Tài liệu hạn chế xem trước, để xem đầy đủ mời bạn chọn Tải xuống
1
/ 55 trang
THÔNG TIN TÀI LIỆU
Thông tin cơ bản
Định dạng
Số trang
55
Dung lượng
3,81 MB
Nội dung
CHƯƠNG 1: CÁC BÀI TOÁN VỀ HÀM SỐ BÀI 1. PHƯƠNG PHÁP HÀM SỐ I. TÍNH ĐƠN ĐIỆU, CỰC TRỊ HÀM SỐ, GIÁ TRỊ LỚN NHẤT & NHỎ NHẤT CỦA HÀM SỐ 1.y=fxab⇔ ( ) x x a b∀ < ∈ ( ) ( ) f x f x< 2.y=fxab⇔ ( ) x x a b∀ < ∈ ( ) ( ) f x f x> 3.y=fxab⇔ƒ′x≥∀x∈abƒ′x= ∈ab 4.y=fxab⇔ƒ′x≤∀x∈abƒ′x= ∈ab 5.Cực trị hàm số: !"# ( ) k x x f x ′ = ⇔ $%& k x 6. '(#)*&+!,&-! • './y=ƒx)01#02ab3"# ( ) n x x a b∈ 45 [ ] ( ) ( ) ( ) ( ) ( ) { } 67 67 8 n x a b f x f x f x f a f b ∈ = [ ] ( ) ( ) ( ) ( ) ( ) { } 6 6 n x a b f x f x f x f a f b ∈ = • 9y=fx2ab3: [ ] ( ) ( ) [ ] ( ) ( ) 6 8 67 x a b x a b f x f a f x f b ∈ ∈ = = • 9y=fx2ab3: [ ] ( ) ( ) [ ] ( ) ( ) 6 8 67 x a b x a b f x f b f x f a ∈ ∈ = = b j j j x x x − ε + ε i i i x x x − ε + ε a x • !;& ( ) f x x= α + β #0< [ ] 8a b (#)*&(#,&(=> a; b ??@ AB9'@ C@ D6EFG?H9IJK9@ AB9'LMN9 GOL@ AB9'LMN9 1.9P-QRS#:ux=vx)!<!<- ( ) y u x= +* ( ) y v x= 2.9P-&QRS#:ux≥vx)! Q=<!RST+*Q= ( ) y u x= UVQW#0 <+*Q= ( ) y v x= 3.9P-&QRS#:ux≤vx)! Q=<!RST+*Q= ( ) y u x= UVQW%R*<+*Q= ( ) y v x= 4.9P-QRS#:ux=m)!<! <-RXy=m+* ( ) y u x= 5.G@Lux≥m>∀x∈?⇔ ( ) ? 6 x u x m ∈ ≥ 6.G@Lux≤m>∀x∈?⇔ ( ) ? 67 x u x m ∈ ≤ 7.G@Lux≥mPx∈?⇔ ( ) ? 67 x u x m ∈ ≥ 8.G@Lux≤mPx∈?⇔ ( ) ? 6 x u x m ∈ ≤ III. Các bài toán minh họa phương pháp hàm số Bài 1. Y<! ( ) Zf x mx mx= + − a.L:mQRS#:ƒx=Px∈283 b.L:m&QRS#:ƒx≤P>∀x∈28[3 c.L:m&QRS#:ƒx≥Px∈ [ ] 8Z− Giải: a.G$QRS#:ƒx=5 ( ) ( ) ( ) ( ) Z Z Z Z f x mx mx m x x g x m x x x = + − = ⇔ + = ⇔ = = = + + − \ƒx=Px∈283: [ ] ( ) [ ] ( ) 8 8 6 67 x x g x m g x ∈ ∈ ≤ ≤ Z ] m⇔ ≤ ≤ b.L∀x∈28[3: ( ) Z f x mx mx= + − ≤ ⇔ ( ) Zm x x+ ≤ ⇔ ( ) [ ] Z 8[ g x m x x x = ≥ ∀ ∈ + [ ] ( ) 8[ 6 x g x m ∈ ⇔ ≥ ^< ( ) ( ) Z g x x = + − .#028[30_⇔ [ ] ( ) ( ) 8[ 6 [ ] x g x g m ∈ = = ≥ c.L+*x∈ [ ] 8Z− : ( ) Z f x mx mx= + − ≥ ⇔ ( ) Zm x x+ ≥ \` ( ) [ ] Z 8Z g x x x x = ∈ − + ab(c.de_5 f9 x = :&QRS#:#V! Zm = ≥ 0+gP f9 ( ] 8Zx∈ :G@L ⇔ ( ) g x m≤ P ( ] 8Zx∈ ( ] ( ) 8Zx Min g x m ∈ ⇔ ≤ ^< ( ) ( ) Z g x x = + − . ( ] 8Z 0_ ( ] ( ) ( ) 8Z Z h x Min g x g m ∈ ⇔ = = ≤ α β b x a v(x) u(x) a b x y = m f9 [ ) 8x∈ − : x x+ < 0G@L ( ) g x m⇔ ≥ P [ ) 8x∈ − [ ) ( ) 8 Max g x m − ⇔ ≥ L ( ) ( ) ( ) [ ] Z 8 x g x x x x − + ′ = ≤ ∀ ∈ − + ^< ( ) g x 0 [ ) ( ) ( ) 8 ZMax g x g m − = − = − ≥ Kết luận:ƒx≥Px∈ [ ] 8Z− ( ] ) 8 Z 8 h m ⇔ ∈ −∞ − +∞ U Bài 2. L:m&QRS#:5 Z Z Z x mx x − − + − < P>∀x≥ Giải:G@L ( ) Z Z [ Z Z mx x x m x f x x x x x ⇔ < − + ∀ ≥ ⇔ < − + = ∀ ≥ L ( ) h h [ [ [ f x x x x x x x x − ′ = + − ≥ − = > ÷ _# ( ) f x d iYGL ( ) ( ) ( ) Z Z Z x f x m x f x f m m ≥ ⇔ > ∀ ≥ ⇔ = = > ⇔ > Bài 3. L:m&QRS#: ( ) [ x x m m m + + − + − > > x∀ ∈ ¡ Giải: \` x t = > : ( ) [ x x m m m + + − + − > > x∀ ∈ ¡ ( ) ( ) ( ) [ [ [ m t m t m t m t t t t⇔ + − + − > ∀ > ⇔ + + > + ∀ > ( ) [ [ t g t m t t t + ⇔ = < ∀ > + + L ( ) ( ) [ [ t t g t t t − − ′ = < + + 0 ( ) g t #0 [ ) 8+∞ _ #_⇔ ( ) ( ) t Max g t g m ≥ = = ≤ Bài 4. L:mQRS#:5 ( ) h [x x x m x x+ + = − + − P Giải: \jcP [x≤ ≤ G$@L ( ) h [ x x x f x m x x + + ⇔ = = − + − Chú ý:9W ( ) f x ′ #7b%&:<(#&QTQ%k=)l Thủ thuật:\` ( ) ( ) Z g x x x x g x x x ′ = + + > ⇒ = + > + ( ) ( ) h [ h [ h x x x h x x x − ′ = − + − > ⇒ = − < − − E_#5 ( ) g x > +!d8 ( ) h x m+!._ ( ) h x > +!d ⇒ ( ) ( ) ( ) g x f x h x = dE_# ( ) f x m= P [ ] ( ) [ ] ( ) ( ) ( ) [ ] ( ) 8[ 8[ 87 8 [ h 8m f x f x f f ⇔ ∈ = = − Bài 5.L:m &QRS#:5 ( ) Z Z Z x x m x x+ − ≤ − − P Giải:\jcP x ≥ 9e.+G@L+* ( ) Z x x+ − > ;Rn &QRS#: ( ) ( ) ( ) Z Z Z f x x x x x m= + − + − ≤ \` ( ) ( ) ( ) Z Z Z 8 g x x x h x x x= + − = + − L ( ) ( ) ( ) Z o 8 Z g x x x x h x x x x x ′ ′ = + > ∀ ≥ = + − + > ÷ − ^< ( ) g x > +!d x∀ ≥ 8 ( ) h x > +!d0 ( ) ( ) ( ) f x g x h x= d x∀ ≥ 4&QRS#: ( ) f x m≤ P ( ) ( ) Z x f x f m ≥ ⇔ = = ≤ Bài 6. L:m ( ) ( ) [ o x x x x m+ − ≤ − + P> [ ] [ox∀ ∈ − Cách 1. G@L ( ) ( ) ( ) [ of x x x x x m⇔ = − + + + − ≤ > [ ] [ox∀ ∈ − ( ) ( ) ( ) ( ) ( ) ( ) [ o [ o x f x x x x x x x x − + ′ = − + + = − + = ⇔ = ÷ + − + − I;Q.0_#67 [ ] ( ) ( ) [o oMax f x f m − = = ≤ Cách 2.\` ( ) ( ) ( ) ( ) [ o [ o h x x t x x + + − = + − ≤ = L [t x x= − + + 4&QRS#:#V! [ ] ( ) [ ] [ 8h [ 8 8ht t m t f t t t m t≤ − + + ∀ ∈ ⇔ = + − ≤ ∀ ∈ L5 ( ) f t t ′ = + > ⇒ ( ) f t d0 ( ) [ ] 8 8hf t m t≤ ∀ ∈ ⇔ [ ] ( ) ( ) 8h 7 h of t f m= = ≤ Bài 7. L:m Z o ] Z x x x x m m + + − − + − ≤ − + > [ ] Zox∀ ∈ − Giải: \` Z o t x x= + + − > ⇒ ( ) ( ) ( ) Z o p Z ot x x x x= + + − = + + − ⇒ ( ) ( ) ( ) ( ) p p Z o p Z o ]t x x x x≤ = + + − ≤ + + + − = ( ) ( ) ( ) ] Z Z o p 8 Z8Z x x x x t t ⇒ + − = + − = − ∈ ab ( ) ( ) ( ) ( ) Z8Z p 8 8 Z8Z 7 Z Z f t t t f t t t f t f ′ = − + + = − < ∀ ∈ ⇒ = = _ ( ) Z8Z 7 Z q f t m m m m m ⇔ = ≤ − + ⇔ − − ≥ ⇔ ≤ − ≥ Bài 8. (Đề TSĐH khối A, 2007) L:mQRS#: [ Z x m x x− + + = − P" Giải: \45 x ≥ $QRS#: [ Z x x m x x − − ⇔ − + = + + \` [ ) [ [ x u x x − = = − ∈ + + 4 ( ) Z g t t t m= − + = L ( ) o Z g t t t ′ = − + = ⇔ = ^<_0= Z m⇔ − < ≤ Bài 9. (Đề TSĐH khối B, 2007):YT#U5q*r m > QRS#: ( ) ] x x m x+ − = − )g>PQeP Giải:\jcP5 x ≥ G$QRS#:5 ( ) ( ) ( ) o x x m x⇔ − + = − ( ) ( ) ( ) o x x m x⇔ − + = − ( ) ( ) ( ) Z Z o Z q 7 o Zx x x m x x x m⇔ − + − − = ⇔ = = + − = _ ( ) g x m⇔ = >Pc<. ( ) 8+∞ L;+;_5 ( ) ( ) Z [ g x x x x ′ = + > ∀ > ^< ( ) g x ! ( ) g x )01+! ( ) ( ) 8 ) x g g x →+∞ = = +∞ 0 ( ) g x m= >P∈ ( ) 8+∞ q;_ m∀ > QRS#: ( ) ] x x m x+ − = − PQeP fss 7f Bài 10. (Đề TSĐH khối A, 2008)L:mQRS#:>P"QeP5 [ [ o ox x x x m+ + − + − = Giải:\` ( ) [ ] [ [ o o 8 8o f x x x x x x= + + − + − ∈ L5 ( ) ( ) ( ) ( ) Z Z [ [ 8o o o f x x x x x x ′ = − + − ∈ ÷ ÷ − − \` ( ) ( ) ( ) ( ) ( ) Z Z [ [ 8 o o o , xu x v x x x x x = − = − ∈ − − ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) o u x v x x u v u x v x x > ∀ ∈ ⇒ = = < ∀ ∈ ( ) ( ) o f x x f x x f ′ > ∀ ∈ ′ ⇒ < ∀ ∈ ′ = 9:GGL@LPQeP ⇔ [ o o Z om+ ≤ < + Bài 11. (Đề TSĐH khối D, 2007): L:mPQRS#:P Z Z Z Z h h x y x y x y m x y + + + = + + + = − Giải:\` 8u x v y x y = + = + ( ) ( ) Z Z Z Z Zx x x x u u x x x x + = + − × + = − +! 8 u x x x v y y x x x y y = + = + ≥ = = + ≥ = 4P#V! ( ) Z Z h h ] Z h u v u v uv m u v u v m + = + = ⇔ = − + − + = − ⇔ u v )!P-QRS#:; ( ) h ]f t t t m= − + = PP ( ) f t m⇔ = P t t ,t 8 t t≥ ≥ xofsf(x) [ o o + I;QG.0-! ( ) f t +* t ≥ −∞ s h f ∞ ( ) f t ′ – – + ( ) f t f ∞ u[ f ∞ 9:.0PP u [ m⇔ ≤ ≤ ∨ ≥ Bài 12.(Đề 1I.2 Bộ đề TSĐH 1987-2001): L:x&QRS#: ( ) < x x y y+ + + ≥ >+* y∀ ∈ ¡ Giải: \` < u y y = + ∈ − G@L ( ) ( ) ( ) ( ) 6 u g u x u x u g u ∈ − ⇔ = + + ≥ ∀ ∈ − ⇔ ≥ ^< ( ) y g u= )!<X+* u ∈ − 0 ( ) 6 u g u ∈ − ≥ ( ) ( ) g x x x x x x g − ≥ − + ≥ ≥ + ⇔ ⇔ ⇔ + + ≥ ≤ − ≥ Bài 13.Y< Z a b c a b c ≥ + + = YT#U5 [a b c abc+ + + ≥ Giải:G\L ( ) ( ) ( ) [ Z [a b c bc abc a a a bc⇔ + + − + ≥ ⇔ + − + − ≥ ( ) ( ) o h f u a u a a⇔ = − + − + ≥ #< ( ) ( ) Z [ b c u bc a + ≤ = ≤ = − 9R ( ) y f u= )!<X+* ( ) 8 Z [ u a ∈ − L ( ) ( ) ( ) ( ) ( ) ( ) Z o h 8 Z [ [ f a a a f a a a= − + = − + ≥ − = − + ≥ 0_# ( ) 8f u ≥ ( ) 8 Z [ u a ∀ ∈ − q;_ [a b c abc+ + + ≥ \XT7._# a b c⇔ = = = Bài 14. (IMO 25 – Tiệp Khắc 1984): Y< a b c a b c ≥ + + = YT#U5 u u ab bc ca abc+ + − ≤ Giải: ( ) ( ) ( ) ( ) ( ) ( ) ( ) a b c a bc a a a bc a a a u f u+ + − = − + − = − + − = \ ( ) ( ) ( ) y f u a u a a= = − + − +* ( ) ( ) [ a b c u bc − + ≤ = ≤ = )!<X+*( #=> ( ) ( ) ( ) u [ u a a f a a + − = − ≤ = < +! ( ) ( ) ( ) ( ) ( ) Z u u [ [ u [ Z Z u f a a a a a− = − + + = − + − ≤ ^< ( ) y f u= )!<X+* ( ) 8 [ u a ∈ − +! ( ) u u f < 8 ( ) ( ) u [ u f a− ≤ 0 ( ) u u f u ≤ \XT7._# Z a b c⇔ = = = Bài 15. YT#U5 ( ) ( ) [a b c ab bc ca+ + − + + ≤ ∀ [ ] a b c∈ Giải: G$&XT+j!;&ab, c ( ) ( ) ( ) [ ] [ f a b c a b c bc a b c= − − + + − ≤ ∀ ∈ \ ( ) y f a= )!<X+* [ ] a∈ 0 ( ) ( ) ( ) { } 67 8 f a f f≤ L ( ) ( ) ( ) ( ) ( ) [ ] [ [8 [ [ [ f b c f bc f a a b c= − − − ≤ = − ≤ ⇒ ≤ ∀ ∈ Bài 16. Y6M5 ( ) ( ) ( ) ( ) [ ] a b c d a b c d a b c d− − − − + + + + ≥ ∀ ∈ Giải: G%k&XT+j!;&a, b, c, d, 5 ( ) ( ) ( ) ( ) [ ] ( ) ( ) ( ) [ ] f a b c d a b c d b c d a b c d = − − − − + − − − + + + ≥ ∀ ∈ \ ( ) [ ] y f a a= ∀ ∈ )!<X0 [ ] ( ) ( ) ( ) { } 6 6 a f a f f ∈ = L ( ) [ ] f b c d b c d= + + + ≥ ∀ ∈ ( ) ( ) ( ) ( ) ( ) ( ) ( ) [ ] ( ) ( ) f b c d b c d g b c d b c d c d = − − − + + + ⇔ = − − − + − − + + \ ( ) [ ] y g b b= ∀ ∈ )!<X0 [ ] ( ) ( ) ( ) { } 6 b g b Min g g ∈ = L ( ) ( ) ( ) ( ) 8 g c d g c d c d cd = + + ≥ = − − + + = + ≥ ⇒ ( ) ( ) [ ] f g b b= ≥ ∀ ∈ q;_ ( ) f a ≥ _Q BÀI 2. TÍNH ĐƠN ĐIỆU CỦA HÀM SỐ A. TÓM TẮT LÝ THUYẾT. 1.y=fxab⇔ƒ′x≥∀x∈abƒ′x= ∈ab 2.y=fxab⇔ƒ′x≤∀x∈abƒ′x= ∈ab Chú ý: L#<RS#:Q$gc/%11. 2.<(!v_w, jcPƒ′x=∈ab CÁC BÀI TẬP MẪU MINH HỌA Bài 1. L:m ( ) ( ) o h Z mx m x m y x + + − − = + #02+∞ Giải: !#02+∞⇔ ( ) u mx mx y x x + + ′ = ≤ ∀ ≥ + ⇔ ( ) u u mx mx m x x x + + ≤ ⇔ + ≤ − ∀ ≥ ⇔ ( ) u u x m x x x − = ≥ ∀ ≥ + ( ) 6 x u x m ≥ ⇔ ≥ L5 ( ) ( ) u x u x x x x + ′ = > ∀ ≥ + ⇒ux#02+∞⇒ ( ) ( ) u 6 Z x m u x u ≥ − ≤ = = Bài 2. L:m ( ) ( ) Z Z [ Z y x m x m x − = + − + + − #0Z Giải. !d#0Z⇔ ( ) ( ) ( ) Z Zy x m x m x ′ = − + − + + ≥ ∀ ∈ ^< ( ) y x ′ )01x=+!x=Z0⇔y′≥∀x∈2Z3 ⇔ ( ) [ ] Z Zm x x x x+ ≥ + − ∀ ∈ ⇔ ( ) [ ] Z Z x x g x m x x + − = ≤ ∀ ∈ + [ ] ( ) Z 67 x g x m ∈ ⇔ ≤ L5 ( ) ( ) [ ] ] Z x x g x x x + + ′ = > ∀ ∈ + ⇒gx#02Z3⇒ [ ] ( ) ( ) Z 67 Z u x m g x g ∈ ≥ = = Bài 3. L:m ( ) ( ) Z Z Z Z m y x m x m x= − − + − + #0 [ ) +∞ Giải: !d [ ) +∞ ⇔ ( ) ( ) Z y mx m x m x ′ = − − + − ≥ ∀ ≥ ⇔ ( ) o m x x x − + ≥ − + ∀ ≥ ⇔ ( ) ( ) o x g x m x x − + = ≤ ∀ ≥ − + L5 ( ) ( ) o Z Z x x g x x x − + ′ = = − + Z o Z o x x x x = = − ⇔ = = + 8 ( ) ) x g x →∞ = LxGGL⇒ ( ) ( ) 67 Z x g x g m ≥ = = ≤ Bài 4. ( ) ( ) ( ) Z u u Zy x mx m m x m m= − − − + + − − [ ) +∞ Giải: !d#0 [ ) +∞ ( ) Z u u y x mx m m x ′ ⇔ = − − − + ≥ ∀ ≥ L ( ) u Z Zm m ′ = − +V ( ) Z Z u [ m = − + > 0 y ′ = P x x< G@Lgx≥SjP')!5 L ( ) y x ′ ≥ > x∀ ≥ ⇔ [ ) G+∞ ⊂ ( ) ( ) h h Z Z Z h o Z m x x y m m m S m m ′ ∆ > − ≤ ≤ ′ ⇔ < ≤ ⇔ = − + + ≥ ⇔ ⇔ − ≤ ≤ < = < Bài 5. L:m ( ) x m x m y x m + − + + = − #0 ( ) +∞ Giải: !#0 ( ) +∞ ⇔ ( ) [ x mx m m y x x m − + − − ′ = ≥ ∀ > − ⇔ ( ) ( ) [ g x x g x x mx m m x m x m ≥ ∀ > = − + − − ≥ ∀ > ⇔ ≤ − ≠ Cách 1:Phương pháp tam thức bậc 2 L5 ( ) m ′ ∆ = + ≥ _#gx=P x x≤ G@Lgx≥SjP')!5 Lgx≥>∀x∈+∞⇔ ( ) G+∞ ⊂ ( ) ( ) o Z Z Z m m x x g m m m m S m ′ ≤ ≤ ∆ ≥ ⇔ ≤ ≤ ⇔ = − + ≥ ⇔ ⇔ ≤ − ≤ − ≥ + = − ≤ Cách 2:Phương pháp hàm số L5g′x=[x−m≥[x−m∀xm⇒gx#02+∞ x x x x xyfYL ^< ( ) ( ) ( ) o Z 6 Z Z x g m m m g x m m m m m ≥ = − + ≥ ≤ − ≥ ⇔ ⇔ ⇔ ≤ − ⇔ ≥ + ≤ ≤ ≤ Bài 6. L:m ( ) ( ) [ h < Z Z y m x m x m m= − + − + − + . x∀ ∈ ¡ Giải:i0=!<( ( ) h [ Z y m x m x ′ ⇔ = − + − ≤ ∀ ∈ ¡ ( ) ( ) [ ] h [ Z 8g u m u m u⇔ = − + − ≤ ∀ ∈ − ^< ( ) [ ] 8y g u u= ∈ − )!<X0_ ( ) ( ) o ] [ Z g m m g m − = − ≤ ⇔ ⇔ ≤ ≤ = − + ≤ Bài 7. L:m! Z [ p y mx x x x= + + + d+*r x∈ ¡ Giải: i0=!<( < < <Z Z y m x x x x ′ ⇔ = + + + ≥ ∀ ∈ ¡ ⇔ ( ) ( ) Z < < [< Z< Z m x x x x x+ + − + − ≥ ∀ ∈ ¡ ( ) [ ] Z [ Z m u u g u u⇔ ≥ − − + = ∀ ∈ − +* [ ] < u x= ∈ − L ( ) ( ) [ 8 g u u u u u u u ′ = − − = − + = ⇔ = − = I;QGGL_#_0=!<(⇔ [ ] ( ) ( ) h 67 o x g u g m ∈ − = − = ≤ Bài 8. Y<! ( ) ( ) ( ) Z Z Z y m x m x m x m= + + − − + + L:mc<.-!%!U[ Giải. ab ( ) ( ) ( ) Z y m x m x m ′ = + + − − + = ^< u Z m m ′ ∆ = + + > 0 y ′ = P x x< 4<.-!%!U[ [ ] 8 8 8 [y x x x x x ′ ⇔ ≤ ∀ ∈ − = m⇔ + > +! [x x− = L [x x− = ⇔ ( ) ( ) ( ) ( ) ( ) [ [ Z o [ m m x x x x x x m m − + = − = + − = + + + ( ) ( ) ( ) ( ) [ Z m m m m⇔ + = − + + + u o Z u o m m m ± ⇔ − − = ⇔ = cnQ+* m + > _# u o o m + = B. ỨNG DỤNG TÍNH ĐƠN ĐIỆU CỦA HÀM SỐ I. DẠNG 1: ỨNG DỤNG TRONG PT, BPT, HỆ PT, HỆ BPT Bài 1. '.QRS#:5 h Z Z [ x x x+ − − + = Giải. \jcP5 Z x ≤ \` ( ) h Z Z [ f x x x x= + − − + = L5 ( ) [ Z h Z Z f x x x x ′ = + + > − ⇒fx#0 ( Z −∞ 6`c(f −=0QRS#:fx=P%_&x=− Bài 2. '.QRS#:5 h Z ]x x x+ = − + + Giải. G&QRS#:⇔ ( ) Z ] hf x x x x= − + + − + = f9 Z x ≤ :fxz⇒+gP f9 Z x > : ( ) Z Z ] h f x x x x x ′ = + − > ∀ > ÷ + + ⇒fx#0 ( ) Z +∞ !f =0>Px= Bài 3. '.&QRS#:5 Z h [ h u u h Z u ]x x x x+ + − + − + − < { Giải. \jcP h u x ≥ \` ( ) Z h [ h u u h Z uf x x x x x= + + − + − + − L5 ( ) ( ) ( ) Z [ h Z [ h u Z h Z u Z h u [ u h f x x x x x ′ = + + + > + × − × − × − ⇒fx#0 ) h u +∞ 6!fZ=]0{⇔fxzfZ⇔xzZ q;_P-&QRS#:t<)! h Z u x≤ < Bài 4. '.@L5 Z h [ Z h u u Z o x x x x x x x x x x+ + + = + + − + − + { Giải. { ( ) ( ) ( ) ( ) ( ) Z h [ Z h u u Z o x x x x x x x f x x x x g x ⇔ = + + + − − − = − + − + = Lfx+!g′x=−ox +x−uz∀x⇒gx 9P-fx=gx)!<!<- ( ) ( ) +!y f x y g x= = ^<fxd8gx.+! ( ) ( ) Zf g= = 0{P%_&x= Bài 5. L:m67 ( ) < < m x x x x x x + + ≤ + + + ∀ { Giải. \` ( ) < < t x x t x x x= + ≥ ⇒ = + = + ⇒ t≤ ≤ ⇒ t≤ ≤ c{ ⇔ ( ) m t t t t + ≤ + + ∀ ∈ ⇔ ( ) t t f t m t t + + = ≥ ∀ ∈ + ⇔ ( ) 6 t f t m ∈ ≥ ^< ( ) ( ) t t f t t + ′ = > + 0ft ⇒ ( ) ( ) Z 6 t f t f ∈ = = ⇒ Z m ≤ ⇒ Z 67 m = Bài 6. '.QRS#: < ] ] < x x x− = < < ] ] < ] ] < x x x x x x x x− = − ⇔ + = + { ab ( ) ] u f u u= + L ( ) ] ) u f u u ′ = + > E_# ( ) f u { ( ) ( ) < < < f x f x x x x⇔ = ⇔ = ⇔ = [ k x k π π ⇔ = + ∈ ¢ Bài 7. L: ( ) x y∈ π ,tP < < Z h x y x y x y − = − + = π Giải. < < < < x y x y x x y y − = − ⇔ − = − ab!`#R ( ) ( ) < f u u u u= − ∈ π L ( ) f u u ′ = + > E_# ( ) f u #0 ( ) π 4 ( ) ( ) [ Z h f x f y x y x y = π ⇔ = = + = π [...]... chn trờn on [a, b] Xột mt phõn hoch bt kỡ ca on [a, b], tc l chia on [a, b] thnh n phn tu ý bi cỏc im chia: a = x0 < x1 < < xn 1 < xn = b Trờn mi on [ xk 1 , xk ] ly bt kỡ im k [ xk 1 , xk ] v gi k = xk xk 1 l di ca [ xk 1 , xk ] Khi ú: n f ( ) k k k =1 = f ( 1 ) 1 + f ( 2 ) 2 + + f ( n ) n gi l tng tớch phõn ca hm f(x) trờn on [a, b] Tng tớch phõn ny ph thuc vo phõn hoch , s khong chia... x 1 x 4 : ( P1 ) f ( x) = 2 x + ( m + 5 ) x 4 ; 1 x 4 : ( P2 ) Gi (P) l th ca y = f(x) (P) = (P1) (P2) khi ú (P) cú 1 trong cỏc hỡnh dng th sau õy P1 P2 A P P1 Honh ca cỏc im c bit P trong 1 th (P): A A P2 5m Honh giao im (P1), (P2) xA = 1; xB = 4 ; Honh nh (P1): xC = B 2 B B C C C Nhỡn vo th ta xột cỏc kh nng sau: 2 Nu xC [xA, xB] m[ 3, 3] thỡ Minf(x) = Min{f(1), f(4)} 3 m... mt bt ng thc rt ni ting trong sut th k 20 Trờn õy l mt cỏch chng minh bt ng thc ny trong 45 cỏch chng minh Bn c cú th xem tham kho y cỏc cỏch chng minh trong cun sỏch: Nhng viờn kim cng trong bt ng thc Toỏn hc ca tỏc gi do NXB Tri thc phỏt hnh thỏng 3/2009 BI 3 GI TR LN NHT, NH NHT CA HM S A GI TR LN NHT, NH NHT CA HM S I TểM TT Lí THUYT 1 Bi toỏn chung: Tỡm giỏ tr nh nht hoc ln nht ca hm s f ( x )... tớch phõn xỏc nh ca k =1 b hm s f(x) trờn on [a, b] v kớ hiu l: f ( x ) dx a Khi ú hm s y = f(x) c gi l kh tớch trờn on [a, b] 2 iu kin kh tớch: Cỏc hm liờn tc trờn [a, b], cỏc hm b chn cú hu hn im giỏn on trờn [a, b] v cỏc hm n iu b chn trờn [a, b] u kh tớch trờn [a, b] 3 í ngha hỡnh hc: b Nu f(x) > 0 trờn on [a, b] thỡ f ( x ) dx l din tớch ca hỡnh thang cong gii hn bi cỏc a ng: y = f(x), x = a,... xn=b a=x0 x 4 Cỏc nh lý, tớnh cht v cụng thc ca tớch phõn xỏc nh: 4.1 nh lý 1: Nu f(x) liờn tc trờn on [a, b] thỡ nú kh tớch trờn on [a, b] 4.2 nh lý 2: Nu f(x), g(x) liờn tc trờn on [a, b] v f(x) g(x),x[a, b] b thỡ a b f ( x ) dx g ( x ) dx Du bng xy ra f(x) g(x), x[a, b] a 4.3 Cụng thc Newton - Leipnitz: Nu f ( x ) dx = F ( x ) + c b thỡ f ( x ) dx = F ( x ) b = F ( b) F ( a ) b f (... x2 Xột 2 kh nng sau: a) Nu m < 1 thỡ x1 x 2 = m + 1 < 0 x1 < 0 < x 2 Bng bin thi n Nhỡn BBT suy ra x CĐ = 0 b) Nu m > 1 thỡ x1 x 2 > 0 3 ( m + 3) < 0 x1 < x 2 < 0 v x1 + x 2 = 4 Bng bin thi n Nhỡn BBT suy ra x CĐ = x 2 < 0 Kt lun: Vy m 1 hm s luụn cú x CĐ 0 xx10x2+f 0+00+f+ CT C CT+ xx1x20+f 0+00+f+ CT C CT+ Bi 4 ( thi TSH khi B 2002) Tỡm m hm s y = mx 4 + ( m 2 9 ) x 2 + 10 cú 3 im cc tr Gii... y = m vi th y = g(x) Nhỡn bng bin thi n suy ra ng thng y = m ct y = g(x) ti ỳng 1 im f ( x ) = 0 cú ỳng 1 nghim Vy hm s y = f (x) khụng th ng thi cú cc i v cc tiu Bi 7 Chng minh rng: f ( x ) = x 4 + px 3 + q 0 x Ă 256q 27 p 4 3 p Gii Ta cú: f ( x ) = 4 x 3 + 3 px 2 = x 2 ( 4 x + 3 p ) = 0 x = v nghim kộp x = 0 4 Do f (x) cựng du vi (4x + 3p) nờn lp bng bin thi n ta cú: 3 p 256q 27 p 4 0... gúc Oxyz Tp hp cỏc im M ( x, y, z ) tho món iu kin x, y, z [ 0,1] nm trong hỡnh lp phng ABCDABCO cnh 1 vi A(0, 1, 1); B(1, 1, 1); C(1, 0, 1); D(0, 0, 1); A(0, 1, 0); B(1, 1, 0); C(1, 0, 0) Mt khỏc do x + y + z = 3 nờn M ( x, y, z ) nm trờn mt phng (P): x + y + z = 3 2 2 Vy tp hp cỏc im M ( x, y, z ) tho món iu kin gi thit nm trờn thit din EIJKLN vi cỏc im E, I, J, K, L, N l trung im cỏc cnh hỡnh lp... x = 0 Bng bin thi n Nhỡn bng bin thi n suy ra: x0+f 0+ f 0 f ( x ) f ( 0 ) = 0 (pcm) a , b, c > 0 Bi 2 Cho CMR: T = 2 a 2 + 2 b 2 + 2 c 2 3 3 2 2 2 2 b +c c +a a +b a + b + c = 1 2 2 2 a b c a b c Ta cú: T = 1 a 2 + 1 b 2 + 1 c 2 = a ( 1 a 2 ) + b ( 1 b 2 ) + c ( 1 c 2 ) Xột hm s f ( x ) = x ( 1 x 2 ) vi x > 0 1 2 Ta cú f ( x ) = 1 3x = 0 x = 3 > 0 2 Nhỡn bng bin thi n f ( x )... bin thi n suy ra: Hm s y = f (x) ch cú cc tiu m khụng cú cc i CT+ g > 0 m = 1 thỡ f (x) cú 3 nghim phõn bit x1 < x 2 < x 3 c) Nu g ( 0) 0 Nhỡn bng bin thi n suy ra: Hm s y = f (x) cú cc i nờn khụng tho món yờu cu bi toỏn Kt lun: m 1 7 , 1 + 3 3 xx1x2x3+f 0+00+f+ 7 { } U 1 CT C CT+ Bi 15 Cho hm s y = f ( x ) = x 4 + ( m + 3) x 3 + 2 ( m + 1) x 2 Chng minh rng: m 1 hm s luụn cú cc i ng thi . _# u o o m + = B. ỨNG DỤNG TÍNH ĐƠN ĐIỆU CỦA HÀM SỐ I. DẠNG 1: ỨNG DỤNG TRONG PT, BPT, HỆ PT, HỆ BPT Bài 1. '.QRS#:5 h Z Z [ x x x+. ( ) ( ) Z u Z f x f x> = > ∀ ∈ − II. DẠNG 2: ỨNG DỤNG TRONG CHỨNG MINH BẤT ĐẲNG THỨC Bài 1.YT#U5 Z Z h Z| Z| h| x