题目
Problem
(i)
f(x)=x−3(2x+5)2x=3
(a) Find f′(x) in the form Q(x)P(x) where P(x) and Q(x) are fully factorised quadratic expressions.
(b) Hence find the range of values of x for which f(x) is increasing.
(6)
(ii)
g(x)=xsin4x0≤x<4π
The curve with equation y=g(x) has a maximum at the point M.
Show that the x coordinate of M satisfies the equation
tan4x+kx=0
where k is a constant to be found.
(5)
题目中文翻译
(i)
f(x)=x−3(2x+5)2x=3
(a) 求 f′(x),并写成 Q(x)P(x) 的形式,其中 P(x) 和 Q(x) 都是完全因式分解后的二次式。
(b) 由此求 f(x) 为增函数时 x 的取值范围。
(ii)
g(x)=xsin4x0≤x<4π
曲线 y=g(x) 在点 M 处取得最大值。
证明点 M 的 x 坐标满足方程
tan4x+kx=0
其中 k 为待求常数。
解答
(i)(a)
解法一
思路
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用商法则求导,再从分子提取公因式。
答题过程
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We have
f(x)=x−3(2x+5)2
Use the quotient rule:
f′(x)=(x−3)2(x−3)dxd(2x+5)2−(2x+5)2dxd(x−3)
Now
dxd(2x+5)2=4(2x+5)
and
dxd(x−3)=1
So
f′(x)=(x−3)24(2x+5)(x−3)−(2x+5)2=(x−3)2(2x+5)[4(x−3)−(2x+5)]=(x−3)2(2x+5)(4x−12−2x−5)=(x−3)2(2x+5)(2x−17)
Therefore
f′(x)=(x−3)2(2x+5)(2x−17)
解法二
思路
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先将分式拆成多项式与一个简单分式,逐项求导后再通分。
答题过程
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Since 2x+5=2(x−3)+11, expanding the square and dividing by x−3 gives
f(x)=4(x−3)+44+x−3121.
Differentiating and putting the result over a common denominator,
f′(x)===4−(x−3)2121(x−3)2[2(x−3)]2−112(x−3)2(2x+5)(2x−17).
(i)(b)
For f(x) to be increasing,
f′(x)>0
From part (a),
f′(x)=(x−3)2(2x+5)(2x−17)
Since
(x−3)2>0
for x=3, the sign of f′(x) depends on
(2x+5)(2x−17)
So we solve
(2x+5)(2x−17)>0
The critical values are
x=−25,x=217
The product is positive outside these two values. Therefore
x<−25orx>217
(ii)
解法一
思路
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对乘积和根式分别用乘积法则与链式法则。极大值点在区间内部,且该点的余弦因子非零。
答题过程
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Let xM be the x-coordinate of M. Since g(0)=0 and g(x)>0 for 0<x<π/4, the maximum occurs at an interior point.
Using the product rule and chain rule,
g′(x)=sin4x+sin4x2xcos4x(0<x<π/4).
At M, set g′(xM)=0 and multiply by sin4xM:
sin4xM+2xMcos4xM=0.
If cos4xM=0, then xM=π/8 and the left-hand side is 1, so this is impossible. Dividing by cos4xM gives
tan4xM+2xM=0,
so k=2.
解法二
思路
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函数在给定区间非负,最大值的位置在平方后不变。先平方,可以避免对根式直接求导。
答题过程
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On the given interval g(x)≥0. Its maximum occurs at the same point as the maximum of
H(x)=[g(x)]2=x2sin4x.
The maximum is interior because H(0)=0 and H(x)>0 for 0<x<π/4. At its x-coordinate xM,
0=H′(xM)=2xMsin4xM+4xM2cos4xM.
As xM>0, divide by 2xM to get sin4xM+2xMcos4xM=0. The cosine factor cannot be zero, since then the sine would be 1. Therefore
tan4xM+2xM=0,k=2.