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LuyÖn thi trªn m¹ng
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C©u I.
1
) §Ó f(x) = 0 cã nghiÖm, ta ph¶i cã :
2
2
2
∆
' = (m +1) − 2(m + 4m + 3) = = −m − 6m − 5 ≥ 0 ⇒ − 5 ≤ m ≤ − 1.
2
) Víi ®iÒu kiÖn trªn, gäi x , x lµ c¸c nghiÖm cña f(x) = 0, x ≤ x . ThÕ th×
1
2
1
2
−
(m +1) − ∆'
−(m +1) + ∆'
x1 =
, x =
.
2
2
2
§
iÒu kiÖn cña bµi to¸n ®−îc nghiÖm nÕu x ≥1, suy ra
2
2
−
(m + 6m + 5) ≥ m + 3.
NÕu m ≤ 3 bꢀt ph−¬ng tr×nh ®−îc nghiÖm. Víi m ≥ − 3, b×nh ph−¬ng hai vÕ, ®i ®Õn
−
6 − 2 2
−6 + 2 2
2
0
≥ 2m +12m +14 ⇒
≤ m ≤
.
2
2
KÕt hîp c¸c ®iÒu kiÖn ta ®−îc :
−
6 + 2 2
−5 ≤ m ≤
.
2
3
) Theo hÖ thøc Vi Ðt
2
2
m + 4m + 3
m + 8m + 7
x x − 2(x + x ) =
+ 2(m +1) =
1
2
1
2
2
2
2
XÐt hµm g(m) = m + 8m + 7 trªn ®o¹n [− 5 ; − 1].
§
å thÞ cña parabol cã ®Ønh t¹i m = − 4, suy ra
o
min g(m) = g(−4) = −9 max g(m) = g(−1) = 0
−5≤m≤−1
−5≤m≤−1
|
g(m) |
9
VËy
5m≤ ma x≤−1| g(m) | = 9 . V× A =
, vËy max A = ®¹t ®−îc khi m = − 4.
−
2
2
C©u II.
1
) Víi k = 3, ta cã hµm sè
2
y = − 2x + 3 x +1
Hµm sè ®−îc x¸c ®Þnh víi mäi x vµ cã ®¹o hµm
2
3
x
3x − 2 x +1
y' = −2 +
=
.
x2 +1
x2 +1
2
2
2
2
Ta cã y' > 0 ⇔ 3x > 2 x +1 , suy ra x > 0, b×nh ph−¬ng hai vÕ th× ®−îc 9x > 4x + 4 ⇒ x >
, tõ ®ã lËp
5
®
−îc b¶ng biÕn thiªn
−
∞
∞
2
x
+ ∞
+ ∞
5
y'
y
−
0
+
+
5
C¸c tiÖm c©n xiªn cña ®å thÞ :