Worked Solutions to the AP-Style Practice SetAP 风格练习题配套解析
Topics 1.1 - 1.16专题 1.1 至 1.16AB
The table gives values of $f(x)$ near $x=2$; estimate $\displaystyle\lim_{x\to 2}f(x)$.下表给出了 $f(x)$ 在 $x=2$ 附近的值，估计 $\displaystyle\lim_{x\to 2}f(x)$。
As $x\to 2^{-}$ the values $4.61,\,4.9601,\,4.996$ climb steadily toward $5$; as $x\to 2^{+}$ the values $5.004,\,5.0401,\,5.41$ come down toward $5$ from above. (M1)当 $x\to 2^{-}$ 时，$4.61,\,4.9601,\,4.996$ 稳步逼近 $5$；当 $x\to 2^{+}$ 时，$5.004,\,5.0401,\,5.41$ 从上方逼近 $5$。(M1)
Both one-sided trends converge to the same value, so the table estimate is $\displaystyle\lim_{x\to 2}f(x)=5$. (A1)两侧趋势收敛于同一值，故由表格估计 $\displaystyle\lim_{x\to 2}f(x)=5$。(A1)
$\displaystyle\lim_{x\to 3}\dfrac{x^2-9}{x-3}=$
Direct substitution gives $\tfrac{0}{0}$, so factor the numerator: $x^{2}-9=(x-3)(x+3)$. (M1)直接代入得 $\tfrac{0}{0}$ 不定式，故因式分解分子：$x^{2}-9=(x-3)(x+3)$。(M1)
$$ \lim_{x\to 3}\frac{(x-3)(x+3)}{x-3}=\lim_{x\to 3}(x+3)=6. $$(A1)
$\displaystyle\lim_{x\to 4}\dfrac{\sqrt{x}-2}{x-4}=$
Multiply numerator and denominator by $\sqrt{x}+2$: (M1)分子分母同乘 $\sqrt{x}+2$：(M1)
$$ \frac{\sqrt{x}-2}{x-4}\cdot\frac{\sqrt{x}+2}{\sqrt{x}+2}=\frac{x-4}{(x-4)(\sqrt{x}+2)}=\frac{1}{\sqrt{x}+2}\longrightarrow\frac{1}{2+2}=\frac{1}{4}. $$(A1)
$\displaystyle\lim_{x\to 0}\dfrac{\sin(5x)}{3x}=$
Write the quotient so the argument of $\sin$ matches its own denominator: (M1)改写商式，使 $\sin$ 的自变量与其分母一致：(M1)
$$ \frac{\sin(5x)}{3x}=\frac{5}{3}\cdot\frac{\sin(5x)}{5x}\longrightarrow\frac{5}{3}\cdot 1=\frac{5}{3}. $$(A1)
$\displaystyle\lim_{x\to 0}\dfrac{1-\cos x}{x^{2}}=$
Multiply by $\dfrac{1+\cos x}{1+\cos x}$: (M1)乘以 $\dfrac{1+\cos x}{1+\cos x}$：(M1)
$$ \frac{1-\cos x}{x^{2}}=\frac{1-\cos^{2}x}{x^{2}(1+\cos x)}=\frac{\sin^{2}x}{x^{2}(1+\cos x)}=\left(\frac{\sin x}{x}\right)^{2}\cdot\frac{1}{1+\cos x}\longrightarrow 1^{2}\cdot\frac{1}{2}=\frac{1}{2}. $$(A1)
If $4-x^{2}\le g(x)\le 4+x^{2}$ for all $x$, find $\displaystyle\lim_{x\to 0}g(x)$.若对所有 $x$ 均有 $4-x^{2}\le g(x)\le 4+x^{2}$，求 $\displaystyle\lim_{x\to 0}g(x)$。
Both bounding functions are polynomials, continuous everywhere, so evaluate their limits at $x=0$ directly: $\displaystyle\lim_{x\to 0}(4-x^{2})=4$ and $\displaystyle\lim_{x\to 0}(4+x^{2})=4$. (M1)两个界函数均为多项式，处处连续，可直接代入求极限：$\displaystyle\lim_{x\to 0}(4-x^{2})=4$，$\displaystyle\lim_{x\to 0}(4+x^{2})=4$。(M1)
Since both bounds tend to $4$, the Squeeze Theorem forces $\displaystyle\lim_{x\to 0}g(x)=4$. (A1)两界均趋于 $4$，由夹逼定理知 $\displaystyle\lim_{x\to 0}g(x)=4$。(A1)
The graph of $f$ is shown; which statement about $x=2$ is true?$f$ 的图像如图所示，关于 $x=2$ 处，下列哪个说法正确？
The left branch rises through the open circle at height $3$, so $\displaystyle\lim_{x\to 2^{-}}f(x)=3$. The right branch begins (also with an open endpoint) at height $1$ and climbs from there, so $\displaystyle\lim_{x\to 2^{+}}f(x)=1$. (M1)左侧分支上升至高度 $3$ 处的空心圆，故 $\displaystyle\lim_{x\to 2^{-}}f(x)=3$。右侧分支从高度 $1$ 处（同样为空心端点）开始并向上延伸，故 $\displaystyle\lim_{x\to 2^{+}}f(x)=1$。(M1)
Since $3\ne 1$, the one-sided limits disagree, so the two-sided limit $\displaystyle\lim_{x\to 2}f(x)$ does not exist: this is statement (B). (A1)因 $3\ne 1$，两个单侧极限不相等，故双侧极限 $\displaystyle\lim_{x\to 2}f(x)$ 不存在，即选项 (B)。(A1)
Let $f(x)=\begin{cases}\dfrac{x^{2}-4}{x-2}, & x\ne 2\\[4pt]k, & x=2\end{cases}$. For what value of $k$ is $f$ continuous at $x=2$?设 $f(x)=\begin{cases}\dfrac{x^{2}-4}{x-2}, & x\ne 2\\[4pt]k, & x=2\end{cases}$，$k$ 取何值时 $f$ 在 $x=2$ 处连续？
For $x\ne 2$, factor: $\dfrac{x^{2}-4}{x-2}=\dfrac{(x-2)(x+2)}{x-2}=x+2$, so $\displaystyle\lim_{x\to 2}f(x)=2+2=4$. (M1)当 $x\ne 2$ 时，因式分解：$\dfrac{x^{2}-4}{x-2}=\dfrac{(x-2)(x+2)}{x-2}=x+2$，故 $\displaystyle\lim_{x\to 2}f(x)=2+2=4$。(M1)
Continuity requires $f(2)=\displaystyle\lim_{x\to 2}f(x)$, so $k=4$. (A1)连续性要求 $f(2)=\displaystyle\lim_{x\to 2}f(x)$，故 $k=4$。(A1)
$g(x)=\dfrac{x+1}{x^{2}+x}$ has which type of discontinuity at $x=0$?$g(x)=\dfrac{x+1}{x^{2}+x}$ 在 $x=0$ 处属于哪种不连续？
Factor: $x^{2}+x=x(x+1)$, so $g(x)=\dfrac{x+1}{x(x+1)}=\dfrac{1}{x}$ for $x\ne 0,-1$. (M1)因式分解：$x^{2}+x=x(x+1)$，故当 $x\ne 0,-1$ 时 $g(x)=\dfrac{x+1}{x(x+1)}=\dfrac{1}{x}$。(M1)
The factor $(x+1)$ cancels, but $x$ itself does not; at $x=0$ the numerator of the reduced form is $1\ne 0$ while the denominator $\to 0$, so $g$ blows up: this is an infinite discontinuity (vertical asymptote at $x=0$), not a removable one. (A1)因子 $(x+1)$ 可以消去，但 $x$ 本身不能。在 $x=0$ 处，约简后分子为 $1\ne 0$，而分母 $\to 0$，故 $g$ 发散：这是无穷不连续（$x=0$ 处为竖直渐近线），而非可去不连续。(A1)
Find $a$ and $b$ so that $f(x)=\begin{cases}2x+a, & x\le 1\\ bx^{2}+3, & 1\lt x\lt 2\\ 4x-b, & x\ge 2\end{cases}$ is continuous everywhere.求 $a$ 和 $b$，使 $f(x)=\begin{cases}2x+a, & x\le 1\\ bx^{2}+3, & 1\lt x\lt 2\\ 4x-b, & x\ge 2\end{cases}$ 处处连续。
At $x=1$: the pieces must agree, so $2(1)+a=b(1)^{2}+3$, i.e. $a-b=1$. (M1)在 $x=1$ 处：两段须相等，故 $2(1)+a=b(1)^{2}+3$，即 $a-b=1$。(M1)
At $x=2$: the pieces must agree, so $b(2)^{2}+3=4(2)-b$, i.e. $4b+3=8-b$, giving $5b=5$, so $b=1$. (M1)在 $x=2$ 处：两段须相等，故 $b(2)^{2}+3=4(2)-b$，即 $4b+3=8-b$，解得 $5b=5$，$b=1$。(M1)
Substituting back into $a-b=1$ gives $a=1+b=2$. (A1)代回 $a-b=1$，得 $a=1+b=2$。(A1)
$\displaystyle\lim_{x\to 2^-}\dfrac{x+3}{x-2}=$
As $x\to 2^{-}$, the numerator $x+3\to 5>0$, a positive constant near this point, while the denominator $x-2\to 0^{-}$ (small and negative, since $x<2$). (M1)当 $x\to 2^{-}$ 时，分子 $x+3\to 5>0$，在该点附近为正常数；分母 $x-2\to 0^{-}$（因 $x<2$，趋近于 0 且为负）。(M1)
A positive number divided by a vanishing negative quantity diverges to $-\infty$. (A1)正数除以趋于零的负量将发散至 $-\infty$。(A1)
$\displaystyle\lim_{x\to\infty}\dfrac{6x^{2}-x}{3x^{2}+4}=$
Divide numerator and denominator by $x^{2}$: (M1)分子分母同除以 $x^{2}$：(M1)
$$ \frac{6-1/x}{3+4/x^{2}}\longrightarrow\frac{6-0}{3+0}=2. $$(A1)
$\displaystyle\lim_{x\to\infty}\dfrac{\sqrt{9x^{4}+1}}{x^{2}-3x}=$
Since $x\to\infty$ means $x^{2}>0$ throughout, $\sqrt{9x^{4}+1}=\sqrt{x^{4}\left(9+\tfrac{1}{x^{4}}\right)}=x^{2}\sqrt{9+\tfrac{1}{x^{4}}}$. (M1)因 $x\to\infty$ 时始终有 $x^{2}>0$，故 $\sqrt{9x^{4}+1}=\sqrt{x^{4}\left(9+\tfrac{1}{x^{4}}\right)}=x^{2}\sqrt{9+\tfrac{1}{x^{4}}}$。(M1)
Divide numerator and denominator by $x^{2}$: (M1)分子分母同除以 $x^{2}$：(M1)
$$ \frac{x^{2}\sqrt{9+1/x^{4}}}{x^{2}(1-3/x)}=\frac{\sqrt{9+1/x^{4}}}{1-3/x}\longrightarrow\frac{\sqrt{9}}{1}=3. $$(A1)
$\displaystyle\lim_{x\to -\infty}\bigl(\sqrt{x^{2}+4x}+x\bigr)=$
Multiply and divide by the conjugate $\sqrt{x^{2}+4x}-x$: (M1)乘除以共轭表达式 $\sqrt{x^{2}+4x}-x$：(M1)
$$ \sqrt{x^{2}+4x}+x=\frac{(x^{2}+4x)-x^{2}}{\sqrt{x^{2}+4x}-x}=\frac{4x}{\sqrt{x^{2}+4x}-x}. $$For $x\to-\infty$, write $\sqrt{x^{2}+4x}=|x|\sqrt{1+4/x}=-x\sqrt{1+4/x}$ since $x<0$ means $|x|=-x$: (M1)当 $x\to-\infty$ 时，写 $\sqrt{x^{2}+4x}=|x|\sqrt{1+4/x}=-x\sqrt{1+4/x}$，因 $x<0$ 时 $|x|=-x$：(M1)
$$ \frac{4x}{-x\sqrt{1+4/x}-x}=\frac{4x}{-x\left(\sqrt{1+4/x}+1\right)}=\frac{-4}{\sqrt{1+4/x}+1}\longrightarrow\frac{-4}{1+1}=-2. $$(A1)
Let $f$ be continuous on $[0,3]$ with $f(0)=-2$ and $f(3)=5$. Which conclusion does the IVT guarantee?设 $f$ 在 $[0,3]$ 上连续，且 $f(0)=-2$，$f(3)=5$。介值定理可保证哪个结论？
The IVT guarantees $f(c)=d$ only for $d$ strictly between $f(0)=-2$ and $f(3)=5$. Checking each option: $0\in(-2,5)$ (works); $6\notin(-2,5)$; $-3\notin(-2,5)$; differentiability is never a conclusion of the IVT. (M1)介值定理仅保证 $d$ 严格介于 $f(0)=-2$ 与 $f(3)=5$ 之间时 $f(c)=d$ 成立。逐一检验：$0\in(-2,5)$（成立）；$6\notin(-2,5)$；$-3\notin(-2,5)$；可导性从来都不是介值定理的结论。(M1)
So the guaranteed conclusion is (A): $f(c)=0$ for some $c\in(0,3)$. (A1)故保证成立的结论是 (A)：存在 $c\in(0,3)$ 使 $f(c)=0$。(A1)
The continuous function $h$ has selected values below. What is the minimum number of real zeros of $h$ on $[1,9]$ guaranteed by the IVT?连续函数 $h$ 的部分值如表所示。介值定理能保证 $h$ 在 $[1,9]$ 上至少有多少个实零点？
The values are $h(1)=-4,\ h(3)=2,\ h(5)=-1,\ h(7)=3,\ h(9)=-5$. Since $h$ is continuous on each subinterval $[1,3],[3,5],[5,7],[7,9]$, the IVT applies wherever consecutive values have opposite signs. (M1)各值为 $h(1)=-4,\ h(3)=2,\ h(5)=-1,\ h(7)=3,\ h(9)=-5$。因 $h$ 在每个子区间 $[1,3],[3,5],[5,7],[7,9]$ 上均连续，只要相邻两值异号，介值定理即可应用。(M1)
Every consecutive pair alternates in sign: $(-,+)$ on $[1,3]$, $(+,-)$ on $[3,5]$, $(-,+)$ on $[5,7]$, $(+,-)$ on $[7,9]$: four sign changes, so the IVT guarantees at least $4$ zeros. (A1)相邻各值符号交替：$[1,3]$ 上为 $(-,+)$，$[3,5]$ 上为 $(+,-)$，$[5,7]$ 上为 $(-,+)$，$[7,9]$ 上为 $(+,-)$：共四次变号，故介值定理保证至少 $4$ 个零点。(A1)
$\displaystyle\lim_{h\to 0}\dfrac{(2+h)^{3}-8}{h}=$
Expand: $(2+h)^{3}=8+12h+6h^{2}+h^{3}$, so the numerator is $12h+6h^{2}+h^{3}=h(12+6h+h^{2})$. (M1)展开：$(2+h)^{3}=8+12h+6h^{2}+h^{3}$，故分子为 $12h+6h^{2}+h^{3}=h(12+6h+h^{2})$。(M1)
$$ \lim_{h\to 0}\frac{h(12+6h+h^{2})}{h}=\lim_{h\to 0}(12+6h+h^{2})=12. $$(A1)
$\displaystyle\lim_{x\to 0}\dfrac{\frac{1}{x+3}-\frac{1}{3}}{x}=$
Combine over a common denominator: $\dfrac{1}{x+3}-\dfrac{1}{3}=\dfrac{3-(x+3)}{3(x+3)}=\dfrac{-x}{3(x+3)}$. (M1)通分：$\dfrac{1}{x+3}-\dfrac{1}{3}=\dfrac{3-(x+3)}{3(x+3)}=\dfrac{-x}{3(x+3)}$。(M1)
$$ \lim_{x\to 0}\frac{1}{x}\cdot\frac{-x}{3(x+3)}=\lim_{x\to 0}\frac{-1}{3(x+3)}=\frac{-1}{3(3)}=-\frac{1}{9}. $$(A1)
Evaluate each limit, showing all algebraic steps: (a) $\displaystyle\lim_{x\to 5}\dfrac{x^{2}-25}{x^{2}-4x-5}$; (b) $\displaystyle\lim_{x\to 9}\dfrac{x-9}{\sqrt{x}-3}$; (c) $\displaystyle\lim_{x\to 0}\dfrac{\sin(3x)}{\tan(2x)}$.计算下列各极限，展示完整代数步骤：(a) $\displaystyle\lim_{x\to 5}\dfrac{x^{2}-25}{x^{2}-4x-5}$；(b) $\displaystyle\lim_{x\to 9}\dfrac{x-9}{\sqrt{x}-3}$；(c) $\displaystyle\lim_{x\to 0}\dfrac{\sin(3x)}{\tan(2x)}$。
$x^{2}-25=(x-5)(x+5)$ and $x^{2}-4x-5=(x-5)(x+1)$. (M1)$x^{2}-25=(x-5)(x+5)$，$x^{2}-4x-5=(x-5)(x+1)$。(M1)
$$ \lim_{x\to 5}\frac{(x-5)(x+5)}{(x-5)(x+1)}=\lim_{x\to 5}\frac{x+5}{x+1}=\frac{10}{6}=\frac{5}{3}. $$(A1)
Write $x-9=(\sqrt{x}-3)(\sqrt{x}+3)$: (M1)写 $x-9=(\sqrt{x}-3)(\sqrt{x}+3)$：(M1)
$$ \lim_{x\to 9}\frac{(\sqrt{x}-3)(\sqrt{x}+3)}{\sqrt{x}-3}=\lim_{x\to 9}(\sqrt{x}+3)=3+3=6. $$(A1)
Since $\tan(2x)=\dfrac{\sin(2x)}{\cos(2x)}$, rewrite the quotient as: (M1)因 $\tan(2x)=\dfrac{\sin(2x)}{\cos(2x)}$，将商式改写为：(M1)
$$ \frac{\sin(3x)}{\tan(2x)}=\frac{\sin(3x)\cos(2x)}{\sin(2x)}=\frac{3}{2}\cdot\frac{\sin(3x)}{3x}\cdot\frac{2x}{\sin(2x)}\cdot\cos(2x)\longrightarrow\frac{3}{2}\cdot 1\cdot 1\cdot 1=\frac{3}{2}. $$(A1)
$f(x)=\begin{cases} \dfrac{x^{2}-x-6}{x-3}, & x<3\\[4pt] ax+b, & 3\le x\le 5\\[4pt] x^{2}-9, & x>5 \end{cases}$
Factor: $x^{2}-x-6=(x-3)(x+2)$, so for $x<3$, $f(x)=\dfrac{(x-3)(x+2)}{x-3}=x+2$. (M1)因式分解：$x^{2}-x-6=(x-3)(x+2)$，故当 $x<3$ 时，$f(x)=\dfrac{(x-3)(x+2)}{x-3}=x+2$。(M1)
Hence $\displaystyle\lim_{x\to 3^{-}}f(x)=3+2=5$. Because $f$ must be continuous at $x=3$, the middle piece must satisfy $f(3)=3a+b=5$: this is the restriction on $a$ and $b$. (A1)故 $\displaystyle\lim_{x\to 3^{-}}f(x)=3+2=5$。因 $f$ 须在 $x=3$ 处连续，中间段须满足 $f(3)=3a+b=5$：这就是对 $a$、$b$ 的限制。(A1)
From (a): $3a+b=5$. (M1)由 (a)：$3a+b=5$。(M1)
At $x=5$, continuity requires $f(5)=5a+b$ to equal $\displaystyle\lim_{x\to 5^{+}}(x^{2}-9)=25-9=16$, so $5a+b=16$. (A1)在 $x=5$ 处，连续性要求 $f(5)=5a+b$ 等于 $\displaystyle\lim_{x\to 5^{+}}(x^{2}-9)=25-9=16$，故 $5a+b=16$。(A1)
Subtracting the equations: $2a=11$, so $a=\dfrac{11}{2}$. (M1)两式相减：$2a=11$，故 $a=\dfrac{11}{2}$。(M1)
Back-substitute: $b=5-3a=5-\dfrac{33}{2}=-\dfrac{23}{2}$. (A1)代回：$b=5-3a=5-\dfrac{33}{2}=-\dfrac{23}{2}$。(A1)
With $a=\tfrac{11}{2}$, the three pieces have derivatives $1$ (from $x+2$, $x<3$), $\tfrac{11}{2}$ (from $ax+b$, $3
At $x=3$: left derivative $=1$, right derivative $=\tfrac{11}{2}$; since $1\ne\tfrac{11}{2}$, $f'$ jumps, so $f$ has a corner (not differentiable) at $x=3$. (A1)在 $x=3$ 处：左导数 $=1$，右导数 $=\tfrac{11}{2}$；因 $1\ne\tfrac{11}{2}$，$f'$ 跳跃，故 $f$ 在 $x=3$ 处出现拐角（不可导）。(A1)
At $x=5$: left derivative $=\tfrac{11}{2}$, right derivative $=2(5)=10$; since $\tfrac{11}{2}\ne 10$, $f'$ again jumps, so $f$ has a corner at $x=5$ as well. (A1)在 $x=5$ 处：左导数 $=\tfrac{11}{2}$，右导数 $=2(5)=10$；因 $\tfrac{11}{2}\ne 10$，$f'$ 再次跳跃，故 $f$ 在 $x=5$ 处同样出现拐角。(A1)
Let $f(x)=\dfrac{2x^{2}-x-6}{x^{2}-4}$.设 $f(x)=\dfrac{2x^{2}-x-6}{x^{2}-4}$。
Factor: $2x^{2}-x-6=(x-2)(2x+3)$ and $x^{2}-4=(x-2)(x+2)$, so $f(x)=\dfrac{2x+3}{x+2}$ for $x\ne 2$. (M1)因式分解：$2x^{2}-x-6=(x-2)(2x+3)$，$x^{2}-4=(x-2)(x+2)$，故当 $x\ne 2$ 时 $f(x)=\dfrac{2x+3}{x+2}$。(M1)
The factor $(x+2)$ does not cancel, and at $x=-2$ the reduced numerator is $2(-2)+3=-1\ne 0$, so $x=-2$ is a vertical asymptote. As $x\to -2^{+}$, the denominator $\to 0^{+}$ with numerator near $-1$, so $f\to-\infty$. (A1)因子 $(x+2)$ 不消去，且在 $x=-2$ 处约简后分子为 $2(-2)+3=-1\ne 0$，故 $x=-2$ 为竖直渐近线。当 $x\to -2^{+}$ 时，分母 $\to 0^{+}$，分子接近 $-1$，故 $f\to-\infty$。(A1)
As $x\to -2^{-}$, the denominator $\to 0^{-}$ with numerator still near $-1$, so $f\to+\infty$. (A1)当 $x\to -2^{-}$ 时，分母 $\to 0^{-}$，分子仍接近 $-1$，故 $f\to+\infty$。(A1)
The factor $(x-2)$ cancels top and bottom, so $f$ is undefined but not asymptotic at $x=2$. (M1)因子 $(x-2)$ 在分子分母中均可消去，故 $f$ 在 $x=2$ 处无定义但不是渐近线。(M1)
The hole sits at the limiting height of the reduced form: (M1)空洞位于约简形式的极限高度处：(M1)
$$ \lim_{x\to 2}\frac{2x+3}{x+2}=\frac{7}{4}. $$So there is a removable discontinuity, a hole, at $\left(2,\tfrac{7}{4}\right)$. (A1)故存在一个可去不连续点，即空洞，位于 $\left(2,\tfrac{7}{4}\right)$。(A1)
Numerator and denominator of the original $f$ both have degree $2$, so divide by $x^{2}$: (M1)原式 $f$ 的分子分母次数均为 $2$，故除以 $x^{2}$：(M1)
$$ \lim_{x\to\pm\infty}\frac{2x^{2}-x-6}{x^{2}-4}=\lim_{x\to\pm\infty}\frac{2-1/x-6/x^{2}}{1-4/x^{2}}=\frac{2}{1}=2. $$Both $\displaystyle\lim_{x\to\infty}f(x)$ and $\displaystyle\lim_{x\to-\infty}f(x)$ equal $2$. (A1) The horizontal asymptote is $y=2$. (A1)$\displaystyle\lim_{x\to\infty}f(x)$ 与 $\displaystyle\lim_{x\to-\infty}f(x)$ 均等于 $2$。(A1) 水平渐近线为 $y=2$。(A1)
A diver's depth $d(t)$ in meters at time $t$ seconds is continuous on $[0,20]$, with the values in the table above.潜水员的深度 $d(t)$（单位：米）在时刻 $t$（单位：秒）处连续，定义在 $[0,20]$ 上，取值如表所示。
| $t$ (s) | 0 | 4 | 10 | 15 | 20 |
|---|---|---|---|---|---|
| $d(t)$ (m) | 0 | 8 | 22 | 18 | 5 |
On $[4,10]\subset[0,20]$, $d$ is continuous (given). (M1)在 $[4,10]\subset[0,20]$ 上，$d$ 连续（已知）。(M1)
$d(4)=8$ and $d(10)=22$, and $15$ lies between $8$ and $22$. (A1)$d(4)=8$，$d(10)=22$，且 $15$ 介于 $8$ 与 $22$ 之间。(A1)
By the IVT, there exists $c\in(4,10)\subset(0,10)$ with $d(c)=15$. (R1)由介值定理，存在 $c\in(4,10)\subset(0,10)$ 使 $d(c)=15$。(R1)
On $[10,15]$, $d(10)=22$ and $d(15)=18$ are both above $15$, so this subinterval alone does not bracket $15$. But on $[15,20]\subset[0,20]$, $d$ is continuous. (M1)在 $[10,15]$ 上，$d(10)=22$ 与 $d(15)=18$ 均大于 $15$，故该子区间本身无法夹住 $15$。但在 $[15,20]\subset[0,20]$ 上，$d$ 连续。(M1)
$d(15)=18$ and $d(20)=5$, and $15$ lies between $5$ and $18$. (A1)$d(15)=18$，$d(20)=5$，且 $15$ 介于 $5$ 与 $18$ 之间。(A1)
By the IVT, there exists $c\in(15,20)\subset(10,20)$ with $d(c)=15$: yes, the IVT is enough, using the subinterval $[15,20]$ rather than the whole of $[10,20]$. (R1)由介值定理，存在 $c\in(15,20)\subset(10,20)$ 使 $d(c)=15$：是的，介值定理足以得出结论，只需使用子区间 $[15,20]$，而非整个 $[10,20]$。(R1)
Every tabulated value of $d$ is at most $22$, so no two consecutive entries bracket $25$: there is no pair $d(t_1)<25
The claim is incorrect: the IVT never guarantees a value outside the range spanned by known outputs, so nothing here certifies $d(t)=25$ anywhere on $[0,20]$. (R1)该说法不正确：介值定理绝不会保证已知输出范围之外的某个值一定被取到，故本题数据无法证明 $[0,20]$ 上任何时刻有 $d(t)=25$。(R1)
Compare each tabulated value to $15$: $d(0)=0$ (below), $d(4)=8$ (below), $d(10)=22$ (above), $d(15)=18$ (above), $d(20)=5$ (below). Crossings are guaranteed only where consecutive values sit on opposite sides of $15$: between $t=4$ and $t=10$, and between $t=15$ and $t=20$. (A1)将各表格值与 $15$ 比较：$d(0)=0$（低于），$d(4)=8$（低于），$d(10)=22$（高于），$d(15)=18$（高于），$d(20)=5$（低于）。只有相邻值分别位于 $15$ 两侧时才保证有穿越：即 $t=4$ 与 $t=10$ 之间，以及 $t=15$ 与 $t=20$ 之间。(A1)
That gives two guaranteed crossings, so the diver is at depth $15$ meters at least $2$ times on $[0,20]$. (R1)共有两次保证穿越，故潜水员在 $[0,20]$ 上至少 $2$ 次处于 $15$ 米深处。(R1)
Let $f(x)=x^{2}\cos\!\bigl(\tfrac{1}{x}\bigr)$ for $x\ne 0$, and define $f(0)=0$.设 $f(x)=x^{2}\cos\!\bigl(\tfrac{1}{x}\bigr)$（$x\ne 0$），并定义 $f(0)=0$。
For all $x\ne 0$, $-1\le\cos\!\left(\tfrac{1}{x}\right)\le 1$. (M1)对所有 $x\ne 0$，$-1\le\cos\!\left(\tfrac{1}{x}\right)\le 1$。(M1)
Multiplying through by $x^{2}\ge 0$ preserves the inequality direction: (M1)乘以 $x^{2}\ge 0$ 不改变不等号方向：(M1)
$$ -x^{2}\le x^{2}\cos\!\left(\frac{1}{x}\right)\le x^{2}. $$Both $-x^{2}\to 0$ and $x^{2}\to 0$ as $x\to 0$. (A1)当 $x\to 0$ 时，$-x^{2}\to 0$ 且 $x^{2}\to 0$。(A1)
Since the trapped quantity $f(x)$ is squeezed between two functions with the same limit, the Squeeze Theorem gives $\displaystyle\lim_{x\to 0}f(x)=0$. (R1)因被夹量 $f(x)$ 夹在两个极限相同的函数之间，由夹逼定理得 $\displaystyle\lim_{x\to 0}f(x)=0$。(R1)
Part (a) shows $\displaystyle\lim_{x\to 0}f(x)=0$, and by definition $f(0)=0$. (A1)由 (a)，$\displaystyle\lim_{x\to 0}f(x)=0$，且由定义 $f(0)=0$。(A1)
Since the limit exists and equals the function's value at that point, $f$ satisfies the definition of continuity at $x=0$. (R1)因极限存在且等于该点处的函数值，$f$ 满足在 $x=0$ 处连续的定义。(R1)
$\displaystyle\frac{f(x)-f(0)}{x-0}=\frac{x^{2}\cos(1/x)-0}{x}=x\cos\!\left(\frac{1}{x}\right)$ for $x\ne 0$. (M1)当 $x\ne 0$ 时，$\displaystyle\frac{f(x)-f(0)}{x-0}=\frac{x^{2}\cos(1/x)-0}{x}=x\cos\!\left(\frac{1}{x}\right)$。(M1)
Since $-1\le\cos(1/x)\le 1$, multiplying by $|x|$ (to keep the inequality valid whether $x$ is positive or negative) gives $-|x|\le x\cos(1/x)\le |x|$, and both bounds $\to 0$ as $x\to 0$, so by the Squeeze Theorem $\displaystyle\lim_{x\to 0}\frac{f(x)-f(0)}{x-0}=0$. (A1)因 $-1\le\cos(1/x)\le 1$，乘以 $|x|$（以保证不论 $x$ 正负不等式均成立）得 $-|x|\le x\cos(1/x)\le |x|$，且当 $x\to 0$ 时两界均 $\to 0$，由夹逼定理得 $\displaystyle\lim_{x\to 0}\frac{f(x)-f(0)}{x-0}=0$。(A1)
This limit is exactly $f'(0)$, so $f$ is differentiable at $x=0$ with $f'(0)=0$: the tangent line to $f$ at the origin is horizontal. (R1)该极限正是 $f'(0)$，故 $f$ 在 $x=0$ 处可导，且 $f'(0)=0$：$f$ 在原点处的切线为水平线。(R1)