※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 (0,2) 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
- ∫120√4−x2dx
- ∫2√30√4−x2dx
- ∫2√30√4+x2dx
- ∫120√4+x2dx
- ∫1201√4−x2dx
- ∫e2+x3dx.
- e2+x3+C
- 2e2+x3+C
- 12e2+x3+C
- 13e2+x3+C
- 3e2+x3+C
- Let f(x)=x2+2x−12x3+1, then f′(1) is
- −1
- 0
- 1
- 2
- None of the above
- Let f(x)=√sec4x, then f′(x) is
- tan4x2√sec4x
- 12√sec4x
- 2√sec4x
- 2√sec4xtan4x
- None of the above
- Find lim.
- 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 (0,2) 而準線為 y=4 的拋物線方程式為 x2=−4(y−3)=12−4y,即 y=12−x24。那麼在 x 軸上方的範圍為 y≥0,即 x2≤12,因此 x∈[−2√3,2√3]。運用曲線弧長公式可知s=∫2√3−2√3√1+y′2(x)dx=2∫2√30√1+(−x2)2dx=∫2√30√4+x2dx
故選(C)。訣竅
運用變數代換法可立刻求解。解法
令 u=2+x3,整理後求導可知 dx=3du,如此所求可改寫並計算如下∫e2+x3dx=∫eu⋅3du=3eu+C=3e2+x3+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f′(x)=(x2+2x−1)′(2x3+1)−(x2+2x−1)(2x3+1)′(2x3+1)2=(2x+2)(2x3+1)−(x2+2x−1)⋅6x2(2x3+1)2
取 x=1 代入有f′(1)=(2⋅1+2)(2⋅13+1)−(12+2⋅1−1)⋅6⋅12(2⋅13+1)2=4⋅3−2⋅632=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f′(x)=ddx(sec4x)12=12(sec4x)−12ddxsec4x=12(sec4x)−12⋅sec4xtan4x⋅4=2sec4xtan4x√sec4x=2√sec4xtan4x
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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