1041微微微甲甲甲01-04班班班期期期中中中考考考解解解答答答和和和評評評分分分標標標準準準 1. (32%) Evaluate the following limits.
(a) (8%) lim
x→−∞(
√
x2+x + x)
(b) (8%) lim
x→0( sin x
x )
1 1−cos x
(c) (8%) lim
x→∞
sin (√1
x2+1)
√
x2+2 −√ x2−1 (d) (8%) lim
x→∞[(
x 1 + x)
x
− 1 e]x
Solution:
(a)
x→−∞lim (
√
x2+x + x) = lim
x→−∞(
√
x2+x + x)
√
x2+x − x
√
x2+x − x (3)
= lim
x→−∞
x
√
x2+x − x
= lim
x→−∞
1
−1
−x
√
x2+x − 1 (as x < 0) (5)
= lim
x→−∞
1
−1
√
x2+x x2 −1
= lim
x→−∞
1
−
√
1 +x1−1
= − 1
2 (8)
(b) Since
x→0lim sin x
x =1, we see that
x→0lim
sin(√1
x2+1)
√
x2+2 −√ x2−1
=lim
x→0
sin(√1
x2+1)
√1 x2+1
⋅lim
x→0
√
x2+2 +√ x2−1
√
x2+1 × 3
→1 ×2
3 (+3)
as x → ∞. Therefore
x→∞lim
sin(√1
x2+1)
√
x2+2 −√
x2−1 = 2
3. (+3)
(c) Rewrite as
( sin x
x )
1
1−cos x =exp( 1
1 − cos xln ∣sin x x ∣).
By l’Hopital theorem, we get
x→0lim 1
1 − cos xln ∣sin x x ∣ =lim
x→0 x sin xx cos x−sin x
x2
sin x
=lim
x→0
x cos x − sin x
x sin2x (3)
=lim
x→0
−x sin x sin2x + 2x sin x cos x
=lim
x→0
−x sin x + 2x cos x
=lim
x→0
−1
3 cos x − 2x sin x= − 1
3. (6)
Since ex is continuous (7), we get
x→0lim( sin x
x )
1
1−cos x =e−1/3. (8)
(d)
x→∞lim (1+11
x
)
x
−1e
1 x
=lim
t→0+
(1+t1 )
1 t −1e
t (x = 1
t) (+2)
=lim
t→0+( 1 1 + t)
1
t −1+tt +ln(1 + t)
t2 (L′Hopital)
= 1 e lim
t→0+
−t + (1 + t) ln(1 + t)
t2(1 + t) (L′Hopital) (+2)
= 1 e lim
t→0+
ln(1 + t)
2t + 3t2 (L′Hopital) (+2)
= 1 e lim
t→0+ 1 1+t
2 + 6t = 1
2e (L′Hopital) (+2)
Page 2 of 12
2. (12%) Let f (x) =
⎧⎪
⎪⎪
⎪⎪
⎪⎪
⎪
⎨
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎩
xαsin ( 1
xβ), x > 0
0, x = 0
sin(xβ)
1 − cos x, x < 0.
(a) For what values of α and β will f (x) be continuous at x = 0?
(b) For what values of α and β will f (x) be differentiable at x = 0?
Solution:
(a) f (x) is continuous at x = 0
Ô⇒ 0 = f (0) = lim
x→0+f (x) = lim
x→0−f (x) (1 point) (1) f (0) = 0
(2) lim
x→0+f (x) = lim
x→0+xαsin ( 1
xβ)(1 point) since
−1 ≤ sin ( 1
xβ) ≤1 ∀β ∈ R and
lim
x→0+−xα=0 = lim
x→0+xα if α > 0 hence (by Squeeze theorem) (1 point)
lim
x→0+f (x) = 0 , if α > 0 (1 point) if α ≤ 0 and β < 0 , then
lim
x→0+f (x) = lim
x→0+xαsin ( 1
xβ) = lim
x→0+
sin x−β
x−β xα−β=0 if α − β > 0 (3)
lim
x→0−f (x) = lim
x→0−
sin xβ 1 − cos x
= lim
x→0−
βxβ−1cos xβ
sin x (by l’Hospital’s rule) (1 point)
= lim
x→0−
x
sin xβxβ−2cos xβ=0 if β > 2 (1point) note that if β = 0 , then
lim
x→0−f (x) = lim
x→0−
sin x0
1 − cos x= lim
x→0−
sin 1 1 − cos x it doesn’t converge
Therefore , f (x) is continuous at x = 0 if α > 0 and β > 2 2.(b) f (x) is differentiable at x = 0 iff f′(0+) =f′(0−)(1 point) Moreover, f (x) is differentiable at x = 0
Ô⇒ f (x) is continuous at x = 0 Ô⇒ α > 0 and β > 2 (1)
f′(0+) = lim
h→0+
f (h) − f (0) h − 0 = lim
h→0+
hαsin (h1β) −0 h
= lim
h→0+hα−1sin 1
hβ 1 point since
−1 ≤ sin ( 1
hβ) ≤1 ∀β ∈ R and
lim
h→0+−hα−1=0 = lim
h→0+hα−1 if α − 1 > 0 hence (by Squeeze theorem) (1 point)
f′(0+) =0 if α > 1 (1 point) if α − 1 ≤ 0 and β < 0 , then
lim
h→0+
f (h) − f (0) h − 0 = lim
h→0+
sin h−β
h−β hα−β−1=0 if α − β − 1 > 0 (2)
f′(0−) = lim
h→0−
f (h) − f (0) h − 0 = lim
h→0− sin hβ 1−cos h
h = lim
h→0−
sin hβ hβ
hβ−1 1 − cos h
= lim
h→0−
hβ−1
1 − cos h= lim
h→0−
(β − 1)hβ−2
sin h (by l’Hospital’s rule)T
= lim
h→0−
h
sin h(β − 1)hβ−3=0 if β − 3 > 0 (1 point) hence
f′(0−) =0 if β > 3 (1 point) note that if β = 1 , then
lim
h→0−
f (h) − f (0) h − 0 ) = lim
h→0−
1 1 − cos h it doesn’t converge
Therefore , f (x) is differentiable at x = 0 if α > 1 and β > 3
Page 4 of 12
3. (8%) Let f (x) be a twice differentiable one-to-one function. Suppose that f (2) = 1, f′(2) = 3, f′′(2) = e. Find d
dxf−1(1) and d2
dx2f−1(1).
Solution:
let y = f (x) , then x = f−1(y) and when x = 2 , y = 1 (1)
dy dx
dx
dy =1 (2 points) implies that
d
dyf−1(y) = 1 f′(x) hence
d
dyf−1(1) = 1 f′(2)=
1
3 (2 points) (2)
d dx(
dy dx
dx dy) =
d
dx(1) Ô⇒ d2y dx2
dx dy + (
dy dx)2
dx
dy =0 (2 points.) i.e.
f′′(2)d
dyf−1(1) + (f′(2))2 d2
dy2f−1(1) = 0 i.e.
e1
3+32 d2
dy2f−1(1) = 0 i.e.
d2
dy2f−1(1) = −e
27 (2 points) [another way]
let g(x) = f−1(x)
since g(f (x)) = x , we have g′(f (x))f′(x) = 1 (2 points) that is ,
g′(1) = g′(f (2)) = 1 f′(2) =
1
3 (2 points) moreover
d
dx[g′(f (x))f′(x)] = d dx(1)
Ô⇒g′′(f (x))[f′(x)]2+g′(f (x))f′′(x) = 0 (2 points) Ô⇒g′′(1) =−e
3 1 32 =
−e
27 (2 points) [another way]
f (f−1(x)) = x (2 points) Ô⇒f′(f−1(x)) d
dxf−1(x) = 1 Ô⇒
d
dxf−1(x) = 1
f′(f−1(x)) (2 points) Ô⇒
d2
dx2f−1(x) = d dx
1 f′(f−1(x))=
−dxdf−1(x)
f′(f−1(x))2 (2 points) Ô⇒
d
dxf−1(1) =1
3 (2 points) and d2
dx2f−1(1) =−e
27 (2 points)
4. (8%) Find the value of the number c such that the families of curves y = (x + α)−1and y = c(x + β)1/3 are orthogonal trajectories, that is, every curve in one family is orthogonal to every curve in the other family.
Solution:
y = (x + α)−1 then y′=
−1
(x + α)2 (1 pt) y = c(x + β)13 then y′=
c
3(x + β)−23 (1 pt) Let point of intersect be (x0, y0)
Orthogonal ⇒ −1 (x0+α)2 ⋅
c
3(x0+β)−23 = −1
⇒c = 3(x0+α)2(x0+β)23 (2 pts) We also have y0=
1
x0+α=c(x + β)13
⇒ 1
c(x0+α) = (x + β)13 (2 pts)
combine with the equation above we have c = 3
c2 ⇒c3=3 ⇒ c =√3
3 (2 pts)
Page 6 of 12
5. (8%) Find the nth derivative of the function f (x) = xn 1 − x.
Solution:
Here are two ways to compute f(n)(x).
First one need to write f (x) = xn 1 − x=
xn−1 1 − x +
1 1 − x
= −(xn−1+xn−2+ ⋯ +1) + 1
1 − x (3 pts)
Note first term become zero after n times of differentiation. (1 pt) (
1
1 − x)′= (−1) ⋅ 1
(1 − x)2 ⋅ (−1) = 1 (1 − x)2 (
1 1 − x)(k)=
k!
(1 − x)k+1 So we have f(n)(x) = n!
(1 − x)n+1. (4 pts) Second way is to apply Leibniz’s rule.
f (x) = xn 1 − x=xn⋅
1 1 − x
Then f(n)(x) = Σni=0Cin (xn)(n−i)⋅ ( 1
1 − x)(n)(4 pts)
=Σni=0 Cin n!
i!xi⋅ (i!) 1 (1 − x)(i+1)
=Σni=0 Cin n! xi
(1 − x)i+1 (4 pts)
6. (8%) Suppose that three points on the parabola y = x2 have the property that their normal lines intersect at a common point. Show that the sum of their x-coordinates is 0.
Solution:
Let (x1, x21), (x2, x22), (x3, x23)be such three points.
If x1x2x3=0, say x3=0. Then the common point is on the y-axis.
The normal lines passing (x1, x21), (x2, x22)are y − x21=
−1 2x1
(x − x1), y − x22=
−1 2x2
(x − x2)
⇒0 = x = −2x1x2(x1+x2)
Since x1x2≠0, we have x1+x2=0. Hence x1+x2+x3=0.
Now if x1x2x3≠0, the normal lines passing (x1, x21), (x2, x22), (x3, x23)are y − x21=
−1 2x1
(x − x1), y − x22=
−1 2x2
(x − x2), y − x23=
−1 2x2
(x − x3)
⇒x = −2x1x2(x1+x2) = −2x2x3(x2+x3) = −2x1x3(x1+x3)
⇒x1(x1+x2) =x3(x2+x3)
⇒x1(x1+x2+x3−x3) =x3(x2+x3+x1−x1)
⇒ (x1−x3)(x1+x2+x3) =0. Hence x1+x2+x3=0.
評分標準
寫出法線方程式並把點帶入得三分 解出交點的 x 座標得兩分 得到 x 座標相加是0得三分
Page 8 of 12
7. (12%) A cone-shaped paper drinking cup is to be made to hold 9 cm3 of water. Find the height and radius of the cup that will use the smallest amount of paper.
Solution:
We have 1
3πr2h = 9, θ = 2πr
√ r2+h2
⇒A =1 2(
√
r2+h2)2 2πr
√
r2+h2 =πr
√ r2+h2 So A(r) = πr
√ r2+ (
27 πr2)2=π
√ r4+
729 π2r2
⇒A′(r) = π( 4r3−π14582r3
2
√
r4+π7292r2
) Let A′(r) = 0 ⇒ 4r3−1458
π2r3 =0 ⇒ r = 3
√6
2π2 Then h = 27
π(√63
2π2)2
=33
√ 2 π. These are answer since for r < 3
√6
2π2, A′(r) < 0 and for r > 3
√6
2π2, A′(r) > 0.
評分標準
列出體積關係式得兩分
算出扇形角度以及半徑各得一分
列出所要求的面積式子得一分 換成同一個變數再一分 對面積式子微分找出 critical number 得兩分 求出 r 和 h 各一分
說明為何是極小值得兩分
8. (12%)
(a) Suppose that f (x) and g(x) are differentiable on open interval containing [a, b] and f (a) > g(a), f (b) > g(b).
Show that if the equation f (x) = g(x) has exactly one solution on [a, b] then at the solution x0∈ [a, b], f (x) and g(x) have the same tangent line.
(Hint: Consider h(x) = f (x) − g(x). Show that h(x) ≥ 0 for all x ∈ [a, b].)
(b) For α > 0, if the equation ex=kxαhas exact one solution on [0, ∞), solve k in terms of α.
Solution:
(a) h(x) = f (x) − g(x) is diff on [a, b]. h(a) > 0, h(b) > 0
If h(¯x) < 0 for some ¯x ∈ (a, b) ((+2): Correct assumption to start with.),
then by the intermediate value thm, there are some x1 ∈ [a, ¯x] and x2 ∈ [¯x, b] s.t. h(x1) =0 = h(x2) i.e.
f (x) = g(x) has at least two solution x1, x2∈ [a, b]. ((+2): Use IVT.)
∴h(x) ≥ 0 ∀ x ∈ [a, b] if f (x) = g(x) has exactly one solution on [a, b] →←
Suppose that r0 is the only root for f (x) = g(x), r0∈ [a, b]. Then h(r0)is a local minimum value. ((+2):
See local minimum),
∵h(x) is diff on [a, b] ∴ h′(r0) =0 ⇒ f′(r0) =g′(r0). ((+2): Use Rolle’s Theorem to conclude.) (b) f (x) = ex, g(x) = kxα.
for x = 0, f (0) = 1 > g(0) = 0 for x large enough f (x) > g(x).
Hence if ex=kxα has exactly one solution on [0, ∞) then at the root x = x0, f (x) and g(x) have the same tangent line.
i.e. if f (x0) =g(x0)then f′(x0) =g′(x0). ((+2): Apply part (a)),
⎧⎪
⎪
⎨
⎪⎪
⎩
ex0=kxα0 −(1)
ex0=kαxα−10 −(2) ⇒ kxα0 =kαxα−10 ⇒ x0=α (1) ⇒ eα=kαα, k = (e
α)α. ((+2): Find correct answer).
Page 10 of 12
9. (20%) Let f (x) = (x3+x2)1/3. (a) Find all asymptotes of f (x).
(b) Find the intervals of increase or decrease.
(c) Find the intervals of concavity.
(d) Find the local maximum and minimum values.
(e) Find the inflection points.
(f) Sketch the graph of y = f (x).
Solution:
(a) Since f (x) is finite for any finite x ∈ R and f (x) → ±∞ as x → ±∞, it does not have any vertical or horizontal asymptotes. However, since
x→±∞lim f (x)
x = lim
x→±∞(1 + 1 x)
1
3 =1 (2%)
and
x→±∞lim (f (x) − 1 ⋅ x) = lim
x→±∞
(x3+x2) −x3 (x3+x2)
2
3 +x(x3+x2)
1 3+x2
= lim
x→±∞
1 (1 +x1)
2
3 + (1 +1x)
1 3+1
= 1
3 (2%)
f has a slant asymptote y = x +1 3. (b)
f (x) = (x3+x2)
1 3
f′(x) = 1
3(x3+x2)−
2
3(3x2+2x) = 3x + 2
3x13(x + 1)23 (2%)
f′(x) > 0 for x ∈ (−∞, −2
3)or (0, ∞), and f′(x) < 0 for x ∈ (−2 3, 0).
⇒f (x) is increasing on (−∞, −2
3)and (0, ∞), decreasing on (−2
3, 0). (3%) (c)
f′′(x) =1 3[−
2
3(x3+x2)−
5
3(3x2+2x)2+ (x3+x2)−
2
3(6x + 2)] = − 2
9x43(x + 1)53 (2%)
f′′(x) > 0 for x ∈ (−∞, −1), and f′′(x) < 0 for x ∈ (−1, 0) or (0, ∞).
⇒f (x) is concave upward on (−∞, −1) , concave downward on (−1, 0) and (0, ∞). (3%) (d) f′(x) goes from positive to negative across x = −2
3 and from negative to positive across x = 0, and f (x) is defined at these points.
⇒f (−2 3) =
√3
4
3 is the local maximum (1%) and f (0) = 0 is the local minimum (1%).
(e) f′′(x) changes sign only across x = −1 and f is continuous at that point.
⇒f (−1) = 0 is the only inflection point (1%).
(f)
Page 12 of 12
