题目
Problem
The diagram shows the curve y = cos x sin 2 x y=\cos x\sqrt{\sin2x} y = cos x sin 2 x 0 ≤ x ≤ 1 2 π 0\le x\le\frac12\pi 0 ≤ x ≤ 2 1 π M M M x = a x=a x = a
(a) Find the exact value of a a a
(6)
(b) The region enclosed between the x x x 2 π 2\pi 2 π x x x
Find the exact volume of the solid generated.
(5)
题目中文翻译
图中显示了曲线 y = cos x sin 2 x y=\cos x\sqrt{\sin2x} y = cos x sin 2 x 0 ≤ x ≤ 1 2 π 0\le x\le\frac12\pi 0 ≤ x ≤ 2 1 π M M M x = a x=a x = a
(a) 求 a a a
(6)
(b) 将 x x x x x x 2 π 2\pi 2 π
求所生成的立体的体积的精确值。
(5)
原卷通用作答要求
Answer all questions. You should use a calculator where appropriate.
You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.
Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.
请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。
解答
(a)
解法一
思路
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用乘积法则与链式法则求导，在内部极值处令导数为零，再化成一个三角方程。端点函数值为零，可确认内部驻点为最大点。
答题过程
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The chain rule gives
d d x sin 2 x = 2 cos 2 x 2 sin 2 x = cos 2 x sin 2 x . \frac{\mathrm{d}}{\mathrm{d}x}\sqrt{\sin2x}
=\frac{2\cos2x}{2\sqrt{\sin2x}}
=\frac{\cos2x}{\sqrt{\sin2x}}. d x d sin 2 x = 2 sin 2 x 2 cos 2 x = sin 2 x cos 2 x . For 0 < x < π / 2 0<x<\pi/2 0 < x < π /2
d y d x = − sin x sin 2 x + cos x cos 2 x sin 2 x . \frac{\mathrm{d}y}{\mathrm{d}x}
=-\sin x\sqrt{\sin2x}
+\frac{\cos x\cos2x}{\sqrt{\sin2x}}. d x d y = − sin x sin 2 x + sin 2 x cos x cos 2 x . At the stationary point x = a x=a x = a
− sin a sin 2 a + cos a cos 2 a = 0. -\sin a\sin2a+\cos a\cos2a=0. − sin a sin 2 a + cos a cos 2 a = 0. By the cosine addition formula,
cos 3 a = 0. \cos3a=0. cos 3 a = 0. Since 0 < a < π / 2 0<a<\pi/2 0 < a < π /2 0 < 3 a < 3 π / 2 0<3a<3\pi/2 0 < 3 a < 3 π /2 3 a = π / 2 3a=\pi/2 3 a = π /2
The curve is positive inside the interval and zero at both endpoints. Thus this unique interior stationary point is the maximum, and
a = π 6 . \boxed{a=\frac\pi6}. a = 6 π .
(b)
解法一
思路
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旋转体积使用 y 2 y^2 y 2 cos 3 x sin x \cos^3x\sin x cos 3 x sin x
答题过程
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The volume of revolution is
V = π ∫ 0 π / 2 y 2 d x = π ∫ 0 π / 2 cos 2 x sin 2 x d x . \begin{align*}
V=&\,\pi\int_0^{\pi/2}y^2\,\mathrm{d}x\\[4mm]
=&\,\pi\int_0^{\pi/2}\cos^2x\sin2x\,\mathrm{d}x.
\end{align*} V = = π ∫ 0 π /2 y 2 d x π ∫ 0 π /2 cos 2 x sin 2 x d x . Using sin 2 x = 2 sin x cos x \sin2x=2\sin x\cos x sin 2 x = 2 sin x cos x
V = 2 π ∫ 0 π / 2 cos 3 x sin x d x = [ − π 2 cos 4 x ] 0 π / 2 = 0 − ( − π 2 ) = π 2 . \begin{align*}
V=&\,2\pi\int_0^{\pi/2}\cos^3x\sin x\,\mathrm{d}x\\
=&\,\bigg[-\frac\pi2\cos^4x\bigg]_0^{\pi/2}\\
=&\,0-\bigg(-\frac\pi2\bigg)\\
=&\,\boxed{\frac\pi2}.
\end{align*} V = = = = 2 π ∫ 0 π /2 cos 3 x sin x d x [ − 2 π cos 4 x ] 0 π /2 0 − ( − 2 π ) 2 π .
解法二
思路
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用降幂和积化和差把平方函数改写成两项正弦，也能直接求积分。
答题过程
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As above,
V = π ∫ 0 π / 2 cos 2 x sin 2 x d x . V=\pi\int_0^{\pi/2}\cos^2x\sin2x\,\mathrm{d}x. V = π ∫ 0 π /2 cos 2 x sin 2 x d x . Since cos 2 x = ( 1 + cos 2 x ) / 2 \cos^2x=(1+\cos2x)/2 cos 2 x = ( 1 + cos 2 x ) /2 sin 2 x cos 2 x = 1 2 sin 4 x \sin2x\cos2x=\tfrac12\sin4x sin 2 x cos 2 x = 2 1 sin 4 x
cos 2 x sin 2 x = 1 2 sin 2 x + 1 4 sin 4 x . \cos^2x\sin2x
=\frac12\sin2x+\frac14\sin4x. cos 2 x sin 2 x = 2 1 sin 2 x + 4 1 sin 4 x . Therefore
V = π [ − 1 4 cos 2 x − 1 16 cos 4 x ] 0 π / 2 = π [ ( 1 4 − 1 16 ) − ( − 1 4 − 1 16 ) ] = π 2 . \begin{align*}
V=&\,\pi\bigg[-\frac14\cos2x-\frac1{16}\cos4x\bigg]_0^{\pi/2}\\
=&\,\pi\bigg[\bigg(\frac14-\frac1{16}\bigg)
-\bigg(-\frac14-\frac1{16}\bigg)\bigg]\\
=&\,\boxed{\frac\pi2}.
\end{align*} V = = = π [ − 4 1 cos 2 x − 16 1 cos 4 x ] 0 π /2 π [ ( 4 1 − 16 1 ) − ( − 4 1 − 16 1 ) ] 2 π . 
