= lim(x + 1) lim x 1 x 1 (x 2 + 1) 2 (for the latter let y = x2 + 1) lim

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1 1061 微乙 班期中考解答和評分標準 1. (10%) (x + 1)( (a) 求 x+1 9). x 1 x 1 tan (π(x )) (b) 求. x (x ) x (a) (5 points) Method without L Hospital rule: (x + 1)( x+1 9) = (x + 1) x+1 x 1 x 1 x 1 x 1 (x + 1) (for the latter let y = x + 1) = (x + 1) y x 1 y y = d(y ) dy y= Method with L Hospital rule: = ln = 18 ln. (5 points) (x + 1)( x+1 9) = x+1 9 x 1 x 1 x 1 x 1 = x+1 ln x x 1 1 = 18 ln. (5 points) 註 : When doing the derivative of ( x +1 ) = x +1 ln x by chain rule, missing any one term of { x +1, ln, x} will cost you points and missing any two terms of { x +1, ln, x} will cost you 5 points. (b) (5 points) tan (π(x )) sin (π(x )) π = x (x ) x x (π(x )) x cos (π(x )) sin (π(x )) π = x (π(x )) x x cos (π(x )) sin y π = y 0 y x x cos (π(x )) = 1 π = π. (5 points) 註 : You will get no points if you write tan(π(x )) sin(π(x )) = 1 or = 1. x x x x (for the former let y = π(x )) Page 1 of 8

2 . (10%) (a) 陳述平均值定理 (b) 證明 tan 1 y tan 1 x y x 對 y x 0 都成立 (a) (5 points) Assume f(x) is defined in [a, b] and satisfies (i) f(x) is continuous in [a, b]; (ii) f(x) is differentiable in (a, b). Then there exists c (a, b) such that f (c) = f(b) f(a). (5 points) b a 註 1: You will get no points if you write f(c) = f(b) f(a). b a 註 : You will get points if you state Rolle s Theorem instead of Mean Value Theorem. (b) (5 points) We may assume y > x since the case y = x is obvious. Define f(t) = tan 1 t in [x, y]. Then f(t) satisfies the assumptions of Mean Value Theorem. Hence there exists c (x, y) such that f (c) = f(y) f(x). y x That is, Since 1 1 and y x > 0, we have 1 + c c = tan 1 y tan 1 x. ( points) y x tan 1 y tan 1 x 註 : For the last points, one should specify inequality. 1 (y x) y x. ( points) 1 + c c 1 instead of 1 0 to obtain the final 1 + c Page of 8

3 . (10%) (a) 令 h(x) = x x, 求 h (). (b) 令 f(x) = x, 求 f (5) (x). (a) h(x) = e x ln x (1%) h (x) = x x (x ln x) (1%) = x x (ln x + 1) (%) h () = 4(ln + 1) (1%) (b) f(x) = e x ln (1%) f (x) = x ln (1%) f (x) = ln ( x ) = (ln ) x f (5) (x) = (ln ) 5 x (%) 4. (10%) 求 d dx (cos tan 1 x). Method 1 Let f(x) = cos x and g(x) = tan 1 x, d dx (cos tan 1 x) = d dx f(g(x)) = f d (g(x)) g(x) (%) dx = sin tan 1 d x dx (tan 1 x) (4%) = sin tan 1 1 x (4%) x + 1 Method Let f(x) = cos tan 1 x and θ = tan 1 x, where θ ( π/, π/). (%) For both cases of θ [0, π/) and θ ( π/, 0], we have: f(x) = cos θ = 1 x + 1 = (x + 1) 1/ (%) f (x) = ( 1 ) (x + 1) / x (%) = x (x + 1) / (%) Page of 8

4 5. (10%) 用線性逼近估計 1 + sin(0.00). 設 f(x) = 1 + sin(x) f cos(x) (x) = 1 + sin(x) ( pt) ( pt) f(x) L(x) = f(a) + f (a)(x a) choose a = 0 (1 pt) ( pt) f(0.00) L(0.00) = f(0) + f (0)(0.00 0) = = (14%) 用隱函數微分求 dy dx 及 d y dx 等號兩邊對 x 微分得到 y dy dx + y + xdy dy + xy + x dx dx x = 0 ( 式 1) dy 將 x = 1, y = 0 帶入上式得 dx = ( pt) 再將, 式 (1) 兩邊對 x 微分得到 在曲線 y + xy + x y x = 1 的點 x = 1, y = 0 上之值 (5 pt) 6y( dy dx ) + y d y dx + dy dx + dy dx + y xd + y + xdy dx dx + xdy dx + x d y 6x = 0 (5 pt) dx dy 將 x = 1, y = 0, dx = d 帶入上式得 y dx = ( pt) Page 4 of 8

5 7. (%) 若 y = f(x) = 9(x ) x. (a) y = f(x) 在 ( 區間 ) 遞增 (%) y = f(x) 在 ( 區間 ) 遞減 (%) (b) y = f(x) 在 ( 區間 ) 凹向上 (%) y = f(x) 在 ( 區間 ) 凹向下 (%) (c) y = f(x) 之極大值 ( 若存在的話 ): ( 座標 ) (%) y = f(x) 之極小值 ( 若存在的話 ): ( 座標 ) (%) (d) y = f(x) 之所有漸近線為 (4%) (e) 畫出 y = f(x) 之圖形 (%) ˆ y = f(x) = 9(x ) x (a) (6 points) Increment: (, 0) (0, ).. Decrement: (, ) (, ). f (x) = 9x + 81 x 4 f (x) = 0 x 4 0 x 0. 9(x + )(x ) = 0 x = or Increment:f (x) < 0 < x < (, 0) (0, ) Decrement: f (x) < 0 x > or x < (, ) (, ) (b) (6points) concave upward: (, 0) (, ).. concave downward: (, ) (0, ). f (x) = 18x 4 x 5 f (x) = 0 x 5 0 x 0. 18(x 18) = 0 x = or concave upward: f (x) > 0 x > or x < (, 0) (, ). concave downward:f (x) > 0 < x < (, ) (0, ). ˆ (a,b)score: answer 1 point separately(a didn t wirte x 0 lose 0.5 point), right differential equations 1 point, how you find interval 1 points. Page 5 of 8

6 (c) (4points) local max: (, ), local mini:(, ) local max: value of f of f(c) f(x) when x is near c. local mini: value of f of f(c) f(x) when x is near c. ˆ score: answer 1 point separately, right concept points. (d) (4 points) asymptote:x = 0, y = 0 9(x Horizontal Asymptotes: y = 0 ) 9 = x x x x + 9( ) x x 9(x Vertical Asymptotes: x = 0 ) 9( ) x 0 x x x f(x) Slant Asymptotes: y = mx + k x x mean y = x = 0) = m x 9(x ) x 4 9( ) (it s x x 4 ˆ score: answer 1 points separately, right concept points. ( notify: if you jump through process lose 1point.) (f) ( points) Plot the f(x) and indicate asymptote, local point and infection point. ˆ ˆ vertical asymptote 0.5point, horiaontal asymptote 0.5point. ˆ infection points 0.5point separately, local points 0.5point spearately. Page 6 of 8

7 8. (1%) 半徑 之圓形紙, 由圓心 O 截去一角 AOB. 黏住 OA, OB 作成錐形杯, 求最大容量 錐之 πr 體積是 h, 其中 r = 錐底半徑,h = 錐高 A B O 已知錐體體積公式 V = πr h 且 r + h = 9, 0 h, 0 r 1. ( 方法一 ) 將 r = 9 h 帶入錐體體積公式得 兩邊對 h 微分得 V (h) = π(9 h )h V (h) = π(9 h ) 因為局部最大值可能發生在微分等於 0 的點, 欲使 V (h) = 0,h 必須滿足 9 h = 0, 因此 h = ±, 但 h = 不合 ( 因為 0 h ) 使用二階檢測來檢查 h = 時是不是局部最大值 (local maximum), 將 V (h) = π(9 h ) 兩 邊在對 h 微分得 V (h) = πh 所以 V ( ) < 0, 因此 V ( ) 是一個局部最大值 最後檢查邊界點,V (0) = 0 且 V () = 0 皆小於 V ( ), 因此 V ( ) = π 為錐杯的最大體積. ( 方法二 ) 將 h = 9 r 帶入錐體體積公式得 兩邊對 r 微分得 V (r) = π (r 9 r V (r) = πr 9 r r 9 r ) = π (18r r ) 因為局部最大值可能發生在微分等於 0 的點, 欲使 V (r) = 0,r 必須滿足 r(18 r ) = 0, 因此 r = ± 6 或 0, 但 r = 6 不合 ( 因為 0 r ) 使用一階檢測來檢查 r = 0 和 6 是否為局部最大值, 由 V (r) 知當 6 r 6 時 V (r) 0, 且當 r 6 時 V (r) 0 因此 r = 6 為局部最大值 最後檢查邊界點,V (0) = 0 且 V () = 0 皆小於 V ( 6), 因此 V ( 6) = π 為錐杯的最大體積 配分方式 :( 本題作法還有很多種, 不管哪種都是以以下原則配分 ) Page 7 of 8

8 1. 正確寫出 V (h) 或 V (r) 得 4 分. 正確地把上面的函數做一次微分得 分. 解出一次微分等於 0 的點且說明負不合得 分, 未說明負不合得 1 分 4. 答案部份包含 4 分, 斟酌扣分但最多扣到 4 分 (a) 未說明為什麼微分等於 0 的點是局部最大值扣 分 (b) 未檢查邊界點扣 1 分 (c) 答案錯誤扣 1 分 Page 8 of 8