※注意：請用 2B 鉛筆作答於答案卡，並先詳閱答案卡上之「畫記說明」。每題 $5$ 分。
- A parabola is drawn having focus $\left(0,2\right)$ and directrix $y=4$. The definite integral representing the arc length of that portion of the parabola on or above the $x$-axis is given by
- $\displaystyle\int_0^{12}\sqrt{4-x^2}dx$
- $\displaystyle\int_0^{2\sqrt3}\sqrt{4-x^2}dx$
- $\displaystyle\int_0^{2\sqrt3}\sqrt{4+x^2}dx$
- $\displaystyle\int_0^{12}\sqrt{4+x^2}dx$
- $\displaystyle\int_0^{12}\frac1{\sqrt{4-x^2}}dx$
- $\displaystyle\int e^{\frac{2+x}3}dx$.
- $e^{\frac{2+x}3}+C$
- $2e^{\frac{2+x}3}+C$
- $\displaystyle\frac12e^{\frac{2+x}3}+C$
- $\displaystyle\frac13e^{\frac{2+x}3}+C$
- $3e^{\frac{2+x}3}+C$
- Let $\displaystyle f\left(x\right)=\frac{x^2+2x-1}{2x^3+1}$, then $f'\left(1\right)$ is
- $-1$
- $0$
- $1$
- $2$
- None of the above
- Let $f\left(x\right)=\sqrt{\sec4x}$, then $f'\left(x\right)$ is
- $\displaystyle\frac{\tan4x}{2\sqrt{\sec4x}}$
- $\displaystyle\frac1{2\sqrt{\sec4x}}$
- $2\sqrt{\sec4x}$
- $2\sqrt{\sec4x}\tan4x$
- None of the above
- Find $\displaystyle\lim_{x\to\sqrt3}5^{\frac1{3-x^2}}$.
- $0$
- Does not exist and neither $\infty$ nor $-\infty$
- $\infty$
- $-\infty$
- None of the above
- Find the slope of the tangent line at the point $\left(1,1\right)$ on the graph of $e^{x-y}=2x^2-y^2$.
- $0$
- $-1$
- $1$
- $2$
- $3$
- Compute the linearization of $f\left(x\right)=\sqrt{x}e^{x-1}$ at $a=1$.
- $\displaystyle L\left(x\right)=\frac32x+\frac12$
- $\displaystyle L\left(x\right)=-\frac32x+\frac12$
- $\displaystyle L\left(x\right)=-\frac32x-\frac12$
- $\displaystyle L\left(x\right)=\frac32x-\frac12$
- None of the above
- Find $\displaystyle\lim_{x\to2}\frac{x^3-8}{x^2-4}$.
- $0$
- $1$
- $2$
- $3$
- None of the above
- If the function $f\left(x\right)$ is differentiable and $f\left(x\right)=\begin{cases}ax^3-6x&\mbox{if}~x\leq1\\bx^2+4&\mbox{if}~x>1\end{cases}$, then $a=$
- $0$
- $1$
- $-14$
- $-24$
- $26$
- Given that the function $f$ is continuous on the interval $\left[1,\infty\right)$, and that $\displaystyle\int_1^x\sqrt{f\left(t\right)}dt=\sqrt{x}$, then $\displaystyle\int_1^{\infty}f^2\left(t\right)dt=$
- $0$
- $\displaystyle\frac1{16}$
- $\displaystyle\frac14$
- $1$
- $\infty$
- $\displaystyle\lim_{x\to1^+}\frac{\exp\left(x^2-1\right)}{x-1}$.
- $0$
- $1$
- $2$
- $\infty$
- None of the above
- $f\left(x\right)=\left(x-1\right)\left(x-2\right)^2\left(x-3\right)^3$. Find the value of $x$ that maximizes $f'\left(x\right)$.
- $0$
- $1$
- $2$
- $3$
- None of the above
- $f\left(x\right)=x^x$. Find $f'\left(1\right)$.
- $0$
- $1$
- $e$
- $\displaystyle\frac1e$
- None of the above
- Compute the area of the region enclosed by the graphs of the given equations: $y=e^x$, $y=e^{-x}$, and $x=\ln3$.
- $1$
- $\displaystyle\frac43$
- $\displaystyle\frac53$
- $2$
- None of the above
- $\displaystyle F\left(x\right)=\int_0^x\left(z-3\right)^2\left(z-4\right)^3\left(z-5\right)^4e^{2z}dz$. Find the value of $x\in\left[0,6\right]$ that minimizes $F\left(x\right)$.
- $3$
- $4$
- $5$
- $6$
- None of the above
- Suppose $\displaystyle\lim_{x\to0^+}f\left(x\right)=A$ and $\displaystyle\lim_{x\to0^-}f\left(x\right)$. Find $\displaystyle\lim_{x\to0^-}f\left(x^2-x\right)$.
- $A$
- $B$
- $A^2-B$
- $A^2+B$
- None of the above
- The integral of a constant is a constant.
- True
- False
- If $f''\left(x\right)=g'\left(x\right)$, then $f'\left(x\right)=g\left(x\right)$.
- True
- False
- If $f''\left(x\right)>0$ for all $x\in\left[a,b\right]$ and $b>a$, then $\max\left[f\left(a\right),f\left(b\right)\right]>f\left(z\right)$ for all $z\in\left(a,b\right)$.
- True
- False
- If $f''\left(x\right)<0$ for all $x\in\left[a,b\right]$ and $b>a$, then there exists $z\in\left[a,b\right]$ such that $\max\left[f\left(a\right),f\left(b\right)\right]<f\left(z\right)$.
- True
- False
訣竅
首先根據題意寫出拋物線方程式，隨後運用曲線弧長公式表達之。解法
焦點在 $\left(0,2\right)$ 而準線為 $y=4$ 的拋物線方程式為 $x^2=-4\left(y-3\right)=12-4y$，即 $\displaystyle y=\frac{12-x^2}4$。那麼在 $x$ 軸上方的範圍為 $y\geq0$，即 $x^2\leq12$，因此 $x\in\left[-2\sqrt3,2\sqrt3\right]$。運用曲線弧長公式可知$\displaystyle s=\int_{-2\sqrt3}^{2\sqrt3}\sqrt{1+y'^2\left(x\right)}dx=2\int_0^{2\sqrt3}\sqrt{1+\left(-\frac{x}2\right)^2}dx=\int_0^{2\sqrt3}\sqrt{4+x^2}dx$
故選(C)。訣竅
運用變數代換法可立刻求解。解法
令 $\displaystyle u=\frac{2+x}3$，整理後求導可知 $dx=3du$，如此所求可改寫並計算如下$\displaystyle\int e^{\frac{2+x}3}dx=\int e^u\cdot3du=3e^u+C=3e^{\frac{2+x}3}+C$
故選(E)。訣竅
運用微分公式求導計算即可。解法
使用微分公式可得$\displaystyle f'\left(x\right)=\frac{\left(x^2+2x-1\right)'\left(2x^3+1\right)-\left(x^2+2x-1\right)\left(2x^3+1\right)'}{\left(2x^3+1\right)^2}=\frac{\left(2x+2\right)\left(2x^3+1\right)-\left(x^2+2x-1\right)\cdot6x^2}{\left(2x^3+1\right)^2}$
取 $x=1$ 代入有$\displaystyle f'\left(1\right)=\frac{\left(2\cdot1+2\right)\left(2\cdot1^3+1\right)-\left(1^2+2\cdot1-1\right)\cdot6\cdot1^2}{\left(2\cdot1^3+1\right)^2}=\frac{4\cdot3-2\cdot6}{3^2}=0$
應選(B)。訣竅
運用連鎖律與基本函數的微分公式計算即可。解法
使用連鎖律求導可知$\displaystyle\begin{aligned}f'\left(x\right)&=\frac{d}{dx}\left(\sec4x\right)^{\frac12}=\frac12\left(\sec4x\right)^{-\frac12}\frac{d}{dx}\sec4x\\&=\frac12\left(\sec4x\right)^{-\frac12}\cdot\sec4x\tan4x\cdot4=\frac{2\sec4x\tan4x}{\sqrt{\sec4x}}=2\sqrt{\sec4x}\tan4x\end{aligned}$
故選(D)。訣竅
仔細極限式中各項的趨近行為即可。解法
由於當 $x>\sqrt3$ 時有 $3-x^2<0$，而當 $x<\sqrt3$ 時有 $3-x^2>0$，故 $\displaystyle\lim_{x\to\sqrt3^\pm}\frac1{3-x^2}=\mp\infty$，進而有 $\displaystyle\lim_{x\to\sqrt3^+}5^{\frac1{3-x^2}}=0$ 與 $\displaystyle\lim_{x\to\sqrt3^-}5^{\frac1{3-x^2}}=\infty$，故選(B)。訣竅
運用隱函數微分求導，隨後代入給定的座標即可求得該點的斜率。解法
使用隱函數微分求導可得$\displaystyle e^{x-y}\left(1-\frac{dy}{dx}\right)=4x-2y\frac{dy}{dx}$
取 $\left(x,y\right)=\left(1,1\right)$ 可得 $\displaystyle1-\left.\frac{dy}{dx}\right|_{\left(x,y\right)=\left(1,1\right)}=4-2\left.\frac{dy}{dx}\right|_{\left(x,y\right)=\left(1,1\right)}$，如此解得 $\displaystyle\left.\frac{dy}{dx}\right|_{\left(x,y\right)=\left(1,1\right)}=3$，故選(E)。訣竅
求導後使用點斜式獲得一階逼近多項式。解法
求導有 $\displaystyle f'\left(x\right)=\frac1{2\sqrt{x}}e^{x-1}+\sqrt{x}e^{x-1}$，那麼在 $x=1$ 處的導數值為 $\displaystyle f'\left(1\right)=\frac32$，如此所求的一階逼近多項式為$\displaystyle L\left(x\right)=f\left(1\right)+f'\left(1\right)\left(x-1\right)=1+\frac32\left(x-1\right)=\frac32x-\frac12$
故選(D)。訣竅
約分後求解即可；亦可使用羅必達法則。解法一
約去因式 $x-2$ 後有$\displaystyle\lim_{x\to2}\frac{x^3-8}{x^2-4}=\lim_{x\to2}\frac{x^2+2x+4}{x+2}=\frac{2^2+2\cdot2+4}{2+2}=3$
故選(D)。解法二
使用羅必達法則可知$\displaystyle\lim_{x\to2}\frac{x^3-8}{x^2-4}=\lim_{x\to2}\frac{3x^2}{2x}=\frac{3\cdot2^2}{2\cdot2}=3$
故選(D)。訣竅
由於可導函數必連續，故由連續性以及左右導數相同可獲得聯立方程式。解法
由連續性可知$\displaystyle b+4=\lim_{x\to1^+}f\left(x\right)=f\left(1\right)=a-6$
再者，由左右導數相同有$\displaystyle\begin{aligned}2b&=\lim_{x\to1^+}\frac{\left(bx^2+4\right)-\left(b+4\right)}{x-1}=\lim_{x\to1^+}\frac{f\left(x\right)-f\left(1\right)}{x-1}\\&=\lim_{x\to1^-}\frac{f\left(x\right)-f\left(1\right)}{x-1}=\lim_{x\to1^-}\frac{\left(ax^3-6x\right)-\left(a-6\right)}{x-1}=\lim_{x\to1^-}\left[a\left(x^2+x+1\right)-6\right]=3a-6\end{aligned}$
因此有聯立方程式如下$\left\{\begin{aligned}&a-b=10\\&3a-2b=6\end{aligned}\right.$
可解得 $a=-14$，故選(C)。訣竅
利用微積分基本定理求得函數後，計算瑕積分求解即可。解法
運用微積分基本定理兩邊取導可知$\displaystyle\sqrt{f\left(x\right)}=\frac1{2\sqrt{x}}$
取四次方有 $\displaystyle f^2\left(x\right)=\frac1{16x^2}$，從而所求的瑕積分之值為$\displaystyle\int_1^{\infty}f^2\left(t\right)dt=\frac1{16}\int_1^{\infty}\frac{dt}{t^2}=\left.-\frac1{16t}\right|_1^{\infty}=\frac1{16}$
應選(B)。訣竅
分析討論分子分母的趨向性。解法
由於分母恆正且趨於零，但分子恆正但不趨於零，故所求發散至正無窮，選(D)。訣竅
為了求一階導函數的極大值，那麼要求二階導函數為零的位置，並用三階導函數判別其特性。解法
求一階導函數可知$\begin{aligned}f'\left(x\right)&=\left(x-2\right)^2\left(x-3\right)^3+2\left(x-1\right)\left(x-2\right)\left(x-3\right)^3+3\left(x-1\right)\left(x-2\right)^2\left(x-3\right)^2\\&=\left[\left(x-2\right)\left(x-3\right)+2\left(x-1\right)\left(x-3\right)+3\left(x-1\right)\left(x-2\right)\right]\left(x-2\right)\left(x-3\right)^2\\&=2\left(3x^2-11x+9\right)\left(x-2\right)\left(x-3\right)^2\end{aligned}$
為了找出 $f'\left(x\right)$ 極大值的 $x$，我們進一步解方程式 $f''\left(x\right)=0$，即$\begin{aligned}f''\left(x\right)&=2\left(6x-11\right)\left(x-2\right)\left(x-3\right)^2+2\left(3x^2-11x+9\right)\left(x-3\right)^2+4\left(3x^2-11x+9\right)\left(x-2\right)\left(x-3\right)\\&=2\left(x-3\right)\left(15x^3-95x^2+195x-129\right)\end{aligned}$
故有 $x=3$ 與另外三個不易表達的實根，然而在 $x=3$ 附近之斜率由負轉正，故達到極小值，因此以上皆非，選(E)。事實上，$\displaystyle\lim_{x\to\infty}f'\left(x\right)=\infty$，故 $f'\left(x\right)$ 本身便無最大值。
訣竅
換底後使用連鎖律求導即可。解法
求導可知$\displaystyle f'\left(x\right)=\frac{d}{dx}x^x=\frac{d}{dx}e^{x\ln x}=e^{x\ln x}\cdot\frac{d}{dx}\left(x\ln x\right)=x^x\left(1+\ln x\right)$
取 $x=1$ 代入有 $f'\left(1\right)$，故選(B)。訣竅
確定曲線交點得積分範圍，將上下曲線相減後取積分可求得面積。解法
容易看出 $y=e^x$ 與 $y=e^{-x}$ 交於 $\left(0,1\right)$，因此所求面積為$\displaystyle A=\int_0^{\ln3}\left(e^x-e^{-x}\right)dx=\left(e^x+e^{-x}\right)\Big|_0^{\ln3}=\left(3+\frac13\right)-2=\frac43$
選(B)。
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