CalcGospel 國際數學圖譜
返回

IAL 2021 June Q7

A Level / Edexcel / P3

IAL 2021 June Paper · Question 7

题目

Problem

Given that

x=6sin⁡22y0<y<π4x=6\sin^2 2y\qquad 0<y<\frac{\pi}{4}

show that

dydx=1ABx−x2\frac{dy}{dx}=\frac{1}{A\sqrt{Bx-x^2}}

where AA and BB are integers to be found.

(5)
题目中文翻译

已知

x=6sin⁡22y0<y<π4x=6\sin^2 2y\qquad 0<y<\frac{\pi}{4}

证明

dydx=1ABx−x2\frac{dy}{dx}=\frac{1}{A\sqrt{Bx-x^2}}

其中 AA 和 BB 为待求整数。

解答

We are given

x=6sin⁡22yx=6\sin^2 2y

Differentiate with respect to yy:

dxdy=6⋅2sin⁡2ycos⁡2y⋅2\frac{dx}{dy}=6\cdot 2\sin2y\cos2y\cdot 2

So

dxdy=24sin⁡2ycos⁡2y\frac{dx}{dy}=24\sin2y\cos2y

Therefore

dydx=124sin⁡2ycos⁡2y\frac{dy}{dx}=\frac{1}{24\sin2y\cos2y}

Now express sin⁡2y\sin2y and cos⁡2y\cos2y in terms of xx.

Since

x=6sin⁡22yx=6\sin^2 2y

we have

sin⁡22y=x6\sin^2 2y=\frac{x}{6}

Because

0<y<π40<y<\frac{\pi}{4}

we have

0<2y<π20<2y<\frac{\pi}{2}

so sin⁡2y>0\sin2y>0 and cos⁡2y>0\cos2y>0.

Thus

sin⁡2y=x6\sin2y=\sqrt{\frac{x}{6}}

Also,

cos⁡22y=1−sin⁡22y\cos^2 2y=1-\sin^2 2y

so

cos⁡22y=1−x6=6−x6\cos^2 2y=1-\frac{x}{6}=\frac{6-x}{6}

Hence

cos⁡2y=6−x6\cos2y=\sqrt{\frac{6-x}{6}}

Therefore

sin⁡2ycos⁡2y=x66−x6=x(6−x)6\sin2y\cos2y =\sqrt{\frac{x}{6}}\sqrt{\frac{6-x}{6}} =\frac{\sqrt{x(6-x)}}{6}

So

dydx=124⋅x(6−x)6\frac{dy}{dx} =\frac{1}{24\cdot \dfrac{\sqrt{x(6-x)}}{6}}

Hence

dydx=14x(6−x)\frac{dy}{dx} =\frac{1}{4\sqrt{x(6-x)}}

Since

x(6−x)=6x−x2x(6-x)=6x-x^2

we get

dydx=146x−x2\frac{dy}{dx}=\frac{1}{4\sqrt{6x-x^2}}

Therefore

A=4,B=6\boxed{A=4,\qquad B=6}