+b+c 3(b ? a) + 4a ? 2b + c 4a ? 2b + c f (?2) = =3+ =3+ ≥3 b?a b?a b?a b?a 6f (0) ? 3f (0) + f (0) b?a 2!??§ C?:? ? ?7 ù

log5 (3x + 4x ) = log4 (5x ? 3x ) )8 ):y = log5 (3x + 4x ) = log4 (5x ? 3x ) Kk 5y = 3x + 4x , 4y = 5x ? 3x ,?\k 5y + 4y = 5x + 4x 5?f (t) = 5t + 4t ?üN4O?ê,df (y ) = f (x) =? x = y ,K 5x = 3x + 4x = 3 4 g (x) = ( )x + ( )x = 1 5 5 1

dug (x) üN4~,…g (2) = 1, C?:?

x = 2??§ ??). 3x + 4x + 5x = 6x

‰Y?x = 3 ~K3:??§ √ √ (3x ? 1)( 9x2 ? 6x + 5 + 1) + (2x ? 3)( 4x2 ? 12x + 13 + 1) = 0 ?ê) ):w,C/k (3x ? 1)[ (3x ? 1)2 + 4 + 1] + (2x ? 3)[ (2x ? 3)2 + 4 + 1] = 0 √ E?êg (x) = t( t2 + 4 + 1),? g (t)???ê,…?üN4O ?ê. ?§? g (3x ? 1) + g (2x ? 3) = 0 =? g (3x ? 1) = ?g (2x ? 3) = g (3 ? 2x) ¤± 4 3x ? 1 = 3 ? 2x =? x = , 5 C?:)?§ 2x + x2 + 1 + w, éêk lg (2x + = lg (2x + ?? ){2: 5? x2 + 1 + ¤± 2x + x2 + 1 + (x2 + 1)2 + 1 ≥ 2x + √ 4x2 + 1 √ 4x2 + 1 > 0 √ √ 4x2 + 1 = 10(x?1)
2

(x2 + 1)2 + 1

4x2 + 1) ? lg [x2 + 1 +

(x2 + 1)2 + 1] = (x ? 1)2

4x2 + 1) + 2x = lg [x2 + 1 +

(x2 + 1)2 + 1] + x2 + 1

(x2 + 1)2 + 1
2

≤1

10(x?1) ≥ 1 ?kx = 1÷v. C?:?? ? )8 √ √ √ 48 2 x2 ? 7 + 3 x2 ? 12 + 4 x2 ? 16 > , (?∞, ?4] x x3 + sin x ? 2a = 0, 4y 3 + 2 1 sin 2y + a = 0 2 (4, +∞)

,?x + 2y 2!e (3x + y )5 + x5 + 4x + y = 0, cos (4x + y ) =? 3!ef (x) = x3 ? 3x2 + 6x ? 6, f (a) = 1, f (b) = ?5,?a + b = 2 ~K4: x ∈ [?1, 1],…?ê 3 + sin x ln g (x) =
2

x2 + 1 e2 , cos (2x) + 2 x+

????M,? ??N ,?êf (x)÷v:f (0) = 0,…?1 ≤ x ≤ 1 k f (x) = ?f (M ) + f (N ) ):w, 1 + 2 cos2 x + sin2 x ln (x + g (x) = 1 + 2 cos2 x w,
3

4 ? 3x + ?‰?.

16 ? 24x + 9x2 ? x3 +

3

4 ? 3x ?

16 ? 24x + 9x2 ? x3

x2 + 1)

= 1 + u(x)

√ sin2 x ln (x + x2 + 1) u(x) = cos 2x + 2 √ √ 16 ? 24x + 9x2 ? x3

???ê. ?g·?2? y= |^ú? (a + b)3 = a3 + b3 + 3ab(a + b) K √ √ y 3 = 4?3x+ 16 ? 24x + 9x2 ? x3 +4?3x? 16 ? 24x + 9x2 ? x3 +3 3 (4 ? 3x)2 ? (16 ? 24x + 9x2 ? x3 )y = √ 3 y 3 = 8 ? 6x + 3 x3 y =? y 3 ? 3xy + 6x ? 8 = 0 (y ? 2)(y 2 + 2y + 4 ? 3x) = 0 5? y 2 + 2y + 4 ? 3x = (y + 1)2 + 3(1 ? x) > 0 ¤± f (x) = 2 =? f (x) = 2x + C du f (0) = 0, =? f (x) = 2x 3
3

4 ? 3x +

16 ? 24x + 9x2 ? x3 +

3

4 ? 3x ?

¤± f (M ) + f (N ) = 2(M + N ) du u(x)max + u(x)min = 0 =? [g (x) ? 1]max + [g (x) ? 1]min = 0 =? M + N = 2 ¤± f (M ) + f (N ) = 4 √ 5

C?:ea5 + a + 1 = 0, e

b + b + 1 = 0, =? a + b =? ? ?tan x + log (3 tan x + 6) = 2 3 ?tan y + 3tan y?1 = 4

?tan x + tan y = C?2:e 2x + 2x = 5, 2y + 2 log2 (y ? 1) = 5 =? x + y =? ~ K5? a, b, x, y ? ? ? ?ê,…f (x) = ax + b÷ v|f (x)| ≤ 1, x ∈ [0, 1], abmax = A, ÷ vx ? √ 1√ y = 2 x ? y ,?x ?‰?. A ):w,a = f (1) ? f (0), b = f (0),K 1 1 ab = f (0)[f (1) ? f (0)] ≤ (f (1))2 ≤ 4 4 ¤± √ √ x?4 y =2 x?y √ Kk x = a2 + b 2 …÷v a2 + b2 ? 4a = 2b ?→ (a ? 2)2 + (b ? 1)2 = 5(a, b ≥ 0) ?K=z:e(a ? 2)2 + (b ? 1)2 = 5(a, b ≥ 0),?a2 + b2 = 4a + 2b = 2(2a + b) ?? d…?? ?k [(a ? 2)2 + (b ? 1)2 ][4 + 1] ≥ [2a ? 4 + b ? 1]2 =? ?5 ≤ 2a + b ? 5 ≤ 5 = 0 ≤ 2a + b ≤ 10 √

y = a,

x ? y = b(a, b ≥ 0)

4

K 0 ≤ 4a + 2b ≤ 20 é??)?U ù ? ‰ Y,? ? ? 3 u v k ? ?a ≥ 0, b ≥ 0, (‰{|^?”{' {

ü.?N???. {üy?:d

√ 4 y+

4(x ? y ) = x

Kk…?? ? √ 4y + (4x ? 4y ) ≤ 4 y + = 4x ? 4y ≤ 5[4y + (4x ? 4y )]

√ √ 4x ≤ x ≤ 20x [4, 20] 1 1 ê.?[f (x)? ]+[f (?x)? ] 2 2

) x ∈ {0}

ax ~K6: f (x) = (a > 0, a = 1), [m]L???L?êm ?? 1 + ax ??

5?f (x) + f (?x) = 1,…|^x ? 1 < [x] ≤ x,K ?ê???{?1, 0} 1 ~K7: ?êf (x) = loga (ax2 ? x + )3?m[1, 2]?? ,??êa ‰?. 2 ):?k 1 x? 2 =? 1 +1 a> 2 x 2x2 x 1 ¤±a > . 2 1 < a < 1?,f (x)?~?ê,Kk 2 5 f (2) > 0 =? a < 8 a > 1,Kf (x)?üN4O.K7k 3 f (1) > 0 =? a > 2 1 5 3 ¤± < a < ?a > 2 8 2 ~ K8: ??? )5?t + t2 ? 3t ? 8 = 0,=t2 ? 2t ? 8 = 0 ?? 5 ? ? êf (x) ? ? ? ?(t, t2 ? 3t ? 8),? ?(2t, t2 + 3t + 6),K ? êy = f (x + 1) + 1

~K9:e?êy = f ):-g (x) =

2x 3x2 + 1

1 ????[ , a],?a ? 4 ????R,…5? g( 1 2x )= = g (x)(x = 0) 3x 1 + 3 x2 1 3× 1 4 4 1 ∈ [ , a] 3 4

2x ,Kg (x) 3x2 + 1

d 1 1 ∈ [ , a] =? 4 4 =

4 Ka ≥ , 3 4 4 1 1 a > ??K?,??e?3x0 > ?f [g (x0 )]k??,K ∈ (0, )??? 3 3 3x0 4 1 4 u ê?U?f (x) k??,¤±a = 4 3 ~K10:? x3 1 (x > 0) +√ f (x) = √ 4 1 + 3 x4 12 ? ?. √ √ √ 4 3x4 + 1 ≥ 2 3x2 , =? 3x4 + 1 ≥ 12x ):5?

f (g (

1 ))k??, 3x0

¤±

1 1 x3 x3 x4 + 1 √ √ √ √ √ f (x) = ≥ + 4 + = 1 + 3 x4 1 + 3 x4 3x4 + 1 3x4 + 1 12 x 1 f (x) ≥ 3 √ 2 3x4 + 1 + √ 3x4 + 1 √ 2 2 ≥ 3

¤±

~K11:???3x ∈ R? ?

?êf (x)? f (f (x)) = 1 + x2 + x4 ? x3 ? x5

,e?3,?‰???~f,e??3,?‰?y?. ): f (1) = a,Kkf (f (1)) = 1 + 1 + 1 ? 1 ? 1 = 1 ¤±f (a) = 1 -x = aKk f (f (a)) = 1 + a2 + a4 ? a3 ? a5 = a = 1 + a2 + a4 ? a3 ? a5 ¤± (a ? 1)(a4 + a2 + 1) = 0

6

¤±a = 1,=f (1) = 1. 2k f (f (x)) = 1 + x2 + x4 ? x3 ? x5 Kk f (x)f (f (x)) = 2x + 4x3 ? 3x2 ? 5x4 -x = 1 =? [f (1)]2 = ?2.w,?÷v. ~K12???êf (x) = cos x + ):w, f (x) = cos x + = = ???f ?p3? √ √ 2 + cos2 x ? 4 2 cos x + 4 sin x + 9 ? 2 √ √ √ cos2 x + 2 2 cos x + 2 + cos2 x ? 4 2 cos x + 4 sin x + 9 ? 2 √ √ √ ( 2 cos x + 1)2 + (sin x ? 0)2 + ( 2 cos x ? 2)2 + (sin x + 2)2 ? 2 ??.? √ √ cos2 x ? 4 2 cos x + 4 sin x + 9 ???.

√ A???é?w,??:P ( 2 cos x, sin x) :F1 (?1, 0), A(2, ?2) ?l?? x2 + y 2 = 1?, = 2 √ √ √ |P F1 | + |P A| = 2 2 ? |P F2 | + |P A| ≤ 2sqrt2 + |AF2 | = 2 2 + 5 √ √ √ 2

¤± √ 2 + 2 10 x = arcsin 9 smax = (|P A| + |P F1 |)max ? .

2=

5+

…=

1! ????R

?êf (x), g (x)?k??ê,?…?êf (x + 1), g ?1 (x + 2)

?”'u??y = xé

?,eg (5) = 2005,?f (6), 2! 8 ?M = {?2, 0, 1}, N = {1, 2, 3, 4, 5},N N f ?ê ±?? f : M ?→ N ? é?? x ∈ M ,? k x +

f (x) + xf (x)??ê,?ù 3! 3 ? ? ? ?m? 4! ?. C ?: ? êf (x) =

? I Xxoy ?,? ? êf (x) = a sin (ax) + cos (ax)(a > 0)3 ? ? ? √ ?”??êg (x) = a2 + 1 ?”¤?¤ ?4?” ?? √ ax2 + bx,? ÷ v e ^? êa,– k?? êb,?f (x)

f (x) =

???????

√ ax2 + bx + c(a < 0)

? ? ? ?D,e ¤ k :(s, f (t)), (s, t ∈ D)

¤??

?/??,?a ? 7

5!??a ∈ (0, 1),?(??êt C?:?(?

?‰?,?

? ?ax2 + ty 2 ≥ (ax + ty )2 é??x, y ?¤á.

?ê(a, b),? é???êx, y ?k (ax2 y 2 + b) + (a(x + y )2 + b) ≥ (ax2 + b)(ay 2 + b)

6! g?êf (x) = ax2 +bx+c(a = 0)÷vf (x?4) = f (2?x),…f (x) ≥ x, x ∈ (0, 2)?,kf (x) ≤ 2 (x + 1) ,…f (x)3R? ? ? ?0, ? ? ? m(m > 1),? ? 3t ∈ R,? ?x ∈ [1, m],? 4 kf (x + t) ≤ x 7! f (x) = ax2 + bx + c(a > b > c) ?”?kü:A(m1 , f (m1 )), B (m2 , f (m2 ),÷v

a2 + a[f (m1 ) + f (m2 )] + f (m1 )f (m2 ) = 0, f (1) = 0 ?? f (mi + k )(i = 1, 2)?– k??? ê ? êk . ü

g ? êf (x) = x2 + ax + b,e ? §f (x) = 0k ü ? ? ê? ?,… ü ? ? 3 ? 1 ê?m,??a, b÷v ^?,? ???3 êk ,k|f (k )| ≤ ¤á. 4 9! f (x) = log2 x, :M (x, y )3y = f (x)? ” ? \$ ? ?,:N (x ? 2, ny )3 ? êy = gn (x)

8! ? ?

?

”?\$?,(n ∈ N ),?y = gn (x) L?? (2):?8?A = {a = |'ux ?§g1 (x) = g2 (x ? 2 + a)k??, a ∈ R} √ √ 5 4 1 2 2 (3): Hn (x) = gn (x) ,?êF (x) = H1 (x)?g1 (x)(0 < a ≤ x ≤ b) ??? log2 , log2 .?a, b 2 b+2 a+2 √ √ 10!???êx, y ÷vx ? 3 x + 1 = 3 y + 2 ? y ,?P = x + y C?:??? êa,? ?3 êb, x ∈ [0, 1]?,?k √ éu¤? 1+x+ √ 1 ? x ≤ 2 ? bxa ?‰?.(aq8cém)

a,(?÷v??? ? ?? êb

x4 + 4x3 + 17x2 + 26x + 106 , |x| ≤ 1 ?? x2 + 2 x + 7 ax6 + bx5 + 4 + (4x2 ? ax2 ? bx)(x2 ? 1) C?:e?êy = ? x6 + 1 3x6 + 15x2 + 2 12! x ∈ (0, 1],??êy = 6 ?? 2x + 15x4 + 3 (x4 ? 6x2 + 1)(x ? x3 ) C?:??êy = ?? (x2 + 1)4 11!??êy =

??1,????6,?a, b

13! ??3R+ ? ?êf (x)?O?ê,…÷v?x > 0, f (x)f f (x) + 8

1 = 1,?f (1) x

C?: f

???R,…f (x) > 0üN4~,…f (x) > f 2 (x)f f (x) ?

1 , x>0k x2 1 x2 = f 2 (1)

?f (1)9?f (x) 14! f (x)???3R? ??ê,…÷v^?,é?? a, b ∈ R,kf (af (b)) = ab,?|f (2013)| 14! a? ? ? ? ê0 < a < 1,f ? ? ? 3[0, 1]? f ?f 1 22 ?ê,…÷v x+y 2 ? ê,÷ vf (0) = 0, f (1) = 1± 9 é ¤

k0 ≤ x ≤ y ≤ 1,k = (1 ? a)f (x) + af (y )

15! f (x)???3(0, ∞)? f (1 ? ?f (x) x2

1 1 1 ) + f (1 + 2 ) lg (x2 ? 2012) = f (1 + 2 ) lg (x2 ? 2013) + 2014 ? 2012 x ? 2013 x ? 2013

16!???êy = f (x), x, y ∈ N + ÷vé??a, b ∈ N + , a = b,?k af (a) + bf (b) > af (b) + bf (a) …é?? n ∈ N ,?kf (f (n)) = 3n (1):?f (1) + f (6) + f (28) (2):-an = f (3n ), n ∈ N + ,?y: n 1 1 1 1 < + + ··· + < 4n + 2 a1 a2 an 4 17!??3R? ?êf (x)÷vf (0) = 0,…é??x, y ∈ (?∞, ?1) (1, ∞),?k 1 1 x+y f( ) + f( ) = f( ) x y 1 + xy … x ∈ (?1, 0), f (x) > 0 ?y: f( 1 1 1 1 ) + f( ) + · · · + f( 2 ) > f( ) 19 29 n + 7n + 11 2

C?: f (x) ????R,…÷v: (1):???(?1, 1),… x > 0, ?1 < f (x) < 0, 9

(2):é???êm, n§kf (m + n) = ?f (0),… ?f (x) üN5. (2):e?êf (x)?3??êg (x),?y:

f (m) + f (n) 1 + f (m)f (n)

1 1 1 1 g( ) + g( ) + · · · + g( 2 ) > g( ) 5 11 n + 3n + 1 2 18! ?êf (x)?3R? ?ê,…÷v f (f (x) ? x2 + x) = f (x) ? x2 + x (1):ef (2) = 3,?f (1),qef (0) = a,?f (a) (2): k…=k???êx0 ,? f (x0 ) = x0 ,??êf (x) 19! f ?????R ?ê,????R,…÷v f (cot (g (x))) = sin 2g (x) + cos 2g (x), 0 < x < π ???34?m[?1, 1]? ?êg (x) = f (x)f (1 ? x) ? ?????. L??.

20! ??êf (x) ????x > 0,…÷ve?^? (1):XJx < y ,Kf (x) < f (y ) 2xy f (x) + f (y ) (2):é¤k ?êx, y ,kf ( )= ,y?:?3?êx0 ,? f (x0 ) < 0 x+y 2 21! N ?g,ê8,R??ê8,S ?÷v±eü?^? ?êf : N ?→ R (1)f (1) = 2 (2):f (n + 1) ≥ f (n) ≥ ??? n f (2n), n = 1, 2, · · · , n+1 g,êM ,? é??f ∈ S ,9??n ∈ N ,?kf (n) < M 8?:

10

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