ข้อสอบคณิต 2 - มีนาคม 2563

$$\begin{gathered} \mathrm{จากสมการพาราโบลา} \\ y = a{\left(x-h\right)}^2+k \\ \mathrm{จะได้} f\left(x\right) = a{\left(x-0\right)}^2+0 = a{x}^2 \\ g\left(x\right) = b{\left(x-1\right)}^2+4 \end{gathered}$$

$$\begin{gathered} \mathrm{จาก} f\left(x\right) \le g\left(x\right) \\ a{x}^2 \le b{\left(x-1\right)}^2+4 \\ a{x}^2-b{\left(x-1\right)}^2-4 \le 0 \end{gathered}$$

$$\begin{gathered} \mathrm{ให้} x = 0 \to a{\left(0\right)}^2-b{\left(0-1\right)}^2-4 = 0 \\ b = -4 \\ g\left(x\right) = -4{\left(x-1\right)}^2+4 \end{gathered}$$

$$\begin{gathered} \mathrm{ให้} x=1 \to a{\left(1\right)}^2-\left(-4\right){\left(1-1\right)}^2-4 = 0 \\ a-4 = 0 \\ a = 4 \\ f\left(x\right) = 4x^2 \\ \mathrm{ดังนั้น} f\left(-1\right)+g\left(-1\right) = 4-12 = -8 \to \mathrm{ตอบ} 1 \end{gathered}$$