微甲

991微甲^01-05班期中考解答和評分標準

1. (14%) Determine whether the following limits exist. If the limit exists, evaluate it. If the limit does not exist, explain why?

(a) lim

x→0

(

sin⁻¹x )(

sin1 x

)

= .

(b) lim

x→1

( x

x− 1− 1 ln x

)

= .

Sol:

(a) Since ¯¯

¯¯sin1 x

¯¯¯¯ ≤ 1, ∀x 6= 0, (2%) we have

−¯¯sin⁻¹x¯¯ ≤(sin⁻¹x)(sin1

x)≤¯¯sin⁻¹x¯¯, ∀x 6= 0. (2%) (missing absolute values: 1%)

Also, lim

x→0¯¯sin⁻¹x¯¯= sin⁻¹(0) = 0. (2%)

By the squeeze theorem, the limit exists and equals 0. (1%) (b) Apply l’Hˆopital’s rule,

xlim→1( x

x− 1− 1

ln x) = lim

x→1

x ln x− (x − 1) (x− 1) ln x

= lim

x→1

ln x ln x + 1−_x¹

= lim

x→1 1 x 1 x+ _x¹2

= 1 2

Note: Correctly using l’Hˆopital’s rule (2%), differentiation calculation (2%), the limit value (3%).

2. (12%) Let f (x) =







sin²ax

x , x > 0

|2x + 1| − |2x − 1| + b cos x, x ≤ 0,

where a, b are constants.

(a) For what values of a and b is f (x) continuous at x = 0?

Answer: a : , b : .

(b) For what values of a and b is f (x) differentiable at x = 0?

Answer: a : , b : .

Sol:

(a) f is continuous at x = 0

⇒ lim

x→0⁺f (x) = lim

x→0⁻f (x) = f (0) (1%)

⇒ lim

x→0⁺

sin²ax

x = lim

x→0⁻|2x + 1| − |2x − 1| + b cos x = 1 − 1 + b cos 0 (1%) If a6= 0

⇒ lim

x→0⁺a²x

(sin ax ax

)2

= lim

x→0⁻(2x + 1)− (1 − 2x) + b cos x = b (2%)

⇒ 0 · 1² = b = b ⇒ b = 0 If a = 0 ⇒ f(x) = 0 for x > 0 (1%)

⇒ lim

x→0⁺0 = b ⇒ b = 0

⇒ a ∈ R, b = 0 (1%)

(b) f is differentiable at 0

⇒ f is continuous at 0 and, by (a), b = 0, f(0) = 0 (1%) (or you may try another way to get this credit)

And by definition

⇒ lim

x→0⁺

f (x)− f(0)

x = lim

x→0⁻

f (x)− f(0)

x (1%)

⇒ lim

x→0⁺

sin²ax

x² = lim

x→0⁻

|2x + 1| − |2x − 1|

x (1%)

If a6= 0

⇒ lim

x→0⁺a²

(sin ax ax

)2

= lim

x→0⁻

4x

x (1%)

⇒ a²· 1² = 4 ⇒ a = ±2 (1%) If a = 0 ⇒ f(x) = 0 for x > 0 (1%)

⇒ lim

x→0⁺0 = 4 →←

⇒ a = ±2, b = 0

3. (8%) The equation of the tangent line of the curve sin⁻¹ (y

x )

= tan⁻¹(xy) at (√

2,

√6 2

)

is .

Sol:

sin⁻¹(y

x) = tan⁻¹xy

Implicit differentiation⇒ 1

√

1− (_x^y)²

· x· y⁰− y · 1

x² = 1

1 + (xy)² · (1 · y + x · y⁰)

Let (x, y) = (√ 2,

√6

2 ) ⇒ 1

√

1− (^√₂³)²

·

√2y⁰ −^√₂⁶

2 = 1

1 + (√ 3)² · (

√6 2 +√

2y⁰)

⇒ y⁰ = 5√ 3

6 =slope at (√ 2,

√6 2 )

⇒ The tangent line : y −

√6

2 = 5√ 3

6 (x−√

2) (or 5√

3x− 6y = 2√ 6) 5 points for the implicit differentiation.

2 points for the computation of y⁰ = 5√ 3 6 . 1 point for the equation of the tangent line.

4. (8%) Let f be a continuous function satisfying

∫ _x

0

f (t) dt +

∫ _x³

0

e^−tf (√³

t) dt = xe^2x+ π^x+ x^π for all x > 0.

The explicit formula for f (x) is .

Sol:∫ _x

0

f (t)dt +

∫ _x³

0

e^−tf (√³

t)dt = xe^2x + π^x+ x^π, x > 0 Differentiate both side ⇒ f(x) + e^−x3f (√³

x³)· 3x² = e^2x + xe^2x · 2 + π^x· ln x + πx^π−1

⇒ f(x)(1 + 3x²e^−x3) = (1 + 2x)e^2x+ π^xln x + πx^π−1

⇒ f(x) = (1 + 2x)e^2x+ π^xln x + πx^π−1 1 + 3x²e^−x3

1 point for d dx

∫ x 0

f (t)dt = f (x).

2 points for d dx

∫ x³ 0

e^−tf (√³

t)dt = e^−x3f (√³

x³)· 3x² = 3x²e^−x3f (x).

1 point for d

dxxe^2x = (1 + 2x)e^2x. 3 points for d

dx(π^x+ x^π) = π^xln x + πx^π−1.

1 point for the final answer f (x) = (1 + 2x)e^2x+ π^xln x + πx^π−1 1 + 3x²e^−x3 . 5. (12%) (a)

∫ cos (ln x)

x dx = .

(b) Let f (x) =





 1

1 + x², −1 ≤ x ≤ 0 sec²x, 0 < x≤ 1.

Then

∫ 1

−1

f (x) dx = .

Sol:

(a) Let u = ln x⇒ du = 1

xdx, (2%) then

∫ cos(ln x)

x dx =

∫

cos udu = sin u + C = sin(ln x) + C, C is a constant. (2%) (b) Since

∫ 1

−1

f (x)dx =

∫ 0

−1

f (x)dx +

∫ 1 0

f (x) dx and

∫ 0

−1

f (x)dx =

∫ 0

−1

1

1 + x²dx = tan⁻¹x¯¯¯⁰

−1(2%) = tan⁻¹0−tan⁻¹(−1) = 0−−π 4 = π

4 (2%) and

∫ 1 0

f (x)dx =

∫ 1 0

sec²xdx = tan x¯¯¯¹

0(2%) = tan 1− tan 0 = tan 1 − 0 = tan 1. (2%) Therefore,

∫ 1

−1

f (x)dx = π

4 + tan 1.

6. (14%) Two different methods are applied to estimate ln (1.2) as follows.

(a) The linear approximation of ln x at x = 1 is .

Use this to estimate ln (1.2).

Answer: ln (1.2)≈ by linear approximation.

(b) Apply Mean Value Theorem to show that 1

1 + x < ln (1 + x)

x < 1 for x > 0.

(c) Use the result in (b) to find an interval in which ln (1.2) is located. Then use the midpoint of this interval to estimate ln (1.2).

Answer: ln (1.2)≈ by mean value theorem.

Sol:

(a) (1) L[ln(x)] = ln(1) + d

dxln(x)¯¯¯

x=1· (x − 1) = x − 1 (3pts) (2) L[ln(1.2)] = 0.2 (3pts)

(b) ln(y + 1) is continuous on [0, x], differentiable on (0.x).

By mean value theorm,

there exists a constant c in (0, x) such that ln(1 + x)− ln(1)

x = 1

1 + c < 1.

Also, 1

1 + c > 1

1 + x, so 1

1 + x < ln(1 + x)

x < 1 (5pts) (c) Using (b), 1

1.2 < ln(1.2)

0.2 < 1, 1

6 < ln(1.2) < 1 5. Take midpoint, ln(1.2)≈ 11

60 (3pts)

7. (12%) Let L be any line through the point (3, 24).

(a) The equation of the line L that cuts off the least area from the first quadrant

is .

The least area is .

(b) The equation of the line L which cuts off shotrest segment by the first quadrant

is .

The length of the shortest segment is .

Sol:

(a) Line equation: y− 24 = −8(x − 3) (3 points) the least area is 144 (3 points)

(b) Line equation: y− 24 = −2(x − 3) (3 points) the shortest segment length is 15√

5 (3 points)

Let y− 24 = m(x − 3) be a line L through (3, 24) with slope m < 0 (No need to consider vertical line m =±∞ and horizontal line m = 0) x-intercept of L: 3− 24

m y-intercept of L: 24− 3m (a) Area A(m) = 1

2(3− 24

m)(24− 3m) = 9

2(16− m − 64 m) = 9

2(16− f(m)) where f (m) = m + 64

m , f⁰(m) = 1− 64

m² = 0, m² = 64, m =±8

Since m < 0 and f⁰(m) > 0 when m <−8 and f⁰(m) < 0 when −8 < m < 0, f has an absolute maximum at m = 8, f (−8) = −16

∴ The least area = 9

2(16− (−16)) = 144 L : y− 24 = −8(x − 3) or 8x + y = 48

(b) Squared segment length L(m) = (3− 24

m)²+ (24− 3m)² = 9[65− 16

m − 16m + 64

m² + m²] = 9[65− g(m)]

where g(m) = 16m + 16

m − m²− 64 m², g⁰(m) = 16− 2m − 16

m² +128

m³ = 16m³− 2m⁴− 16m + 128

m³ = −2

m³[(m³+ 8)(m− 8)], m < 0

Since g⁰(m) > 0 when m <−2 and g⁰(m) < 0 when −2 < m < 0, m = −2 The length of the shortest segment = √

(3 + 12)² + (24 + 6)² = 15√ 5 L : y− 24 = −2(x − 3) or y + 2x = 30

• Four answers, 3 points each. But if the answer is correct but the procedure is wrong, no points will be credited.

* For the area in part (a), 2 points will be credited if your answer differs from the correct answer only by a simple factor (e.g. if your answer is 288).

* If your answer of any line equation is in the form of y− a

x− b = m, 1 point will be deducted.

• Partial credits will be given only if the answers are all wrong:

Parameterization of x-intercept and y-intercept: 1 point each, not given twice for (a) and (b).

Formulation of area (in terms of a single parameter) in (a): 1 point.

Formulation of segment length (in terms of a single parameter) in (b): 1 point.

8. (20%) Let f (x) =







|x|^x, x6= 0 1, x = 0.

(a) Find the horizontal asymptotes if exist. Ans: .

(b) f⁰(x) = .

(c) y = f (x) is increasing on interval(s) .

y = f (x) is decreasing on interval(s) .

(d) f⁰⁰(x) = .

(e) y = f (x) is concave upward on interval(s) .

y = f (x) is concave downward on interval(s) . (Hint: f⁰⁰(x) = 0 ⇔ x = −1)

(f) If the extreme values exist,

f (x) has a local maximum: at x = .

f (x) has a local minimum: at x = .

(g) f (x) has inflection point(s): .

(h) Sketch the graph of y = f (x).

Sol:

(a) lim

x→∞|x|^x =∞^∞=∞,. . . (1 point)

x→−∞lim |x|^x = lim

x→−∞e^{x ln|x|} = e^{limx→−∞x ln|x|}

= e^−∞·∞= e^−∞= 0,. . . (1 point) Therefore, y = 0 is a horizontal asymptotes of f (x).

(b) When x6= 0, y = |x|^x ln y = ln|x|^x = x ln|x|

y⁰

y = ln|x| + x

|x|

d

dx|x| = ln |x| + 1

⇒ y⁰ = f⁰(x) =|x|^x(ln|x| + 1) for x 6= 0 When x = 0,

x→0limf (x) = lim

x→0|x|^x = lim

x→0e^{x ln|x|}

= e^{limx→0x ln|x|} = e^limx→0

ln|x|

1x

= e

limx→0 x1

− 1x2 = e^limx→0−x

= e⁰ = 1 = f (0).

Therefore, f (x) is continuous at 0.

Since f⁰(x) =|x|^x(ln|x| + 1) → −∞ as x → 0, So, f⁰(0) doesn‘t exist.

Thus, f⁰(x) =|x|^x(ln|x| + 1), when x 6= 0, . . . (2 point) f⁰(x) =−∞, when x = 0. . . (1 point) (c) f⁰(x) > 0⇒ x > e⁻¹ or x <−e⁻¹

⇒ f(x) is increasing on (−∞, −e⁻¹)∪ (e⁻¹,∞). . . (1 point)

f⁰(x) < 0⇒ −e⁻¹ < x < e⁻¹

⇒ f(x) is decreasing on (−e⁻¹, e⁻¹). . . (1 point) (d) f⁰(x) =|x|^x(ln|x| + 1) for x 6= 0

⇒ f⁰⁰(x) =|x|^x(ln|x| + 1)²+ 1

x|x|^x for x6= 0. . . (2 point) (e) f⁰⁰(x) > 0⇒ x > 0 or x < −1

⇒ f(x) is concave upward on (−∞, −1) ∪ (0, ∞). . . (2 point) f⁰⁰(x) < 0⇒ −1 < x < 0

⇒ f(x) is concave downward on (−1, 0).. . . (1 point) (f) From (c),

f (x) has a local maximum f (−1

e) = e^1e at x =−1

e. . . (2 point) f (x) has a local minimum f (1

e) = e^−1e at x = 1

e. . . (2 point) (g) From (e),

f (x) has inflection points (−1, 1), (0, 1).. . . (1 point) (h) 1. 畫出漸近線^{y = 0}與^lim

x→∞|x|^x =∞. . . (1 point) 2. 標出極值點與函數在^{(0, 1)}的鉛直切線. . . (1 point) 3. 正確的畫出函數的上凹與下凹. . . (1 point)