ข้อสอบ A-Level คณิตประยุกต์ 1 ปี 2560 (วิชาสามัญเก่า)

$$\begin{gathered} \mathrm{จากโจทย์} f\left(x\right) \mathrm{เป็นฟังก์ชันพหุนามดีกรีสาม} \\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{ซึ่งมีค่าวิกฤตที่} x=4 \mathrm{และ} x=-4 \\ \mathrm{แสดงว่า} f\text{'}\left(x\right) \mathrm{ต้องเป็นฟังก์ชันพหุนามดีกรี} 2 \\ \mathrm{จะได้} f\text{'}\left(x\right) = a\left(x-4\right)\left(x+4\right) ; \mathrm{โดย} a\ne 0 \\ = a\left(x^2-16\right) ; \mathrm{ดิฟ} \mathrm{ทั้ง} 2 \mathrm{ข้าง} \\ f\text{'}\text{'}\left(x\right) = a\left(2x\right) \\ = 2ax \end{gathered}$$

$$\begin{gathered} \underline{\mathrm{หา} f\left(x\right)} \\ f\left(x\right) = \int f\text{'}\left(x\right)dx \\ = \int a\left(x^2-16\right)dx \\ = \frac{a{x}^3}{3}-16ax+c \end{gathered}$$

$$\begin{gathered} \underline{\mathrm{พิจารณาข้อความ}} \\ \left(ก\right) f\text{'}\text{'}\left(-4\right)·f\text{'}\text{'}\left(4\right) < 0 \\ \mathrm{จะได้} 2a\left(-4\right)·2a\left(4\right) < 0 \\ -64a^2 < 0 ; \mathrm{โดย} a^2>0 \mathrm{เสมอ} \\ \mathrm{แสดงว่า} -64a^2 < 0 \mathrm{เป็นจริง} \\ \mathrm{ดังนั้น} f\text{'}\text{'}\left(-4\right)·f\text{'}\text{'}\left(4\right) < 0 \mathrm{ข้อ} \left(ก\right) \mathrm{ถูก} \end{gathered}$$

$$\begin{gathered} \left(ข\right) f\left(4\sqrt{3}\right) = 2f\left(0\right) \hspace{0.17em}\hspace{0.17em}\hspace{0.17em} ; \mathrm{โดย} f\left(x\right) = \frac{a{x}^2}{3}-16ax+c \\ \frac{a{\left(4\sqrt{3}\right)}^3}{3}-16a\left(4\sqrt{3}\right)+c = 2\left[\frac{a{\left(0\right)}^3}{3}-16a\left(0\right)+c\right] \\ \cancel{64\sqrt{3}a}-\cancel{64\sqrt{3}a}+c = 2\left(0-0+c\right) \\ c = 2c \mathrm{ไม่เป็นจริง} \\ \mathrm{ดังนั้น} f\left(4\sqrt{3}\right) \ne 2f\left(0\right) \mathrm{ข้อ} \left(ข\right) \mathrm{ผิด} \end{gathered}$$

$$\begin{gathered} \left(ค\right) 𝑓\left(-4\right)+𝑓\left(4\right)=2𝑓\left(0\right) \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} ; \mathrm{โดย} 𝑓\left(x\right)=\frac{a{x}^3}{3}-16ax+c \\ \left[\frac{a{\left(-4\right)}^3}{3}-16a\left(-4\right)+c\right]+\left[\frac{a{\left(4\right)}^3}{3}-16a\left(4\right)+c\right]=2\left[\frac{a{\left(0\right)}^3}{3}-16a\left(0\right)+c\right] \\ \left[\cancel{\frac{-a{\left(4\right)}^3}{3}}+\bcancel{64a}+c\right]+\left[\cancel{\frac{a{\left(4\right)}^3}{3}}-\bcancel{64a}+c\right]=2\left(0+0+c\right) \end{gathered}$$
$$\begin{gathered} c+c=2c \\ 2c=2c \mathrm{เป็นจริง} \\ \mathrm{ดังนั้น} 𝑓\left(-4\right)+𝑓\left(4\right)=2𝑓\left(0\right) \mathrm{ข้อ} \left(ค\right) \mathrm{ถูก} \\ \left(ง\right) \mathrm{ค่าเฉลี่ยเลขคณิตของ} f\left(-2\right), f\left(-1\right) \\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}, f\left(0\right), f\left(1\right), f\left(2\right) \mathrm{เท่ากับ} f\left(0\right) \\ \frac{f\left(-2\right)+f\left(-1\right)+f\left(0\right)+f\left(1\right)+f\left(2\right)}{5} = f\left(0\right) \end{gathered}$$
$$\begin{gathered} \frac{\left[f\left(-2\right)+f\left(2\right)\right]+\left[f\left(-1\right)+f\left(1\right)\right]+f\left(0\right)}{5} = f\left(0\right) \to \boxed{1} \\ \mathrm{จาก} \left(ค\right) \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}f\left(-4\right)+f\left(4\right)=2c \mathrm{แสดงว่า} f\left(a\right)+f\left(-a\right)=2c \\ \mathrm{จาก} \left(ค\right) \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}f\left(0\right)=c \\ \mathrm{จาก} \boxed{1} \mathrm{จะได้} \frac{2c+2c+c}{5}=c \\ \frac{5c}{5}=c \\ c=c \mathrm{เป็นจริง} \\ \mathrm{ดังนั้น} \mathrm{ค่าเฉลี่ยเลขคณิต} \mathrm{เท่ากับ} f\left(0\right) \mathrm{ข้อ} \left(ง\right) \mathrm{ถูก} \end{gathered}$$