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CIE 9709 2024 June Paper 33 Q7

A Level / CIE / P3

CIE 9709 2024 June Paper 33 Paper · Question 7

题目

Problem

Let f(x)=8x3+54x2−17x−21\mathrm{f}(x)=8x^3+54x^2-17x-21.

(a) Show that x+7x+7 is a factor of f(x)\mathrm{f}(x).

(1)

(b) Find the quotient when f(x)\mathrm{f}(x) is divided by x+7x+7.

(2)

(c) Hence solve the equation

8cos⁡3θ+54cos⁡2θ−17cos⁡θ−21=0,8\cos^3\theta+54\cos^2\theta-17\cos\theta-21=0,

for 0∘≤θ≤360∘0^\circ\le\theta\le360^\circ.

(3)
题目中文翻译

设 f(x)=8x3+54x2−17x−21\mathrm{f}(x)=8x^3+54x^2-17x-21。

(a) 证明 x+7x+7 是 f(x)\mathrm{f}(x) 的一个因式。

(1)

(b) 求 f(x)\mathrm{f}(x) 除以 x+7x+7 所得的商。

(2)

(c) 由此解方程

8cos⁡3θ+54cos⁡2θ−17cos⁡θ−21=0,8\cos^3\theta+54\cos^2\theta-17\cos\theta-21=0,

其中 0∘≤θ≤360∘0^\circ\le\theta\le360^\circ。

(3)
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

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因式定理要求验证 f(−7)=0f(-7)=0,需展示代入后的各项及相消。

答题过程

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By the factor theorem,

f(−7)= 8(−7)3+54(−7)2−17(−7)−21= −2744+2646+119−21= 0.\begin{align*} f(-7)=&\,8(-7)^3+54(-7)^2-17(-7)-21\\[4mm] =&\,-2744+2646+119-21\\[4mm] =&\,0. \end{align*}

Therefore x+7 is a factor of f(x)\boxed{x+7\text{ is a factor of }f(x)}.

(b)

解法一

思路

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利用已知因式做多项式除法,再将所得二次式分解。

答题过程

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The successive terms in the quotient are 8x28x^2, −2x-2x and −3-3. The subtractions are

 f(x)−8x2(x+7)= −2x2−17x−21, (−2x2−17x−21)−(−2x)(x+7)= −3x−21, (−3x−21)−(−3)(x+7)=0.\begin{align*} &\,f(x)-8x^2(x+7)\\[4mm] =&\,-2x^2-17x-21,\\[4mm] &\,(-2x^2-17x-21)-(-2x)(x+7)\\[4mm] =&\,-3x-21,\\[4mm] &\,(-3x-21)-(-3)(x+7)=0. \end{align*}

Thus the quotient is 8x2−2x−3\boxed{8x^2-2x-3}.

Factorising this quadratic,

8x3+54x2−17x−21= (x+7)(8x2−2x−3),8x2−2x−3= 8x2+4x−6x−3= 4x(2x+1)−3(2x+1)= (4x−3)(2x+1).\begin{align*} 8x^3+54x^2-17x-21 =&\,(x+7)(8x^2-2x-3),\\[4mm] 8x^2-2x-3 =&\,8x^2+4x-6x-3\\[4mm] =&\,4x(2x+1)-3(2x+1)\\[4mm] =&\,(4x-3)(2x+1). \end{align*}

Hence

f(x)=(x+7)(4x−3)(2x+1).\boxed{f(x)=(x+7)(4x-3)(2x+1)}.

(c)

解法一

思路

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将前一小题的分解中的 xx 换成 cos⁡θ\cos\theta,逐个解因子为零;排除超出余弦值域的根。

答题过程

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Using part (b),

(cos⁡θ+7)(4cos⁡θ−3)(2cos⁡θ+1)=0.(\cos\theta+7)(4\cos\theta-3)(2\cos\theta+1)=0.

The possibility cos⁡θ=−7\cos\theta=-7 is impossible since −1≤cos⁡θ≤1-1\le\cos\theta\le1.

For cos⁡θ=3/4\cos\theta=3/4,

θ=41.4096…∘or360∘−41.4096…∘.\theta=41.4096\ldots^\circ \quad\text{or}\quad 360^\circ-41.4096\ldots^\circ.

For cos⁡θ=−1/2\cos\theta=-1/2,

θ=120∘or240∘.\theta=120^\circ\quad\text{or}\quad240^\circ.

Thus, in the given interval,

θ=41.4∘, 120∘, 240∘, 318.6∘.\boxed{\theta=41.4^\circ,\ 120^\circ,\ 240^\circ,\ 318.6^\circ}.