微乙

991微乙^01-05班期中考解答和評分標準

1. (10%) 求^{(a) lim}

x→1

√x³− 1 + x²− x

x− 1 。 ^{(5%) (b) lim}

n→∞

√4n⁴+ 3n + 1

√3

8n⁶+ 3n⁵− 2 。 ^(5%) Sol:

(a)

xlim→1

√x³− 1 + x²− x

x− 1 = lim

x→1

x³− 1 + x²− x² (x− 1)(√

x³− 1 + x²+ x)

= lim

x→1

x³− 1 (x− 1)(√

x³− 1 + x²+ x)

= lim

x→1

(x− 1)(x²+ x + 1) (x− 1)(√

x³− 1 + x²+ x)

= lim

x→1

x²+ x + 1

√x³− 1 + x² + x

= 1 + 1 + 1

√1− 1 + 1 + 1

= 3 2 (b)

nlim→∞

√4n⁴+ 3n + 1

√3

8n⁶+ 3n⁵− 2 = lim

n→∞

√4n⁴+ 3n + 1×_n¹²

√3

8n⁶+ 3n⁵− 2 ×_n¹2

= lim

n→∞

√

4 + _n³3 +_n¹4 3

√

8 + _n³ − _n²6

=

√4

√3

8

= 1 評分標準^:

1. 使用羅必達者需先證明 2. 沒寫 ^limit 者扣 ³分 3. (b)使用估計者 ⁰分 2. (15%) 給定 ^{x3+ xy + 2y3 = 4}

(a) 求過點 ^{P = (1, 1)} 之切線方程式 。 ^(10%) (b) 求在點 ^{P = (1, 1)} 之 ^d2y

dx² 。 ^(5%)

Sol:

(a) d

dx(x³+ xy + 2y³) = 3x²+ y + xy⁰+ 6y²y⁰ = 0 —– (1)

⇒ 3 + 1 + y⁰+ 6y⁰ = 0 ⇒ y⁰ = −4 7 Therefore the equation of tangent line is:

y− 1

x− 1 = −4 7 (b) d

dx( d

dx) = 6x + y⁰+ y⁰+ xy⁰⁰+ 12y(y⁰)²+ 6y²y⁰⁰ = 0 —– (2)

⇒ 6 − 8

7 + y⁰⁰+ 1216

49 + 6y⁰⁰ ⇒ y⁰⁰= −430 343 評分標準 ^:

如果 ⁽¹⁾ 錯了^, 但是後面的邏輯正確有六分。

如果 ⁽¹⁾ 寫對後面沒寫^, 那有一分。

(2) 寫對有三分^,微分錯誤扣一分。 底下計算錯誤扣一分。

(b) 小題如果是微分的觀念錯誤⁽例如對 ^x2 微分⁾則不適用上述標準^, 原則上只給一分。

3. (10%) 求 ^d

dx(cos x)^{sin x} 。 Sol:

d

dx(cos x)^{sin x}

= d

dxe^{sin x}·ln (cos x) (this for three points)

= e^{sin x}·ln (cos x)·(

cos x· ln (cos x) + sin x ·(−sinx cos x

))

(this for three points respectively)

= cos x^{sin x}· (cos x · ln (cos x) − sin x · tan x) (all of the calculus are correct, get ten points) 4. (10%) 若^{y = αx + 2} 與^{y = 2 ln x} 相切 ^, 求^α 之值以及切點的座標 。

Sol:

The tangent point has the following constraints.

(the same slope & the same point) 2

x = α

αx + 2 = 2 ln x (4 pts)

Solve the equations, we get

(x, y) = (e², 4) (3 pts)

α = 2e⁻² (3 pts)

5. (10%) 估計 ¹

1 + ln 0.997 。 Sol:

Let

f (x) = 1

1 + ln(1 + x), (2%)

we need to evaluate f (−0.003) by linear approximation near x = 0.

f⁰(x) = −1/(1 + x)

(1 + ln(1 + x))² (2%) f⁰(0) = −1

(1 + ln 1)² =−1 (2%) f (a) = f (0) + f⁰(0)a + ag(a), with lim

a→0g(a) = 0 (2%)

f (−0.003) = 1 + (−1)(−0.003) + (−0.003)g(−0.003) ≈ 1 + (−1)(−0.003) = 1.003 (2%) 6. (10%) 求函數 ^{f (x) = x6} − 3x² 在 ^[−2, 3] 上的最大值與最小值 。

Sol:

(1) Differential (3%)

f (x) = x⁶− 3x²

f⁰(x) = 6x⁵− 6x = 6x(x²+ 1)(x + 1)(x− 1) (2) find Candidate points (5%)

f⁰(x) = 0 x = 0, 1,−1 the candidate points are

x = 0, 1,−1x = −2, 3 (boundary points)

(3) find the Max and min by first-order test or second-order test (2%) (i) first-order test:

−2 < x < −1 → f⁰(x) < 0

−1 < x < 0 → f⁰(x) > 0 0 < x < 1 → f⁰(x) < 0 1 < x < 3 → f⁰(x) > 0 f (1), f (−1) are local min ,and f(0) is local Max

compare with the boundary points

f(1)=f(-1)=-2 are minimum, and f(3)=702 is maximum

(i) second-order test:

f⁰⁰(x) = 30x⁴− 6 f⁰⁰(−1) = 24 > 0 f⁰⁰(0) =−6 < 0

f⁰⁰(1) = 24 > 0 f (1), f (−1) are local min ,and f(0) is local Max compare with the boundary points

f(1)=f(-1)=-2 are minimum, and f(3)=702 is maximum 7. (25%) 若y = f (x) = x(x + 5)

x− 4

(a) y = f (x) 在 ⁽區間⁾遞增 。 ^(3%)

y = f (x) 在 ⁽區間⁾遞減 。 ^(3%)

(b) y = f (x) 之極大值 ^{: (}座標⁾。 ^(3%)

y = f (x) 之極小值 ^{: (}座標⁾。 ^(3%)

(c) y = f (x) 在 ⁽區間⁾ 凹向上 。^(3%)

y = f (x) 在 ⁽區間⁾ 凹向下 。^(3%)

(d) y = f (x) 所有的漸近線為

。 ^(4%) (e) 畫出 ^{y = f (x)} 之圖形 。^(3%)

Sol: Let y = f (x) = x(x + 5) x− 4 . (a) Write

f (x) = x + 9 + 36

x− 4, (1)

and compute

f⁰(x) = 1 + (−1)(36)(x − 4)⁻²

= 1− 36

(x− 4)² = (x− 10)(x + 2) (x− 4)² . So f⁰(x) = 0 if and only if x = −2 or x = 10. We see that





f⁰(x) > 0 if x <−2 or x > 10,

f⁰(x) < 0 if − 2 < x < 10 and x 6= 4.

Therefore, f (x) is increasing on (−∞, −2) ∪ (10, ∞) and decreasing on (−2, 4) ∪ (4, 10).

(b) Since f (x) is increasing for x <−2 and decreasing for −2 < x < 4, f(x) must have a local maximum at x =−2 with value f(−2) = 1. Similarly, f(x) is decreasing for 4 < x < 10 and increasing for x > 10, f (x) must have a local minimum at x = 10 with value f (10) = 25.

Consequently, f (x) has a local maximum at (−2, 1) and a local minimum at (10, 25).

(c) To find concavities of f (x), we compute

f⁰⁰(x) = −(36)(−2)(x − 4)⁻³ = 72 (x− 4)³.

Since f⁰⁰(x) never vanish for x 6= 4, so f(x) has no point of inflection. Moveover, f⁰⁰(x) > 0 for x > 4 and f⁰⁰(x) < 0 for x < 4. Hence, f (x) is concave up on (4,∞) and concave down on (−∞, 4).

(d) f (x) is undefined at x = 4 and |f(x)| → ∞ as x → 4, so that y = f(x) has a vertical asymptote x = 4.

By (1), observe that for large |x|, f(x) is close to x + 9. In other words,

|x|→∞lim [(f (x)− (x + 9)] = lim

|x|→∞

36 x− 4 = 0.

Thus, y = x + 9 is an oblique asymptote for y = f (x).

(e) The figure shown below is the graph of y = f (x). The blue lines are asymptotes.

Note that the unit length of x-axis and y-axis are different.

第⁷題評分標準如下^:

(a) (遞增部份⁾ 算出 ^f0(x)得¹ 分 ^,兩個區間 ⁽−∞, −2) 、 ^(10,∞)各 ¹分 。 (遞減部份⁾ 寫成⁽−2, 4) ∪ (4, 10) 或⁽−2, 10) 都算對^, 得³ 分 。

(b) x 座標和^y 座標寫出其中一個就給分 。 若極大極小寫顛倒 ^, 就都不給分。

(c) 算出 ^f00(x) 得² 分^, 兩個區間 ^(4,∞)、 ⁽−∞, 4) 各 ²分 ^,寫顛倒不給分 。

(d) 寫出 ^{x = 4} 得² 分^; 寫出 y = x + 9 , 或計算過程寫出 f (x) = x + 9 + 36(x− 4)⁻¹ 式 子者 ^, 得² 分 。

(e) 分成 ³ 種程度給分^:

(得 ³ 分⁾ 位置 、 趨勢非常接近正確圖形 ^,有標出兩條漸近線 ^,兩個極值標示正確 。 (得 ² 分⁾ 與正確圖形相去不遠 ^, 但有小錯誤 ^,未標示兩條漸近線或標示錯誤 。 (得 ⁰ 分⁾ 與正確圖形的趨勢相差太大^, 錯誤太多 。

(其他⁾ 答案是區間的項目^, 寫成不等式也可以 。 另外 ^, 有沒有包含端點都無所謂 。 8. (10%) 甲車由原點^{(0, 0)}以每小時 ³⁰公里的速度往 ^x軸正向行駛^, 乙車由點 ^{(0, 13) (}以公里

為單位⁾ 以每小時 ²⁰公里的速度往 ^y 軸負向行駛 。 求兩車最近之距離為多少公里^? Sol:

At time t, the position of car A is (30t, 0), and the position of car B is (0, 13−20t). Then, the distance of A and B is √

(30t− 0)²+ [0− (13 − 20t)]² (3pts)

⇒ AB =√

(30t− 0)²+ [0− (13 − 20t)]²

=√

1300t²− 520t + 169

=

√

1300(t²− 2 5t + 1

25) + 169− 1300 25

=

√

1300(t− 1

5)²+ 117

≥√

117 (7pts)

Hence, we have the minimal distance of A and B is √

117. And it happens at 1 5 hour, which is 12 minutes after start.