已知函数
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的图象过坐标原点O,且在点(-1,f(-1))处的切线的斜率是-5.
(Ⅰ)求实数b,c的值;
(Ⅱ)求f(x)在区间[-1,2]上的最大值;
(Ⅲ)对任意给定的正实数a,曲线y=f(x)上是否存在两点P、Q,使得△POQ是以O为直角顶点的直角三角形,且此三角形斜边中点在y轴上?说明理由.
考点分析:
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长沙市某棚户区改造建筑用地平面示意图如图所示.经规划调研确定,棚改规划建筑用地区域近似地为半径是R的圆面.该圆面的内接四边形ABCD是原棚户建筑用地,测量可知边界AB=AD=4万米,BC=6万米,CD=2万米.
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在△ABC中,角A、B、C的对边分别为a、b、c,向量
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=(a-c,sinC-sinB),满足|
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+
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|=|
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|.
(Ⅰ)求角B的大小;
(Ⅱ)设
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),
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=(2k,cos2A) (k>1),
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•
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有最大值为3,求k的值.
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在平面直角坐标系中,角α,β的始边为x轴的非负半轴,点P(1,2cos
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=-1.
(1)求cos2θ;
(2)求P,Q的坐标并求cos(α-β)的值.
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已知数列{a
n}的前n项和S
n=2-a
n,
(1)求数列{a
n}的通项公式;
(2)求数列{S
n}的前项和.
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已知数列{a
n}的通项公式为a
n=(2n-1)•2
n,我们用错位相减法求其前n项和S
n:由S
n=1×2+3×2
2+5×2
3+…(2n-1)•2
n得2S
n=1×2
2+3×2
3+5×2
4+…(2n-1)•2
n+1,两式项减得:-S
n=2+2×2
2+2×2
3+…+2×2
n-(2n-1)•2
n+1,求得S
n=(2n-3)•2
n+1+6.类比推广以上方法,若数列{b
n}的通项公式为b
n=n
2•2
n,
则其前n项和T
n=
.
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