(9%) Find d dx(sec x)x, −π 2 &lt

1021微微微甲甲甲07-11班班班期期期中中中考考考解解解答答答 1. (9%) Find d

dx(sec x)^x, −π

2 < x < π 2. Solution:

(sec x)^x = e^{x ln sec x} (2%) d

dx(sec x)^x = d

dxe^{x ln sec x} = e^{x ln sec x} d

dx(x ln sec x) (3%) d

dx(sec x)^x = (sec x)^x



ln sec x + xsec x tan x sec x



= (sec x)^x(ln x + x tan x) (4%)

2. (9%) Evaluate ˆ 1

0

(2 + x)² 1 + x² dx.

Solution:

(2 + x)² = x²+ 4x + 4 ⇒ ˆ 1

0

(2 + x)² 1 + x² dx =

ˆ 1 0

1 + 4x + 3 1 + x²dx ˆ ₁

0

(2 + x)² 1 + x² dx =

ˆ ₁

0

1dx + 4 ˆ ₁

0

x

1 + x²dx + 3 ˆ ₁

0

1

1 + x²dx (3%) ˆ 1

0

1dx = x   

1

0 = 1 (2%)

4 ˆ ₁

0

x

1 + x²dx = 2 ln(1 + x²)   

1 0

= 2 ln 2 (2%)

3 ˆ 1

0

1

1 + x²dx = 4 arctan x   

1

0 = 3 arctan 1 = 3 · π 4 = 3

4π (2%) ˆ ₁

0

(2 + x)²

1 + x² dx = 1 + 2 ln 2 + 3 4π

3. (9%) (a) Find lim

t→0⁺

t − ln(1 + t) t² (b) Use (a) to find lim

t→0⁺

pt − ln(1 + t) t

Solution:

(a)5 points

this is indeterminate form of type 0

0 (1 point) lim

t→0⁺

t − ln (1 + t)

t² = lim

t→0⁺

1 −_1+t¹

2t = lim

t→0⁺

1

2(1 + t) = 1 2 (process 2 points,answer 2 points)

(b)4 points

because f (x) =√

x is continous at x = 1/2 and lim

t→0⁺

t − ln (1 + t)

t² = 1

2 so

lim

t→0⁺

pt − ln (1 + t)

t = lim

t→0⁺

rt − ln (1 + t)

t² =

r lim

t→0⁺

t − ln (1 + t)

t² =

r1 2 (process 2 points,answer 2 points)

4. (9%) Evaluate lim

x→∞(2^x+ 3^x+ 5^x)^x1. Solution:

解法一

2^x+ 3^x+ 5^x = 5^x



1 + 2 5

x

+ 3 5

x

(2分)

(2^x+ 3^x+ 5^x)^1x = 5



1 + 2 5

x

+ 3 5

x¹_x

(3分)

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ln lim

x→∞



1 + 2 5

x

+ 3 5

x¹_x!

= lim

x→∞

ln 1 + ²₅
x

+ ³₅
x

x = 0

⇒ lim

x→∞



1 + 2 5

x

+ 3 5

x_x¹

= e⁰ = 1 或直接說

x→∞lim



1 + 2 5

x

+ 3 5

x_x¹

= 1⁰ = 1

(3分)

x→∞lim(2^x+ 3^x+ 5^x)^1x = 5 × 1 = 5 (1 分) 解法二

ln(2^x+ 3^x+ 5^x)^1x = ln(2^x+ 3^x+ 5^x)

x (2分)

x→∞lim

ln(2^x+ 3^x+ 5^x) x

l⁰Hospital

======= lim

x→∞

(ln 2)2^x+(ln 3)3^x+(ln 5)5^x 2^x+3^x+5^x

1 (3分)

= lim

x→∞

(ln 2) ²₅
x

+ (ln 3) ³₅
x

+ (ln 5)

2 5

x

+ ³₅
x

+ 1 (2分)

= ln 5 (1分)

x→∞lim(2^x+ 3^x+ 5^x)^1x = e^{ln 5} = 5 (1 分) 解法三

When x ≥ 1

2^x+ 3^x+ 5^x < 5^x+ 5^x+ 5^x= 3 × 5^x (2分) 5^x < 2^x+ 3^x+ 5^x (2分)

Because

x→∞lim(5^x)^1x = 5 (1分)

x→∞lim(3 × 5^x)^1x = ( lim

x→∞3^1x) × 5 = 1 × 5 = 5 (3分) By Squeeze Theorem

x→∞lim(2^x+ 3^x+ 5^x)^1x = 5 (1 分)

5. (9%) Find the linear approximation of the function g(x) = sin⁻¹ x − 1

x + 1



− tan⁻¹ √ x

at the point x = 3

Solution:

Linear approximation at x = 3 is

g(x) ≈ g(3) + g⁰(3) · (x − 3) (1%)

d

dx(sin⁻¹(x − 1

x + 1)) = 1 q

1 − (^x−1_x+1)²

· d

dx(x − 1 x + 1)

= 1

√x(x + 1) (1% for the chain rule term, 3% in total) d

dx(tan⁻¹(√

x)) = 1

1 + (√

x)² · d dx(√

x)

= 1

2√

x(x + 1) (1% for the chain rule term, 3% in total)

⇒ g⁰(x) = 1 2√

x(x + 1) and g⁰(3) = 1 8√

3 (1%)

g(3) = sin⁻¹(1

2) − tan⁻¹(√ 3)

= π 6 − π

3 = −π

6 (1%)

⇒ g(x) ≈ −π 6 + 1

8√

3(x − 3)

6. (10%) Let y = f (x) satisfy x³+ 2xy + y³ = 13. Find y⁰ and y⁰⁰ at the point x = 1, y = 2.

Solution:

x³+ 2xy + y³ = 13 Implicit differentiation gives

3x²+ 2y + 2xy⁰+ 3y²y⁰ = 0 (∗) 4分

y⁰ = −3x² − 2y 2x + 3y² At (x, y) = (1, 2), y⁰ = −3 − 4

2 + 12 = −1

2 2 分

Differentiate (*) once more we have

6x + 2y⁰+ 2y⁰+ 2xy⁰⁰+ 6y(y⁰)²+ 3y²y⁰⁰= 0 2 分 At (x, y) = (1, 2),

6 + 4



−1 2



+ 2y⁰⁰+ 6 · 2 ·



−1 2

2

+ 3 · 4y⁰⁰ = 0 4 + 3 + 14y⁰⁰= 0 y⁰⁰ = −1

2 2分

7. (10%) Evaluate lim

x→0⁺

ˆ _x²

0

e^t√ t sin√

tdt + x cos x − x

x³ .

Solution:

lim

x→0⁺

´x² 0 e^t√

t sin√

tdt + x cos x − x x³

 0 0



= lim

x→0⁺

(e^x2x sin x) · 2x + (cos x − x sin x − 1) 3x²

= lim

x→0⁺

(2x²e^x2sin x) − x sin x + (cos x − 1) 3x²

= lim

x→0⁺

 2

3e^x2sin x − x sin x

3x² +cos x − 1 3x²



= lim

x→0⁺

2

3e^x2sin x − lim

x→0⁺

1 3

sin x

x + lim

x→0⁺

1 3

cos x − 1 x²

= 0 − 1 3 +1

3 · −1 2



= −1 2.

Grading Policy:

Identify  0 0



type: 3 points d

dx ˆ x²

0

e^t√ t sin√

tdt: 2 points (x cos x − x)⁰: 2 points

Three limits: each get 1 point

8. (15%) The minute hand (分針) on a clock is 8cm long and the hour hand (時針) is 4cm long. How fast is the distance between the tips of the hands changing at two o’clock? Give your answer in the unit cm/hour.

Solution:

Let θ be the angle of minute hand and hour hand.

Let X be the distance of minute hand and hour hand.

By cosine theorem:[3 pts]

X² = 8²+ 4² − 2 × 8 × 4 × cos θ At 2 o’clock. θ = π

3 and X = 4√

3. [2 pts]

The angle vector of minute hand is 2π and the angle vector of hour hand is π

6 [4 pts]

Moreover, the angle vector of θ is π

6 − 2π = −11π

6 . [2 pts]

By diffrerntating: [2 pts]

2XdX

dt = 64 sin θdθ dt To end up, dX

dt = −22π

3 [2 pts].

9. (20%) Let y = f (x) = x − x²

6 − 2 ln x

3 , x > 0.

Answer the following questions and give your reasons (including computations). Put ”None” in the blank if the item asked does not exist.

(a) (5%) Find the interval(s) on which f is increasing.

Answer:

(b) (4%) Find the local maximal point(s) and minimal point(s) of f , if any.

Answer: local maximal point(s) (x, y) = local minimal point(s) (x, y) =

(c) (5%) Find the interval(s) on which f is concave up.

Answer:

(d) (2%) Find the inflection point(s) if any.

Answer:

(e) (4%) Sketch the graph of f . Indicate all information in (a)-(d).

Solution:

a. f⁰(x) > 0

⇒ 1 − x 3 − 2

3x > 0

⇒ x − x² 3 −2

3 > 0, (x > 0)

⇒ 1 < x < 2

(b) by (a)local maximal point (x, y) = (2,4 − 2ln2

3 )

local minimal point (x, y) = (1,5 6)

(c) f⁰⁰(x) > 0

⇒ −1 3 + 2

3x² > 0, (x > 0)

⇒ 0 < x <√ 2

(d) by (c) the inflection point (x, y) = (√ 2,√

2 −1 − ln2

3 )

(e) f is concave up on(0,√

2), f is concave down on (√

2, ∞),and lim

x→0f (x) = ∞.

If you are totally right of (a),(b),(c),(d),I think you will get the full scores of (e).

Sketching the graph of f by yourself.