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LuyÖn thi trªn m¹ng – Phiªn b¶n 1.0
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C©u I.
) Khi m = 0 hµm cã d¹ng y = x − 3x − 9x . §Ò nghÞ b¹n ®äc tù kh¶o s¸t vµ vÏ ®å thÞ.
3
2
1
2
) Khi ®ã ®iÓm x ph¶i lµ ®iÓm uèn cña ®å thÞ. V× vËy ta buéc cho y''(x ) = 0 sÏ ®−îc x ⇔
2
2
2
6
x − 6 = 0 ⇒ x = 1.
2
y(x ) = y(1) = 0 ⇔ −11 + m = 0 ⇒ m = 11. Víi m = 11 hµm
2
cã d¹ng :
3
2
2
3
) y = x − 3x − 9x +11= (x −1)(x − 2x −11) . Khi ®ã ®å thÞ sÏ c¾t trôc hoµnh t¹i ba ®iÓm
x =1− 2 3 ; x = 1 ; x =1+ 2 3 .
1
2
3
C©u II.
1
) T×m c¸c gi¸ trÞ cña x ∈ (0 ; 2π) tháa m·n ph−¬ng tr×nh
sin3x − sinx
=
− cos2x
sin2x + cos2x .
1
ViÕt l¹i ph−¬ng tr×nh : 2
cos2xsinx
= 2cos 2x −
π
.
2
sinx
4
π
Víi 0 < x < π th× cã : cos2x = cos 2x −
.
4
π
9π
Gi¶i ra sÏ ®−îc x =
vµ x =
.
1
2
1
6
16
π
Víi π < x < 2π th× cã : cos2x = −cos 2x −
.
4
Gi¶i ra sÏ ®−îc :
2
1π
29π
x3 =
vµ x =
.
4
1
6
16
2
) Gäi giao cña hai trung tuyÕn
lµ G.
Ta cã :
2 2 2 2
3AG) + a = 2(c + b )
(
2 2 2 2
(
3BG) + b = 2(c + a )
Tõ ®ã :
2
2
2
2
2
9
(AG + BG ) = 4c + a + b ,
2
2
2
AG + BG = AB ⇔ AA ⊥ BB .
1
1
2
2
2
2
VËy AA ⊥ BB ⇔ 9c = 4c + a + b ⇔
1
1
2
2
2
2
⇔
a + b = 5c ⇔ 2abcosC = 4c ⇔
2
abcosC 4c2
⇔
=
⇔ 2cotgC =
chC
absinC
4
(h cotgA + h cotgB)
C C
= 4(cotgA + cotgB)
=
hC