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(2010•南通模拟)设fn(x)=12+rcosx+r2cos2x+r3cos4x+…+rn-1cos2n-2x(n≥2).(1)证明:对任意x∈R,当|r|≤12时,rcosx+r2cos2x≥-38;(2)证明:当|r|≤12,f2n+1(x)对任意x∈R和自然数n(n≥2)都有f2n+1

题目详情
(2010•南通模拟)设fn(x)=
1
2
+rcosx+r2cos2x+r3cos4x+…+rn-1cos2n-2x(n≥2).
(1)证明:对任意x∈R,当|r|≤
1
2
时,rcosx+r2cos2x≥-
3
8

(2)证明:当|r|≤
1
2
,f2n+1(x)对任意x∈R和自然数n(n≥2)都有f2n+1(x)>0.
▼优质解答
答案和解析
(1)1°当r=0时,显然0≥-
3
8

2°当r≠0时,设φ(x)=rcosx+r2cos2x=r2(2cos2x-1)+rcosx
=2r2(cosx+
1
4r
)2-
1
8
-r2≥-
1
8
-r2≥-
1
8
-(
1
2
)2=-
3
8
.(|r|≤
1
2


(2)当|r|≤
1
2
时,∀x∈R,∀n∈N*(n≥2),f2n+1=
1
2
+rcosx+r2cos2x+r3cos4x+r4cos8x++r2n-1cos22(n-1)x+r2ncos22n-1x
=
1
2
+φ(x)+r2φ(4x)++r2(n-1)•φ(4n-1x)
1
2
-
3
8
(1+r2++r2(n-1))
1
2
-
3
8
(1+
1
4
++
1
4n-1
)
=
1
2
-
3
8
1-
1
4n
1-
1
4
=
1
2•4n
=
1
22n+1
>0.