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IAL 2021 June Q9

A Level / Edexcel / P3

IAL 2021 June Paper · Question 9

题目

Problem

(a) Express 12sin⁡x−5cos⁡x12\sin x-5\cos x in the form Rsin⁡(x−α)R\sin(x-\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 give the value of α\alpha in radians, to 3 decimal places.

(3)

The function gg is defined by

g(θ)=10+12sin⁡(2θ−π6)−5cos⁡(2θ−π6)θ>0g(\theta)=10+12\sin\left(2\theta-\frac{\pi}{6}\right)-5\cos\left(2\theta-\frac{\pi}{6}\right)\qquad \theta>0

Find

(b) (i) the minimum value of g(θ)g(\theta)

(ii) the smallest value of θ\theta at which the minimum value occurs.

(3)

The function hh is defined by

h(β)=10−(12sin⁡β−5cos⁡β)2h(\beta)=10-(12\sin\beta-5\cos\beta)^2

(c) Find the range of hh.

(2)
题目中文翻译

(a) 将 12sin⁡x−5cos⁡x12\sin x-5\cos x 化为 Rsin⁡(x−α)R\sin(x-\alpha) 的形式，其中 RR 和 α\alpha 为常数，且 R>0, 0<α<π2R>0,\ 0<\alpha<\dfrac{\pi}{2}。写出 RR 的精确值，并将 α\alpha 的弧度值精确到小数点后 3 位。

函数 gg 定义为

g(θ)=10+12sin⁡(2θ−π6)−5cos⁡(2θ−π6)θ>0g(\theta)=10+12\sin\left(2\theta-\frac{\pi}{6}\right)-5\cos\left(2\theta-\frac{\pi}{6}\right)\qquad \theta>0

求

(b) (i) g(θ)g(\theta) 的最小值；

(ii) 取得该最小值时最小的 θ\theta 值。

函数 hh 定义为

h(β)=10−(12sin⁡β−5cos⁡β)2h(\beta)=10-(12\sin\beta-5\cos\beta)^2

(c) 求 hh 的值域。

解答

(a)

We want

12sin⁡x−5cos⁡x=Rsin⁡(x−α)12\sin x-5\cos x=R\sin(x-\alpha)

Expand the right hand side:

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

Compare coefficients:

Rcos⁡α=12,Rsin⁡α=5R\cos\alpha=12,\qquad R\sin\alpha=5

Therefore

R2=122+52=169R^2=12^2+5^2=169

so

R=13R=13

Also,

tan⁡α=512\tan\alpha=\frac{5}{12}

Since

0<α<π20<\alpha<\frac{\pi}{2}

we get

α=arctan⁡512=0.395\alpha=\arctan\frac{5}{12}=0.395

to 3 decimal places.

Thus

12sin⁡x−5cos⁡x=13sin⁡(x−0.395)\boxed{12\sin x-5\cos x=13\sin(x-0.395)}

where

R=13\boxed{R=13}

(b)

Using part (a),

12sin⁡u−5cos⁡u=13sin⁡(u−α)12\sin u-5\cos u=13\sin(u-\alpha)

where

α=arctan⁡512\alpha=\arctan\frac{5}{12}

For

u=2θ−π6u=2\theta-\frac{\pi}{6}

we get

g(θ)=10+13sin⁡(2θ−π6−α)g(\theta)=10+13\sin\left(2\theta-\frac{\pi}{6}-\alpha\right)

(i)

The minimum value of sin⁡\sin is −1-1.

Therefore

gmin⁡=10+13(−1)g_{\min}=10+13(-1)

So

gmin⁡=−3\boxed{g_{\min}=-3}

(ii)

The minimum occurs when

sin⁡(2θ−π6−α)=−1\sin\left(2\theta-\frac{\pi}{6}-\alpha\right)=-1

The smallest positive solution occurs when

2θ−π6−α=3π22\theta-\frac{\pi}{6}-\alpha=\frac{3\pi}{2}

Therefore

2θ=3π2+π6+α2\theta=\frac{3\pi}{2}+\frac{\pi}{6}+\alpha

So

θ=12(3π2+π6+α)\theta=\frac12\left(\frac{3\pi}{2}+\frac{\pi}{6}+\alpha\right)

Using

α=0.394791…\alpha=0.394791\ldots

we get

θ=2.82366…\theta=2.82366\ldots

Hence

θ=2.82\boxed{\theta=2.82}

to 3 significant figures.

(c)

From part (a),

12sin⁡β−5cos⁡β=13sin⁡(β−α)12\sin\beta-5\cos\beta=13\sin(\beta-\alpha)

So

h(β)=10−(13sin⁡(β−α))2h(\beta)=10-\left(13\sin(\beta-\alpha)\right)^2

Therefore

h(β)=10−169sin⁡2(β−α)h(\beta)=10-169\sin^2(\beta-\alpha)

Since

0≤sin⁡2(β−α)≤10\leq \sin^2(\beta-\alpha)\leq 1

we have

0≤169sin⁡2(β−α)≤1690\leq 169\sin^2(\beta-\alpha)\leq 169

Therefore

10−169≤h(β)≤1010-169\leq h(\beta)\leq 10

Hence the range is

−159≤h(β)≤10\boxed{-159\leq h(\beta)\leq 10}