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IAL 2025 Oct Q4

A Level / Edexcel / P3

IAL 2025 Oct Paper · Question 4

题目

Problem

In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.

(a)

f(x)=3sin⁡2x−3cos⁡2xf(x)=\sqrt{3}\sin 2x-3\cos 2x

Express f(x)f(x) in the form Rsin⁡(2x−α)R\sin(2x-\alpha), where RR and α\alpha are constants, R>0R>0 and 0<α<π20<\alpha<\dfrac{\pi}{2}.

Give the exact value of RR and the exact value of α\alpha.

(3)

(b)

g(x)=18f(3x)+43x>0g(x)=\frac{18}{f(3x)+4\sqrt{3}}\qquad x>0

Using the answer to part (a), find

(i) the exact minimum value of g(x)g(x),

(ii) the smallest value of xx for which this minimum value occurs.

You must make your method clear.

(3)
题目中文翻译

本题必须写出全部解题步骤。

不接受完全依赖计算器技术的解法。

(a)

f(x)=3sin⁡2x−3cos⁡2xf(x)=\sqrt{3}\sin 2x-3\cos 2x

将 f(x)f(x) 写成 Rsin⁡(2x−α)R\sin(2x-\alpha) 的形式，其中 R,αR,\alpha 为常数，且 R>0, 0<α<π2R>0,\ 0<\alpha<\dfrac{\pi}{2}。

写出 RR 的精确值与 α\alpha 的精确值。

(b)

g(x)=18f(3x)+43x>0g(x)=\frac{18}{f(3x)+4\sqrt{3}}\qquad x>0

利用(a)的结果，求

(i) g(x)g(x) 的精确最小值；

(ii) 取得该最小值时最小的 xx 值。

必须清楚写出你的方法。

解答

(a)

We want

3sin⁡2x−3cos⁡2x=Rsin⁡(2x−α)\sqrt3\sin2x-3\cos2x=R\sin(2x-\alpha)

Expand the right side:

Rsin⁡(2x−α)=Rsin⁡2xcos⁡α−Rcos⁡2xsin⁡αR\sin(2x-\alpha)=R\sin2x\cos\alpha-R\cos2x\sin\alpha

Compare coefficients:

Rcos⁡α=3R\cos\alpha=\sqrt3

and

Rsin⁡α=3R\sin\alpha=3

Square and add:

R2=(3)2+32=3+9=12R^2=(\sqrt3)^2+3^2=3+9=12

So

R=23R=2\sqrt3

Also,

tan⁡α=Rsin⁡αRcos⁡α=33=3\tan\alpha=\frac{R\sin\alpha}{R\cos\alpha} =\frac{3}{\sqrt3} =\sqrt3

Since 0<α<π20<\alpha<\frac{\pi}{2},

α=π3\alpha=\frac{\pi}{3}

Therefore

f(x)=23sin⁡(2x−π3)\boxed{f(x)=2\sqrt3\sin\left(2x-\frac{\pi}{3}\right)}

where

R=23,α=π3\boxed{R=2\sqrt3,\quad \alpha=\frac{\pi}{3}}

(b)

Using part (a),

f(3x)=23sin⁡(6x−π3)f(3x)=2\sqrt3\sin\left(6x-\frac{\pi}{3}\right)

So

g(x)=1823sin⁡(6x−π3)+43g(x)=\frac{18}{2\sqrt3\sin\left(6x-\frac{\pi}{3}\right)+4\sqrt3}

For g(x)g(x) to be as small as possible, its denominator must be as large as possible.

The maximum value of

sin⁡(6x−π3)\sin\left(6x-\frac{\pi}{3}\right)

is 11.

Therefore the maximum denominator is

23+43=632\sqrt3+4\sqrt3=6\sqrt3

So the exact minimum value of g(x)g(x) is

1863=3\frac{18}{6\sqrt3}=\sqrt3

Hence

minimum value of g(x)=3\boxed{\text{minimum value of }g(x)=\sqrt3}

For this minimum to occur,

sin⁡(6x−π3)=1\sin\left(6x-\frac{\pi}{3}\right)=1

The smallest positive solution occurs when

6x−π3=π26x-\frac{\pi}{3}=\frac{\pi}{2}

So

6x=π2+π36x=5π6x=5π36\begin{aligned} 6x&=\frac{\pi}{2}+\frac{\pi}{3} \\ 6x&=\frac{5\pi}{6} \\ x&=\frac{5\pi}{36} \end{aligned}

Therefore

x=5π36\boxed{x=\frac{5\pi}{36}}