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IAL 2022 Oct Q8

A Level / Edexcel / P3

IAL 2022 Oct Paper · Question 8

题目

Problem

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

Solutions relying entirely on calculator technology are not acceptable.

(a) Express 8sin⁡x−15cos⁡x8\sin x-15\cos x in the form Rsin⁡(x−α)R\sin(x-\alpha), where R>0R>0 and 0<α<π20<\alpha<\dfrac{\pi}{2}.

Give the exact value of RR, and give the value of α\alpha, in radians, to 4 significant figures.

(3)
f(x)=1541+16sin⁡x−30cos⁡xx>0f(x)=\frac{15}{41+16\sin x-30\cos x}\qquad x>0

(b) Find

(i) the minimum value of f(x)f(x)

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

(4)

(c) State the yy coordinate of the minimum points on the curve with equation

y=2f(x)−5x>0y=2f(x)-5\qquad x>0
(1)

(d) State the smallest value of xx at which a maximum point occurs for the curve with equation

y=−f(2x)x>0y=-f(2x)\qquad x>0
(1)
题目中文翻译

本题中你必须写出解题过程的所有步骤。

不接受完全依赖计算器技术的解法。

(a) 将 8sin⁡x−15cos⁡x8\sin x-15\cos x 写成 Rsin⁡(x−α)R\sin(x-\alpha) 的形式，其中 R>0R>0 且 0<α<π20<\alpha<\dfrac{\pi}{2}。

写出 RR 的精确值，并将 α\alpha 的值（弧度制）精确到小数点后 4 位。

f(x)=1541+16sin⁡x−30cos⁡xx>0f(x)=\frac{15}{41+16\sin x-30\cos x}\qquad x>0

(b) 求

(i) f(x)f(x) 的最小值；

(ii) 使该最小值出现的最小 xx 值。

(c) 写出曲线

y=2f(x)−5x>0y=2f(x)-5\qquad x>0

上极小点的 yy 坐标。

(d) 写出曲线

y=−f(2x)x>0y=-f(2x)\qquad x>0

上极大点首次出现时的最小 xx 值。

解答

(a)

We want

8sin⁡x−15cos⁡x=Rsin⁡(x−α)8\sin x-15\cos x=R\sin(x-\alpha)

Using

sin⁡(x−α)=sin⁡xcos⁡α−cos⁡xsin⁡α\sin(x-\alpha)=\sin x\cos\alpha-\cos x\sin\alpha

we get

Rsin⁡(x−α)=Rcos⁡αsin⁡x−Rsin⁡αcos⁡xR\sin(x-\alpha) =R\cos\alpha\sin x-R\sin\alpha\cos x

Compare coefficients with

8sin⁡x−15cos⁡x8\sin x-15\cos x

So

Rcos⁡α=8R\cos\alpha=8

and

Rsin⁡α=15R\sin\alpha=15

Therefore

R=82+152=17R=\sqrt{8^2+15^2}=17

Also,

tan⁡α=158\tan\alpha=\frac{15}{8}

Hence

α=tan⁡−1158=1.0808…\alpha=\tan^{-1}\frac{15}{8}=1.0808\ldots

So

8sin⁡x−15cos⁡x=17sin⁡(x−α)\boxed{8\sin x-15\cos x=17\sin(x-\alpha)}

where

R=17,α=1.081\boxed{R=17,\qquad \alpha=1.081}

to 4 significant figures.

(b)(i)

The denominator of f(x)f(x) is

41+16sin⁡x−30cos⁡x41+16\sin x-30\cos x

This can be written as

41+2(8sin⁡x−15cos⁡x)41+2(8\sin x-15\cos x)

Using part (a),

41+16sin⁡x−30cos⁡x=41+34sin⁡(x−α)41+16\sin x-30\cos x =41+34\sin(x-\alpha)

The minimum value of f(x)f(x) occurs when the denominator is as large as possible.

Since

sin⁡(x−α)≤1\sin(x-\alpha)\le1

the maximum denominator is

41+34=7541+34=75

Therefore

fmin⁡=1575=15f_{\min}=\frac{15}{75}=\frac15

So the minimum value is

15\boxed{\frac15}

(b)(ii)

The minimum occurs when

sin⁡(x−α)=1\sin(x-\alpha)=1

So the smallest positive value satisfies

x−α=π2x-\alpha=\frac{\pi}{2}

Thus

x=π2+αx=\frac{\pi}{2}+\alpha

Using

α=1.0808…\alpha=1.0808\ldots

we get

x=2.6516…x=2.6516\ldots

Therefore

x=2.65\boxed{x=2.65}

to 3 significant figures.

(c)

For

y=2f(x)−5y=2f(x)-5

the minimum yy value occurs when f(x)f(x) is minimum.

Since

fmin⁡=15f_{\min}=\frac15

we get

ymin⁡=2(15)−5y_{\min}=2\left(\frac15\right)-5

So

y=−235\boxed{y=-\frac{23}{5}}

(d)

For

y=−f(2x)y=-f(2x)

a maximum occurs when f(2x)f(2x) is minimum.

From part (b), the smallest input giving the minimum of ff is

2.6516…2.6516\ldots

So

2x=2.6516…2x=2.6516\ldots

and hence

x=1.3258…x=1.3258\ldots

Therefore the smallest value of xx is

1.33\boxed{1.33}

to 3 significant figures.