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LuyÖn thi trªn m¹ng – Phiªn b¶n 1.0
_________________________________________________________ C©u I.
1) y' = mx m−1(4 − x)2 − 2(4 − x)x m = = x m −1 (4 − x)[4m − (m + 2)x] .
a) XÐt tr−êng hîp m ≥ 2. Khi ®ã ph−¬ng tr×nh y' = 0 cã ba nghiÖm x1 = 0 , x2 = x3 = 4 .
NÕu m − 1 ch½n (tøc m = 3, 5, 7, ...) th× y' sÏ cïng dÊu víi (4 − x) [4m − (m + 2)x] vµ do ®ã : y min (4) = 0 vµ y max (x2 ) =
m m 4m + 4 (m + 2)m +2
= M.
NÕu m - 1 lÎ (tøc m = 2, 4, 6, ...) th× dÊu cña y' lµ dÊu cña
x(4 − x)[4m − (m + 2) x]
LËp b¶ng xÐt dÊu sÏ cã kÕt qu¶
y min (0) = 0 ; y max (x2 ) = M , y min (4) = 0
b) §Ò nghÞ b¹n ®äc tù lµm cho tr−êng hîp m = 1 (y = x(4 − x)2 ) .
2) Kh¶o s¸t, vÏ ®å thÞ hµm sè y = x(4 − x)2
dµnh cho b¹n ®äc. C©u II.
1) x2 − 2(cosB + cosC)x + 2(1 − cosA) ≥ 0 . (1) ∆ ' = (cosB + cosC)2 − 2(1 − cosA) = C+ B 2 B−C A = 4 cos2 cos − 4sin2 = 2 2 2 A B−C = 4sin2 cos2 − 1 ≤ 0 2 2
VËy (1) ®óng víi mäi x.
sin x + cosx 10 = sin x cosx 3 §Æt t = cosx + sin x(− 2 ≤ t ≤ 2) (2)
2) cosx + sin x +
th× t 2 = 1 + 2sin x cosx vµ ta ®−îc t + §Æt ®iÒu kiÖn t ≠ ±1 sÏ tíi
2t
10 t2 − 1 3 =
3t 3 − 10t 2 + 3t + 10 = 0
tøc lµ : 1 + a + b + c + ab + ac + bc ≥ 0 (2) Céng (1) vµ (2) ta cã : abc + 2 (1 + a + b + c + ac + bc + ac) ≥ 0. (t − 2)(3t 2 − 4t − 5) = 0 . hay Ph−¬ng tr×nh nµy cã ba nghiÖm t1 = 2 ; t 2 =
2 − 19 2 + 19 ; t3 = 3 3
4m vµ m+2
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LuyÖn thi trªn m¹ng – Phiªn b¶n 1.0
_________________________________________________________ ChØ cã t 2 lµ thÝch hîp. Thay vµo (2) ta cã ph−¬ng tr×nh π 2 − 19 cos x − = . 4 3 2
§Æt cos α =
2 − 19 3 2
th× ®−îc hai hä nghiÖm : x1 =
π π + α + 2kπ ; x2 = − α + 2mπ 4 4
C©u III.
1) §Æt ®iÒu kiÖn x - a ≠ 0 ; x + a ≠ 0 th× (1) ®−îc biÕn ®æi vÒ d¹ng : x[a − 1)x + a 2 + a + 2b] = 0 (2) Víi ∀a, b (2) ®Òu cã nghiÖm x1 = 0 . Gi¶i (a − 1)x + a 2 + a + 2b = 0 . NÕu a ≠ 1 cã nghiÖm x2 =
a 2 + a + 2b 1− a
NÕu a = 1 ta cã : 0x = − 2(1 + b). (3) Víi b ≠ − 1 th× (3) v« nghiÖm ; víi b = -1 th× (3) nghiÖm ®óng víi ∀x. KiÓm tra x2 cã tháa m·n ®iÒu kiÖn x2 ≠ ±a ? x2 ≠ a ⇔
a 2 + a + 2b ≠ a ⇔ a 2 + a + 2b ≠ 1− a
≠ a − a 2 ⇔ 2(a 2 + b) ≠ 0 ⇔ b ≠ −a 2 x 2 ≠ −a ⇔
a 2 + a + 2b ≠ −a ⇔ a 2 + a + 2b ≠ a 2 − a ⇔ b ≠ −a . 1− a
KÕt luËn : víi b ≠ −1 , (1) cã nghiÖm duy nhÊt x1 = 0 . NÕu a = 1 th× : víi b = − 1, (1) cã nghiÖm lµ ∀x ≠ ± 1.
NÕu a ≠ 1 ; 0 th× : 2 víi b ≠ −a , b ≠ - a, (1) cã hai nghiÖm x1 = 0, a 2 + a + 2b x = 2 1− a víi b = −a 2 hoÆc b = - a th× (1) cã mét nghiÖm x1 = 0 .
NÕu a = 0 th× (1) cã mét nghiÖm x2 = 2b nÕu b ≠ 0 ; (1) sÏ v« nghiÖm nÕu b = 0. 2) V× a 2 + b2 + c2 = 1 nªn - 1 ≤ a, b, c ≤ 1. Do ®ã 1 + a ≥ 0 , 1 + b ≥ 0, 1 + c ≥ 0 ⇒ (1 + a) (1 + b) (1 + c) ≥ 0 ⇒ ⇒ 1 + a + b + c + ab + ac + bc + abc ≥ 0. (1) MÆt kh¸c : a 2 + b2 + c2 + a + b + c + ab + ac + bc =
(1 + a + b + c)2 ≥0, 2
www.khoabang.com.vn LuyÖn thi trªn m¹ng – Phiªn b¶n 1.0 ________________________________________________________________________________ C©u IVa. 1) Víi x > 0 ta cã F(x) = x - ln(1 + x) Þ F’(x) = 1 -
1 x ; = 1 + x 1 + x
víi x < 0 ta cã F(x) = - x - ln(1 - x) Þ F’(x) = - 1 +
1 x . = 1 - x 1 - x
Tõ ®ã suy ra víi x ¹ 0 F’(x) =
x . 1 + | x|
Ta chØ cßn ph¶i chøng minh r»ng F’(0) = 0. Qu¶ vËy 1 1 (F( ∆x) - F(0)) = lim ∆ x → 0 ∆x → 0 ∆x ∆x ∆ ln(1 + x) = lim 1 = 0, ∆x →0 ∆x F’(0) = lim
v× lim
∆x → 0
( ∆x - ln(1 + ∆x)) =
ln(1 + ∆x) = 1. ∆x
e
2) I = ∫ xln2xdx. 1
u = ln x §æt dv = xdx 2
⇒
ln x du = 2 dx x 1 v = x 2, 2
1 e suy ra I = x 2 ln 2 x 2 1 §Ó tÝnh J, ®Æt du = ux u = ln x x ⇒ 1 dv = xdx v = 2
e
∫ 1
e e2 xlnxdx = - J, víi J = ∫ xlnxdx. 2 1
www.khoabang.com.vn LuyÖn thi trªn m¹ng – Phiªn b¶n 1.0 ________________________________________________________________________________ suy ra J =
e 1 2 1 e 1 e2 . x ln x − ∫1 xdx = − 2 1 2 2 2 4( e − 1)
VËy 1 I = (e2 - 1). 4 C©u Ivb. 1) V× K lµ trung®iÓm cña SC, nªn theo h×nhbªn, trong tam gi¸c SAC, SO vµ AK lµ hai ®ûêng trungtuyÕn c¾t nhau t¹i trängt©m H, vËy SH 2 = . SO 3 Theo h×nh bªn , ta cã dt(SNH) =
SN SH . . dt(SDO) = SD SO
=
SN 2 1 SH SM . . dt(SDB),dt(SHM) = . . dt(SOB) SD 3 2 SO SB
=
2 SM 1 . . dt (SDB). 3 SB 2
§ång thêi dt(SNH) + dt(SHM) = dt(SNM) =
Tõ c¸c hÖ thøc trªn, suy ra Û
SN SM dt(SDB). . SD SB
1 SN 1 SM SN SM . = . . + SD SD 3 SD 3 SB
SB SD + = 3. SM SN SM SN 1 1 2) §Æt = y, theo hÖ thøc trªn ta cã + = x, = 3. §ång thêi, do ý nghÜa h×nh häc, ph¶i cã 0 < x £ 1, SB SD x y
0 < y £ 1. V× 1 1 x , = 3 ⇒ y = y x 3x - 1 x nªn 0 < ≤ 1 3x - 1 1 Þ ≤ x ≤ 1. 2 0<x£1
www.khoabang.com.vn LuyÖn thi trªn m¹ng – Phiªn b¶n 1.0 ________________________________________________________________________________ Ta cã theo h×nh bªn V 1 = V SAMN + V SMNK , VSAMN =
SM SN 1 . .VSABD = xyV, SB SD 2
V SMNK =
SM SN SK 1 . . . VSBDC = xyV SB SD SC 4 V1 3 3x 2 1 = xy = ≤ x ≤ 1. V 4 4(3x - 1) 2
suy ra
1 3x 2 3x(3x - 2) f(x) = cã ®¹o hµm f’(x) = , do vËy trªn ®o¹n ; 1 cã b¶ng biÕn thiªn 2 4(3x - 1) 4(3x - 1) 2
Hµm sè
1 2
x
f’ f
1
-
0
3 8
+ 3 8
1 3 VËy víi
V 1 3 1 ≤ x ≤ 1 th× ≤ 1 ≤ . 3 V 8 2