题目
Problem
The diagram shows the graph of y = 5 sin 2 x cos 2 x y=5\sin2x\cos^2x y = 5 sin 2 x cos 2 x 0 ≤ x ≤ 1 2 π 0\le x\le\frac12\pi 0 ≤ x ≤ 2 1 π M M M
(a) Find the exact x x x M M M
(6)
(b) By using the substitution u = cos x u=\cos x u = cos x x x x x = 0 x=0 x = 0 x = 1 4 π x=\frac14\pi x = 4 1 π x = 1 4 π x=\frac14\pi x = 4 1 π
(5)
题目中文翻译
图中显示了 y = 5 sin 2 x cos 2 x y=5\sin2x\cos^2x y = 5 sin 2 x cos 2 x 0 ≤ x ≤ 1 2 π 0\le x\le\frac12\pi 0 ≤ x ≤ 2 1 π M M M
(a) 求 M M M x x x
(6)
(b) 使用换元 u = cos x u=\cos x u = cos x x x x x = 0 x=0 x = 0 x = 1 4 π x=\frac14\pi x = 4 1 π x = 1 4 π x=\frac14\pi x = 4 1 π
(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
展开
用乘积法则求导，再提出内部区间恒正的 cos 2 x \cos^2x cos 2 x sin 2 x \sin^2x sin 2 x
答题过程
展开
Using the product rule,
d y d x = 10 cos 2 x cos 2 x − 10 sin 2 x sin x cos x = 10 cos 2 x ( cos 2 x − 2 sin 2 x ) = 10 cos 2 x ( 1 − 4 sin 2 x ) . \begin{align*}
\frac{\mathrm{d}y}{\mathrm{d}x}
=&\,10\cos2x\cos^2x\\[4mm]
&\,\hspace{2pt}-10\sin2x\sin x\cos x\\[4mm]
=&\,10\cos^2x(\cos2x-2\sin^2x)\\
=&\,10\cos^2x(1-4\sin^2x).
\end{align*} d x d y = = = 10 cos 2 x cos 2 x − 10 sin 2 x sin x cos x 10 cos 2 x ( cos 2 x − 2 sin 2 x ) 10 cos 2 x ( 1 − 4 sin 2 x ) . For an interior stationary point, cos x > 0 \cos x>0 cos x > 0
1 − 4 sin 2 x = 0 ⟹ sin x = 1 2 . 1-4\sin^2x=0
\quad\Longrightarrow\quad \sin x=\frac12. 1 − 4 sin 2 x = 0 ⟹ sin x = 2 1 . Thus x = π / 6 x=\pi/6 x = π /6
x M = π 6 . \boxed{x_M=\frac\pi6}. x M = 6 π .
解法二
思路
展开
先降幂并积化和差，把乘积化成两项正弦，再求导。所得方程是关于 cos 2 x \cos2x cos 2 x
答题过程
展开
Using cos 2 x = ( 1 + cos 2 x ) / 2 \cos^2x=(1+\cos2x)/2 cos 2 x = ( 1 + cos 2 x ) /2
y = 5 2 sin 2 x ( 1 + cos 2 x ) = 5 2 sin 2 x + 5 2 sin 2 x cos 2 x = 5 2 sin 2 x + 5 4 sin 4 x . \begin{align*}
y=&\,\frac52\sin2x(1+\cos2x)\\[4mm]
=&\,\frac52\sin2x+\frac52\sin2x\cos2x\\[4mm]
=&\,\frac52\sin2x+\frac54\sin4x.
\end{align*} y = = = 2 5 sin 2 x ( 1 + cos 2 x ) 2 5 sin 2 x + 2 5 sin 2 x cos 2 x 2 5 sin 2 x + 4 5 sin 4 x . Hence
d y d x = 5 cos 2 x + 5 cos 4 x . \frac{\mathrm{d}y}{\mathrm{d}x}=5\cos2x+5\cos4x. d x d y = 5 cos 2 x + 5 cos 4 x . At an interior stationary point, put c = cos 2 x c=\cos2x c = cos 2 x cos 4 x = 2 c 2 − 1 \cos4x=2c^2-1 cos 4 x = 2 c 2 − 1
2 c 2 + c − 1 = 0 ⟹ ( 2 c − 1 ) ( c + 1 ) = 0. 2c^2+c-1=0
\quad\Longrightarrow\quad (2c-1)(c+1)=0. 2 c 2 + c − 1 = 0 ⟹ ( 2 c − 1 ) ( c + 1 ) = 0. The root c = − 1 c=-1 c = − 1 x = π / 2 x=\pi/2 x = π /2 c = 1 / 2 c=1/2 c = 1/2 x = π / 6 x=\pi/6 x = π /6
The curve is positive inside the interval and zero at both endpoints, so its unique interior stationary point is the maximum:
x M = π 6 . \boxed{x_M=\frac\pi6}. x M = 6 π .
(b)
解法一
思路
展开
先用二倍角把函数化成正弦乘余弦三次方，再按题目指定换元。上下限顺序会倒转。
答题过程
展开
The curve is nonnegative on [ 0 , π / 4 ] [0,\pi/4] [ 0 , π /4 ]
A = ∫ 0 π / 4 5 sin 2 x cos 2 x d x = 10 ∫ 0 π / 4 sin x cos 3 x d x . \begin{align*}
A=&\,\int_0^{\pi/4}5\sin2x\cos^2x\,\mathrm{d}x\\[4mm]
=&\,10\int_0^{\pi/4}\sin x\cos^3x\,\mathrm{d}x.
\end{align*} A = = ∫ 0 π /4 5 sin 2 x cos 2 x d x 10 ∫ 0 π /4 sin x cos 3 x d x . Let u = cos x u=\cos x u = cos x d u = − sin x d x \mathrm{d}u=-\sin x\,\mathrm{d}x d u = − sin x d x u = 1 u=1 u = 1 u = 2 / 2 u=\sqrt2/2 u = 2 /2
Therefore
A = − 10 ∫ 1 2 / 2 u 3 d u = − 5 2 [ ( 2 2 ) 4 − 1 ] = − 5 2 ( 1 4 − 1 ) = 15 8 . \begin{align*}
A=&\,-10\int_1^{\sqrt2/2}u^3\,\mathrm{d}u\\
=&\,-\frac52\bigg[\bigg(\frac{\sqrt2}{2}\bigg)^4-1\bigg]\\
=&\,-\frac52\bigg(\frac14-1\bigg)\\
=&\,\boxed{\frac{15}{8}}.
\end{align*} A = = = = − 10 ∫ 1 2 /2 u 3 d u − 2 5 [ ( 2 2 ) 4 − 1 ] − 2 5 ( 4 1 − 1 ) 8 15 . 
