Worked Solutions to the AP-Style Practice SetAP 风格练习题配套解析
Topics 2.1 - 2.10考点 2.1 至 2.10AB
The average rate of change of $f(x)=x^{2}+2x$ on $[1,3]$ is$f(x)=x^{2}+2x$ 在 $[1,3]$ 上的平均变化率为
The average rate of change of $f$ on $[1,3]$ is $\dfrac{f(3)-f(1)}{3-1}$, where $f(3)=3^{2}+2(3)=15$ and $f(1)=1^{2}+2(1)=3$. (M1)$f$ 在 $[1,3]$ 上的平均变化率为 $\dfrac{f(3)-f(1)}{3-1}$，其中 $f(3)=3^{2}+2(3)=15$，$f(1)=1^{2}+2(1)=3$。(M1)
So the average rate of change is $\dfrac{15-3}{2}=6$. (A1)故平均变化率为 $\dfrac{15-3}{2}=6$。(A1)
$\displaystyle\lim_{h\to 0}\dfrac{(3+h)^{2}-9}{h}$ equals$\displaystyle\lim_{h\to 0}\dfrac{(3+h)^{2}-9}{h}$ 等于
This expression matches $\lim_{h\to 0}\frac{f(3+h)-f(3)}{h}$ for $f(x)=x^{2}$, so it equals $f'(3)$. (M1)此式与 $\lim_{h\to 0}\frac{f(3+h)-f(3)}{h}$（取 $f(x)=x^{2}$）形式一致，故等于 $f'(3)$。(M1)
Expanding directly, $(3+h)^{2}-9=6h+h^{2}$, so the quotient is $6+h\to 6$ as $h\to 0$. (A1)直接展开，$(3+h)^{2}-9=6h+h^{2}$，故商式为 $6+h\to 6$（当 $h\to 0$）。(A1)
If $f$ is differentiable at $x=a$, which of the following must be true?若 $f$ 在 $x=a$ 处可微，则下列哪项必然成立？
Differentiability at $x=a$ requires the limit $\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}$ to exist. (M1)在 $x=a$ 处可微要求极限 $\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}$ 存在。(M1)
A standard theorem states that if this limit exists, $f$ must be continuous at $x=a$: differentiability is a strictly stronger condition than continuity. Options (B), (C), (D) can all fail (e.g. $f(x)=x^{3}$ is differentiable at $0$ with no extremum, $f(a)$ need not be $0$, and $f''(a)$ need not exist). (A1)一条标准定理指出，若该极限存在，则 $f$ 在 $x=a$ 处必连续：可微性严格强于连续性。选项 (B)(C)(D) 均可能不成立（例如 $f(x)=x^{3}$ 在 $0$ 处可微但无极值，$f(a)$ 未必为 $0$，$f''(a)$ 也未必存在）。(A1)
Which of the following is NOT differentiable at $x=0$?下列哪个函数在 $x=0$ 处不可微？
$f(x)=x^{2}$, $\sin x$, and $x^{3}$ all have one-sided difference quotients that agree at $x=0$, giving $f'(0)=0,1,0$ respectively. (M1)$f(x)=x^{2}$、$\sin x$、$x^{3}$ 在 $x=0$ 处的单侧差商均相等，分别给出 $f'(0)=0,1,0$。(M1)
For $f(x)=|x|$: $\lim_{h\to 0^{-}}\frac{|h|}{h}=-1$ while $\lim_{h\to 0^{+}}\frac{|h|}{h}=1$. Since these disagree, $f'(0)$ does not exist: this is the corner at the origin. (A1)对 $f(x)=|x|$：$\lim_{h\to 0^{-}}\frac{|h|}{h}=-1$，而 $\lim_{h\to 0^{+}}\frac{|h|}{h}=1$。二者不相等，故 $f'(0)$ 不存在：这正是原点处的拐角。(A1)
If $f(x)=5x^{4}-3x^{2}+7$, then $f'(x)=$若 $f(x)=5x^{4}-3x^{2}+7$，则 $f'(x)=$
$\frac{d}{dx}[5x^{4}]=20x^{3}$, $\frac{d}{dx}[-3x^{2}]=-6x$, and $\frac{d}{dx}[7]=0$ since the derivative of a constant is zero. (M1)$\frac{d}{dx}[5x^{4}]=20x^{3}$，$\frac{d}{dx}[-3x^{2}]=-6x$，且 $\frac{d}{dx}[7]=0$，因常数的导数为零。(M1)
Summing: $f'(x)=20x^{3}-6x$. (A1)相加：$f'(x)=20x^{3}-6x$。(A1)
$\dfrac{d}{dx}\bigl[\sin x - \cos x\bigr]=$
$\frac{d}{dx}[\sin x]=\cos x$, and $\frac{d}{dx}[-\cos x]=-(-\sin x)=\sin x$. (M1)$\frac{d}{dx}[\sin x]=\cos x$，且 $\frac{d}{dx}[-\cos x]=-(-\sin x)=\sin x$。(M1)
Adding: $\frac{d}{dx}[\sin x-\cos x]=\cos x+\sin x$. (A1)相加：$\frac{d}{dx}[\sin x-\cos x]=\cos x+\sin x$。(A1)
$\dfrac{d}{dx}\bigl[\,e^{x}\ln x\,\bigr]=$
With $u=e^{x}$, $v=\ln x$: $u'=e^{x}$ and $v'=\frac{1}{x}$. (M1)设 $u=e^{x}$，$v=\ln x$：$u'=e^{x}$，$v'=\frac{1}{x}$。(M1)
$\frac{d}{dx}[e^{x}\ln x]=u'v+uv'=e^{x}\ln x+e^{x}\cdot\frac{1}{x}=e^{x}\ln x+\frac{e^{x}}{x}$. (A1)$\frac{d}{dx}[e^{x}\ln x]=u'v+uv'=e^{x}\ln x+e^{x}\cdot\frac{1}{x}=e^{x}\ln x+\frac{e^{x}}{x}$。(A1)
If $h(x)=x^{2}\cos x$, then $h'(x)=$若 $h(x)=x^{2}\cos x$，则 $h'(x)=$
With $u=x^{2}$, $v=\cos x$: $u'=2x$ and $v'=-\sin x$. (M1)设 $u=x^{2}$，$v=\cos x$：$u'=2x$，$v'=-\sin x$。(M1)
$h'(x)=u'v+uv'=2x\cos x+x^{2}(-\sin x)=2x\cos x-x^{2}\sin x$. (A1)$h'(x)=u'v+uv'=2x\cos x+x^{2}(-\sin x)=2x\cos x-x^{2}\sin x$。(A1)
If $g(x)=\dfrac{x}{x^{2}+1}$, then $g'(x)=$若 $g(x)=\dfrac{x}{x^{2}+1}$，则 $g'(x)=$
With $u=x$, $v=x^{2}+1$: $u'=1$ and $v'=2x$. (M1)设 $u=x$，$v=x^{2}+1$：$u'=1$，$v'=2x$。(M1)
$g'(x)=\dfrac{u'v-uv'}{v^{2}}=\dfrac{(1)(x^{2}+1)-x(2x)}{(x^{2}+1)^{2}}=\dfrac{x^{2}+1-2x^{2}}{(x^{2}+1)^{2}}=\dfrac{1-x^{2}}{(x^{2}+1)^{2}}$. (A1)$g'(x)=\dfrac{u'v-uv'}{v^{2}}=\dfrac{(1)(x^{2}+1)-x(2x)}{(x^{2}+1)^{2}}=\dfrac{x^{2}+1-2x^{2}}{(x^{2}+1)^{2}}=\dfrac{1-x^{2}}{(x^{2}+1)^{2}}$。(A1)
The tangent line to $f(x)=x^{3}-2x$ at $x=1$ has equation$f(x)=x^{3}-2x$ 在 $x=1$ 处的切线方程为
$f(1)=1^{3}-2(1)=-1$, and $f'(x)=3x^{2}-2$ gives $f'(1)=3-2=1$. (M1)$f(1)=1^{3}-2(1)=-1$，且 $f'(x)=3x^{2}-2$ 给出 $f'(1)=3-2=1$。(M1)
Point-slope form: $y-(-1)=1(x-1)$, i.e. $y=x-2$. (A1)点斜式：$y-(-1)=1(x-1)$，即 $y=x-2$。(A1)
Let $f(x)=\dfrac{x\sin x}{e^{x}}$. Which differentiation rules are required?设 $f(x)=\dfrac{x\sin x}{e^{x}}$，求导需要用到哪些法则？
$f(x)=\dfrac{x\sin x}{e^{x}}$ is a quotient of $(x\sin x)$ over $e^{x}$, so the quotient rule is needed for the outer division. (M1)$f(x)=\dfrac{x\sin x}{e^{x}}$ 是 $(x\sin x)$ 除以 $e^{x}$ 的商，故外层的除法需要商法则。(M1)
The numerator $x\sin x$ is itself a product of two non-constant functions, so the product rule is also needed to differentiate it. (A1)分子 $x\sin x$ 本身是两个非常数函数之积，故对其求导还需要乘积法则。(A1)
Selected values of $f,g,f',g'$ at $x=2$ are given. If $h(x)=\dfrac{f(x)}{g(x)}$, then $h'(2)=$下表给出了 $f,g,f',g'$ 在 $x=2$ 处的部分取值。若 $h(x)=\dfrac{f(x)}{g(x)}$，则 $h'(2)=$
From the table at $x=2$: $f=3$, $g=4$, $f'=5$, $g'=-1$. (M1)由表格，在 $x=2$ 处：$f=3$，$g=4$，$f'=5$，$g'=-1$。(M1)
$h'(x)=\dfrac{f'(x)g(x)-f(x)g'(x)}{[g(x)]^{2}}$, so $h'(2)=\dfrac{(5)(4)-(3)(-1)}{4^{2}}=\dfrac{20+3}{16}=\dfrac{23}{16}$. (M1)$h'(x)=\dfrac{f'(x)g(x)-f(x)g'(x)}{[g(x)]^{2}}$，故 $h'(2)=\dfrac{(5)(4)-(3)(-1)}{4^{2}}=\dfrac{20+3}{16}=\dfrac{23}{16}$。(M1)
(A1)
The normal line to $y=\sqrt{x}$ at $x=4$ has slope$y=\sqrt{x}$ 在 $x=4$ 处的法线斜率为
$y'=\dfrac{1}{2\sqrt{x}}$, so at $x=4$ the tangent slope is $\dfrac{1}{2\sqrt{4}}=\dfrac{1}{4}$. (M1)$y'=\dfrac{1}{2\sqrt{x}}$，故在 $x=4$ 处切线斜率为 $\dfrac{1}{2\sqrt{4}}=\dfrac{1}{4}$。(M1)
The normal line is perpendicular to the tangent, so its slope is the negative reciprocal: $-\dfrac{1}{1/4}=-4$. (A1)法线与切线垂直，故其斜率为负倒数：$-\dfrac{1}{1/4}=-4$。(A1)
If $f(x)=\sin(2x)$, then $f^{(4)}(x)=$若 $f(x)=\sin(2x)$，则 $f^{(4)}(x)=$
$f'(x)=2\cos(2x)$ and $f''(x)=-4\sin(2x)$, each application of the chain rule multiplies by the constant $2$. (M1)$f'(x)=2\cos(2x)$，$f''(x)=-4\sin(2x)$，每次应用链式法则都乘以常数 $2$。(M1)
$f'''(x)=-8\cos(2x)$ and $f^{(4)}(x)=16\sin(2x)$. (M1)$f'''(x)=-8\cos(2x)$，$f^{(4)}(x)=16\sin(2x)$。(M1)
(A1)
A particle's position is $s(t)=t^{3}-6t^{2}+9t$. The acceleration at $t=3$ is质点的位置为 $s(t)=t^{3}-6t^{2}+9t$，在 $t=3$ 时的加速度为
$v(t)=s'(t)=3t^{2}-12t+9$, and $a(t)=v'(t)=6t-12$. (M1)$v(t)=s'(t)=3t^{2}-12t+9$，$a(t)=v'(t)=6t-12$。(M1)
$a(3)=6(3)-12=6$. (A1)$a(3)=6(3)-12=6$。(A1)
The graph of $f$ consists of two line segments meeting at a sharp corner at $x=3$. Which statement about $f'$ is true?$f$ 的图像由两段折线组成，在 $x=3$ 处形成尖角。关于 $f'$ 的哪项叙述是正确的？
Since the two segments meet at a sharp corner rather than blending smoothly, they have different slopes on either side of $x=3$, so the one-sided limits of $f'$ as $x\to 3^{-}$ and $x\to 3^{+}$ disagree. (M1)两段折线在尖角处相接而非平滑过渡，故在 $x=3$ 两侧斜率不同，即 $x\to 3^{-}$ 与 $x\to 3^{+}$ 时 $f'$ 的单侧极限不相等。(M1)
Because the two segments still meet at the same point (no gap in the graph), $f$ remains continuous at $x=3$; but $f'$ itself jumps there, so $f'(3)$ does not exist. This is statement (B). (A1)因两段折线仍在同一点相接（图像无缺口），故 $f$ 在 $x=3$ 处仍连续；但 $f'$ 在该处跳跃，故 $f'(3)$ 不存在，即选项 (B)。(A1)
$\dfrac{d}{dx}\bigl[\,3^{x}\,\bigr]=$
For $a>0$, $\dfrac{d}{dx}[a^{x}]=a^{x}\ln a$, which follows from rewriting $a^{x}=e^{x\ln a}$ and applying the chain rule. (M1)对 $a>0$，$\dfrac{d}{dx}[a^{x}]=a^{x}\ln a$，可由改写 $a^{x}=e^{x\ln a}$ 并应用链式法则得出。(M1)
With $a=3$: $\dfrac{d}{dx}[3^{x}]=3^{x}\ln 3$. (A1)取 $a=3$：$\dfrac{d}{dx}[3^{x}]=3^{x}\ln 3$。(A1)
Let $f(x)=x\,e^{-x}$. The value of $x>0$ at which the tangent line to $y=f(x)$ is horizontal is closest to设 $f(x)=x\,e^{-x}$，使 $y=f(x)$ 的切线为水平线的 $x>0$ 的值最接近
By the product rule, $f'(x)=e^{-x}+x(-e^{-x})=e^{-x}(1-x)$. (M1)由乘积法则，$f'(x)=e^{-x}+x(-e^{-x})=e^{-x}(1-x)$。(M1)
Since $e^{-x}>0$ for all $x$, $f'(x)=0$ exactly when $1-x=0$, i.e. $x=1$, so the closest choice is $1.0$. (A1)因对所有 $x$ 均有 $e^{-x}>0$，故 $f'(x)=0$ 当且仅当 $1-x=0$，即 $x=1$，最接近的选项为 $1.0$。(A1)
$\displaystyle\lim_{h\to 0}\dfrac{\sin\!\left(\tfrac{\pi}{3}+h\right)-\sin\!\left(\tfrac{\pi}{3}\right)}{h}=$
The expression matches $\lim_{h\to 0}\frac{f(\frac{\pi}{3}+h)-f(\frac{\pi}{3})}{h}$ for $f(x)=\sin x$, so it equals $f'(\pi/3)=\cos(\pi/3)$. (M1)此式与 $\lim_{h\to 0}\frac{f(\frac{\pi}{3}+h)-f(\frac{\pi}{3})}{h}$（取 $f(x)=\sin x$）形式一致，故等于 $f'(\pi/3)=\cos(\pi/3)$。(M1)
$\cos(\pi/3)=\dfrac{1}{2}$. (A1)$\cos(\pi/3)=\dfrac{1}{2}$。(A1)
$\displaystyle\lim_{h\to 0}\dfrac{e^{2+h}-e^{2}}{h}=$
This matches $\lim_{h\to 0}\frac{f(2+h)-f(2)}{h}$ for $f(x)=e^{x}$, i.e. $f'(2)$. (M1)此式与 $\lim_{h\to 0}\frac{f(2+h)-f(2)}{h}$（取 $f(x)=e^{x}$）形式一致，即 $f'(2)$。(M1)
Since $\frac{d}{dx}[e^{x}]=e^{x}$, $f'(2)=e^{2}$. (A1)因 $\frac{d}{dx}[e^{x}]=e^{x}$，故 $f'(2)=e^{2}$。(A1)
Consider $f(x)=x^{2/3}$. Which statement about $f$ at $x=0$ is true?设 $f(x)=x^{2/3}$，关于 $f$ 在 $x=0$ 处的哪项叙述是正确的？
$f'(x)=\dfrac{2}{3}x^{-1/3}=\dfrac{2}{3\sqrt[3]{x}}$, which is undefined at $x=0$. (M1)$f'(x)=\dfrac{2}{3}x^{-1/3}=\dfrac{2}{3\sqrt[3]{x}}$，在 $x=0$ 处无定义。(M1)
As $x\to 0^{+}$, $\sqrt[3]{x}\to 0^{+}$ so $f'(x)\to+\infty$; as $x\to 0^{-}$, $\sqrt[3]{x}\to 0^{-}$ so $f'(x)\to-\infty$: the one-sided limits of $f'$ are infinite with opposite signs. (M1)当 $x\to 0^{+}$ 时，$\sqrt[3]{x}\to 0^{+}$，故 $f'(x)\to+\infty$；当 $x\to 0^{-}$ 时，$\sqrt[3]{x}\to 0^{-}$，故 $f'(x)\to-\infty$：$f'$ 的两个单侧极限均为无穷大，且符号相反。(M1)
$f$ itself is continuous at $x=0$ (its value there is $0$), so this is a cusp, not a vertical tangent (which needs both one-sided limits infinite with the same sign) and not a discontinuity. (A1)$f$ 本身在 $x=0$ 处连续（该处值为 $0$），故这是一个尖点，而非竖直切线（后者要求两个单侧极限均为同号无穷大）也非间断点。(A1)
Let $f(x)=x^{2}-4x+1$.设 $f(x)=x^{2}-4x+1$。
By definition, $f'(x)=\lim_{h\to 0}\dfrac{f(x+h)-f(x)}{h}$. Expand $f(x+h)=(x+h)^{2}-4(x+h)+1=x^{2}+2xh+h^{2}-4x-4h+1$. (M1)由定义，$f'(x)=\lim_{h\to 0}\dfrac{f(x+h)-f(x)}{h}$。展开 $f(x+h)=(x+h)^{2}-4(x+h)+1=x^{2}+2xh+h^{2}-4x-4h+1$。(M1)
Subtracting $f(x)=x^{2}-4x+1$: $f(x+h)-f(x)=2xh+h^{2}-4h=h(2x+h-4)$. (M1)减去 $f(x)=x^{2}-4x+1$：$f(x+h)-f(x)=2xh+h^{2}-4h=h(2x+h-4)$。(M1)
Dividing by $h$ and letting $h\to 0$: $f'(x)=\lim_{h\to 0}(2x+h-4)=2x-4$. (A1)除以 $h$ 并令 $h\to 0$：$f'(x)=\lim_{h\to 0}(2x+h-4)=2x-4$。(A1)
$f'(3)=2(3)-4=2$. (A1)$f'(3)=2(3)-4=2$。(A1)
This means that at $x=3$, $f$ is instantaneously increasing at a rate of $2$ units of $f$ per unit of $x$: it is the slope of the curve $y=f(x)$ at that point. (R1)这表示在 $x=3$ 处，$f$ 的瞬时变化率为每单位 $x$ 增加 $2$ 个单位的 $f$：即曲线 $y=f(x)$ 在该点处的斜率。(R1)
$f(3)=3^{2}-4(3)+1=9-12+1=-2$, and the slope is $f'(3)=2$. (M1)$f(3)=3^{2}-4(3)+1=9-12+1=-2$，斜率为 $f'(3)=2$。(M1)
Point-slope form: $y-(-2)=2(x-3)$, i.e. $y=2x-8$. (A1)点斜式：$y-(-2)=2(x-3)$，即 $y=2x-8$。(A1)
The table gives values of $f,g,f',g'$ at $x=1$ and $x=3$.下表给出了 $f,g,f',g'$ 在 $x=1$ 和 $x=3$ 处的值。
| $x$ | $f$ | $g$ | $f'$ | $g'$ |
|---|---|---|---|---|
| $1$ | $2$ | $4$ | $-1$ | $3$ |
| $3$ | $5$ | $-2$ | $2$ | $1$ |
By the product rule, $\dfrac{d}{dx}[f(x)g(x)]=f'(x)g(x)+f(x)g'(x)$. (M1)由乘积法则，$\dfrac{d}{dx}[f(x)g(x)]=f'(x)g(x)+f(x)g'(x)$。(M1)
At $x=1$: $f'(1)g(1)+f(1)g'(1)=(-1)(4)+(2)(3)=-4+6=2$. (A1)在 $x=1$ 处：$f'(1)g(1)+f(1)g'(1)=(-1)(4)+(2)(3)=-4+6=2$。(A1)
By the quotient rule, $\dfrac{d}{dx}\!\left[\dfrac{f(x)}{g(x)}\right]=\dfrac{f'(x)g(x)-f(x)g'(x)}{[g(x)]^{2}}$. (M1)由商法则，$\dfrac{d}{dx}\!\left[\dfrac{f(x)}{g(x)}\right]=\dfrac{f'(x)g(x)-f(x)g'(x)}{[g(x)]^{2}}$。(M1)
At $x=3$: $\dfrac{(2)(-2)-(5)(1)}{(-2)^{2}}=\dfrac{-4-5}{4}=-\dfrac{9}{4}$. (A1)在 $x=3$ 处：$\dfrac{(2)(-2)-(5)(1)}{(-2)^{2}}=\dfrac{-4-5}{4}=-\dfrac{9}{4}$。(A1)
By the product rule with $u=x$, $v=g(x)$: $h'(x)=g(x)+x\,g'(x)$. (M1)由乘积法则，设 $u=x$，$v=g(x)$：$h'(x)=g(x)+x\,g'(x)$。(M1)
At $x=3$: $h(3)=3\cdot g(3)=3(-2)=-6$ and $h'(3)=g(3)+3g'(3)=-2+3(1)=1$. (M1)在 $x=3$ 处：$h(3)=3\cdot g(3)=3(-2)=-6$，$h'(3)=g(3)+3g'(3)=-2+3(1)=1$。(M1)
Point-slope form: $y-(-6)=1(x-3)$, i.e. $y=x-9$. (A1)点斜式：$y-(-6)=1(x-3)$，即 $y=x-9$。(A1)
A particle moves along the $x$-axis with position $s(t)=t^{3}-9t^{2}+15t+4$ for $t\ge 0$ (seconds, meters).质点沿 $x$ 轴运动，位置函数为 $s(t)=t^{3}-9t^{2}+15t+4$，$t\ge 0$（单位：秒，米）。
Applying the power rule to $s(t)=t^{3}-9t^{2}+15t+4$: (M1)对 $s(t)=t^{3}-9t^{2}+15t+4$ 应用幂法则：(M1)
$$ v(t)=s'(t)=3t^{2}-18t+15. $$(A1)
$$ a(t)=v'(t)=6t-18. $$(A1)
Set $v(t)=0$: $3t^{2}-18t+15=0$, i.e. $t^{2}-6t+5=0$, which factors as $(t-1)(t-5)=0$. (M1)令 $v(t)=0$：$3t^{2}-18t+15=0$，即 $t^{2}-6t+5=0$，因式分解为 $(t-1)(t-5)=0$。(M1)
So $t=1$ and $t=5$; both lie in the domain $t\ge 0$, so the particle is at rest at $t=1$ and $t=5$. (A1)故 $t=1$ 与 $t=5$；两者均在定义域 $t\ge 0$ 内，故质点在 $t=1$ 和 $t=5$ 时静止。(A1)
Test a point in each interval determined by the rest times: $v(0)=15>0$; $v(2)=3(4)-36+15=-9<0$; $v(6)=3(36)-108+15=15>0$. (M1)在由静止时刻划分的各区间内各取一点检验：$v(0)=15>0$；$v(2)=3(4)-36+15=-9<0$；$v(6)=3(36)-108+15=15>0$。(M1)
So $v(t)>0$ on $[0,1)$ and $(5,\infty)$, and $v(t)<0$ on $(1,5)$. (A1)故 $v(t)>0$ 在 $[0,1)$ 与 $(5,\infty)$ 上成立，$v(t)<0$ 在 $(1,5)$ 上成立。(A1)
Since velocity is positive exactly where the particle moves in the positive direction, the particle moves right on $[0,1)$ and $(5,\infty)$, and moves left on $(1,5)$. (R1)因速度为正恰对应质点向正方向运动，故质点在 $[0,1)$ 和 $(5,\infty)$ 上向右运动，在 $(1,5)$ 上向左运动。(R1)
At $t=2$: $v(2)=-9$ (from part (c)) and $a(2)=6(2)-18=-6$. (M1)在 $t=2$ 时：$v(2)=-9$（由 (c)），$a(2)=6(2)-18=-6$。(M1)
Both $v(2)$ and $a(2)$ are negative, so velocity and acceleration have the same sign. (A1)$v(2)$ 与 $a(2)$ 均为负，故速度与加速度同号。(A1)
When velocity and acceleration share a sign, speed is increasing; when they have opposite signs, speed is decreasing. Since both are negative here, the particle is speeding up at $t=2$. (R1)当速度与加速度同号时，速率增大；异号时，速率减小。此处两者均为负，故质点在 $t=2$ 时正在加速。(R1)
Let $f(x)=\begin{cases} x^{2}+ax, & x\le 1\\ bx+c, & x>1 \end{cases}$.设 $f(x)=\begin{cases} x^{2}+ax, & x\le 1\\ bx+c, & x>1 \end{cases}$。
Continuity at $x=1$ requires $\lim_{x\to 1^{-}}f(x)=\lim_{x\to 1^{+}}f(x)$. (M1)$f$ 在 $x=1$ 处连续要求 $\lim_{x\to 1^{-}}f(x)=\lim_{x\to 1^{+}}f(x)$。(M1)
The left piece gives $\lim_{x\to 1^{-}}(x^{2}+ax)=1+a$ and the right piece gives $\lim_{x\to 1^{+}}(bx+c)=b+c$, so the condition is $1+a=b+c$. (A1)左段给出 $\lim_{x\to 1^{-}}(x^{2}+ax)=1+a$，右段给出 $\lim_{x\to 1^{+}}(bx+c)=b+c$，故条件为 $1+a=b+c$。(A1)
Differentiability at $x=1$ additionally requires the one-sided derivatives to agree: $\frac{d}{dx}[x^{2}+ax]=2x+a$ gives a left-hand derivative of $2+a$ at $x=1$, and $\frac{d}{dx}[bx+c]=b$ gives a right-hand derivative of $b$. (M1)在 $x=1$ 处可微还要求两个单侧导数相等：$\frac{d}{dx}[x^{2}+ax]=2x+a$ 给出 $x=1$ 处左导数为 $2+a$，$\frac{d}{dx}[bx+c]=b$ 给出右导数为 $b$。(M1)
So the condition is $2+a=b$. (A1)故条件为 $2+a=b$。(A1)
Since $f(1)=1^{2}+a(1)=1+a$ and this is given to equal $3$: (M1)因 $f(1)=1^{2}+a(1)=1+a$，且已知其等于 $3$：(M1)
$1+a=3$, so $a=2$. (A1)$1+a=3$，故 $a=2$。(A1)
From the differentiability condition in (b): (M1)由 (b) 中的可微性条件：(M1)
$b=2+a=2+2=4$. (A1)$b=2+a=2+2=4$。(A1)
Substituting $a=2$ and $b=4$ into the continuity condition from (a), $1+a=b+c$ gives $3=4+c$, so $c=-1$. (A1)将 $a=2$、$b=4$ 代入 (a) 中的连续性条件 $1+a=b+c$，得 $3=4+c$，故 $c=-1$。(A1)
With $a=2, b=4$: for $x\le 1$, $f'(x)=2x+a=2x+2$; for $x>1$, $f'(x)=b=4$. (M1)取 $a=2,b=4$：当 $x\le 1$ 时，$f'(x)=2x+a=2x+2$；当 $x>1$ 时，$f'(x)=b=4$。(M1)
So $f'(x)=\begin{cases}2x+2, & x\le 1\\ 4, & x>1\end{cases}$, and $f'(1)=2(1)+2=4$, consistent with the right-hand value, confirming differentiability. (A1)故 $f'(x)=\begin{cases}2x+2, & x\le 1\\ 4, & x>1\end{cases}$，且 $f'(1)=2(1)+2=4$，与右侧值一致，验证了可微性。(A1)
Let $f(x)=x^{3}+px^{2}+qx$.设 $f(x)=x^{3}+px^{2}+qx$。
Applying the power rule to $f(x)=x^{3}+px^{2}+qx$: (M1)对 $f(x)=x^{3}+px^{2}+qx$ 应用幂法则：(M1)
$f'(x)=3x^{2}+2px+q$. (A1)$f'(x)=3x^{2}+2px+q$。(A1)
Since $(1,2)$ lies on the graph: $f(1)=1+p+q=2$, giving $p+q=1$. (M1)因 $(1,2)$ 在图像上：$f(1)=1+p+q=2$，得 $p+q=1$。(M1)
Since the tangent is horizontal at $x=1$: $f'(1)=3+2p+q=0$, giving $2p+q=-3$. (M1)因在 $x=1$ 处切线为水平线：$f'(1)=3+2p+q=0$，得 $2p+q=-3$。(M1)
Subtracting the first equation from the second: $(2p+q)-(p+q)=-3-1$, so $p=-4$. (M1)第二式减去第一式：$(2p+q)-(p+q)=-3-1$，故 $p=-4$。(M1)
Back-substituting: $q=1-p=1-(-4)=5$. (A1)代回：$q=1-p=1-(-4)=5$。(A1)
With $p=-4,q=5$: $f(-1)=(-1)^{3}-4(-1)^{2}+5(-1)=-1-4-5=-10$. (M1)取 $p=-4,q=5$：$f(-1)=(-1)^{3}-4(-1)^{2}+5(-1)=-1-4-5=-10$。(M1)
$f'(-1)=3(-1)^{2}-8(-1)+5=3+8+5=16$. (M1)$f'(-1)=3(-1)^{2}-8(-1)+5=3+8+5=16$。(M1)
Point-slope form: $y-(-10)=16(x-(-1))$, i.e. $y=16x+6$. (A1)点斜式：$y-(-10)=16(x-(-1))$，即 $y=16x+6$。(A1)
Set $f'(x)=6$: $3x^{2}-8x+5=6$, i.e. $3x^{2}-8x-1=0$. (M1)令 $f'(x)=6$：$3x^{2}-8x+5=6$，即 $3x^{2}-8x-1=0$。(M1)
By the quadratic formula: $x=\dfrac{8\pm\sqrt{64+12}}{6}=\dfrac{8\pm\sqrt{76}}{6}=\dfrac{8\pm 2\sqrt{19}}{6}=\dfrac{4\pm\sqrt{19}}{3}$. (M1)由求根公式：$x=\dfrac{8\pm\sqrt{64+12}}{6}=\dfrac{8\pm\sqrt{76}}{6}=\dfrac{8\pm 2\sqrt{19}}{6}=\dfrac{4\pm\sqrt{19}}{3}$。(M1)
(A1)