1031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和和和評評評分分分標標標準準準 1. (10%) Find the limit lim
x→0
a + x a − x
1/x
, a > 0.
Solution:
(Method 1) First we observe that
x→0lim
a + x a − x
1x
= lim
x→0eln(a+xa−x)1x = lim
x→0e1xln(a+xa−x) = elimx→01xln(a+xa−x) (2 points)
Note that 1
xln a + x a − x
is of the form 0
0, hence by L’Hopital’s rule (1 point), we have
(1) lim
x→0
1
xln a + x a − x
= lim
x→0
ln (a + x) − ln (a − x) x
= lim
x→0 1
a+x(a + x)0−a−x1 (a − x)0
1 (2 points)
= lim
x→0 1
a+x −a−x−1 1
= lim
x→0
1
a + x+ 1
a − x (1 points)
= 2
a (1 points)
(2) lim
x→0
1
xln a + x a − x
= lim
x→0
ln
a+x a−x
x
= lim
x→0
a+x a−x
−1
·dxd
a+x a−x
1 (2 points)
= lim
x→0
a − x
a + x· (a − x) · 1 − (−a) (a + x) (a − x)2
= lim
x→0
a − x + a + x (a + x) (a − x)
= lim
x→0
2a
a2− x2 (1 points)
= 2
a (1 points)
(3) lim
x→0
1
xln a + x a − x
= lim
x→0
ln (a + x) − ln (a − x) x
= lim
x→0
ln (a + x) − ln a + ln a − ln (a − x)
x (2 points)
= lim
x→0
ln (a + x) − ln a
x + lim
x→0
ln (a − x) − ln a
−x
= lim
x→0
ln (a + x) − ln a
x + lim
y→0
ln (a + y) − ln a
y (1 points)
= 2 d dtln t
t=a
= 2
a (1 points) Hence
x→0lim
a + x a − x
1x
= elimx→0x1ln(a+xa−x) = e2a. (3 points)
(Method 2)
First we observe that a + x
a − x = 1 + 2x
a − x. (2 points)
x→0lim
a + x a − x
1x
= lim
x→0
1 + 2x a − x
1x
By letting t = 2x
a − x, t → 0 as x → 0. (3 point) Hence
x→0lim
1 + 2x a − x
1x
= lim
x→0
1 + 2x a − x
a−x2x a−x2
(2 points)
= lim
t→0(1 + t)1t·a−x2 = lim
t→0
h(1 + t)1tia−x2
Note that we have lim
x→0f (x)g(x) = lim
x→0f (x)limx→0g(x) provided the limits exists.
lim
x→0
h(1 + t)1tia−x2
=h lim
t→0(1 + t)1tilimx→0 2
a−x = e2a (3 points)
(Method 3) First we let t = 1
x. (2 points)
x→0lim
a + x a − x
x1
= lim
t→∞
a +1t a −1t
t
= lim
t→∞
at + 1 at − 1
t
= lim
t→∞
1 + 2
at − 1
t
(3 points)
≈ lim
t→∞
1 + 2
at
t
(2 points)
= lim
t→∞
1 +
2 a
t
t
= ea2 (3 points)
(We should prevent using this unrigorous calculation.) (Method 4)
By letting t = 1
x. (2 points)
x→0lim
a + x a − x
1x
= lim
x→0
1 +xa 1 −xa
1x
= lim
x→0
1 +xax1
1 −xax1 (3 points)
= lim
1 + 1/at t
2. (10%) Find the limit lim
x→0
ln(cos(ax))
ln(cos(bx)), where a, b are constant.
Solution:
We found this problem is an indeterminate form.
Consequently, we think about the L’Hospital’s Rule first.
In this rule, there are three assumptions should be satisfied:
(1)The denominator and the numerator should be differentiable on an open interval that contains 0.
(2) The differentiation of the denominator can NOT equal to zero on an open interval that contains 0.
(ln cos(ax))0 = 1
cos(ax)· (− sin(ax)) · a (ln cos(bx))0= 1
cos(bx)· (− sin(bx)) · b (− sin(ax))0= − cos(ax) · a
(− sin(bx))0 = − cos(bx) · b
Therefore, the origimal can be rewritten as follows by using L’Hospital’s Rule twice:
x→0lim cos bx
cos ax·− cos ax · a
− cos bx · b ·a b =a2
b2 .
If students use L’Hospital’s Rule first time correctly, and they can get 4 points;
If students use L’Hospital’s Rule second time correctly, and they can get another 4 points;
If students calculate the limitation under x approaches to zero correctly, and they can get the other 2 points.
Solution 2
By using L’Hospital’s Rule, the original equation can be written as follows:
lim
x→0
− sin(ax) cos(ax) · a
− sin(bx) cos(bx) · b
= lim
x→0
a tan(ax) b tan(bx)
= lim
x→0
a sec2(ax) · a b sec2(bx) · b
=a2 b2.
Other proper methods are permitted to solve this problem.
3. (12%) Choose the point P on the line segment AB so as (a) to maximize the angle θ; (b) to minimize the angle θ.
x P
A B
5
2
3
Solution:
θ = π − tan−1(5/x) − tan−1( 2
3 − x) (4pts) θ0=
5 x2
1 +x252 +
2 (3−x)2
1 +(3−x)4 2
= 5
x2+ 25− 2 x2− 6x + 13 θ0= 0 =⇒ x2− 10x + 5 = 0 x = 5 − 2√
5(since 5 + 2√
5 > 3)(4pts) θ > 0 when x ∈ (0, 5 − 2√
5) θ < 0 when x ∈ (5 − 2√
5, 3) Thus x = 5 − 2√
5 is a local maximum also a maximum since θ is differentiable on (0,3) (2pts)
Check endpoints for minimum.
x = 0 then θ = π/2 − tan−1(2 3) x = 3 then θ = π/2 − tan−1(5
3)
Since arctangent is increasing ,θ has a minimum when x = 3 (2pts) Answer:(a)x = 5 − 2√
5(b)x = 3
4. (12%) A container in the shape of an inverted cone has height 16 cm and radius 5 cm at the top. It is partially filled with a liquid that oozes(滲出) through the sides at a rate proportional to the area of the container that is in contact with the liquid. If we pour the liquid into the container at a rate of 2 cm3/min, then the height of the liquid decreases at a rate of 0.3 cm/min when the height is 10 cm. If our goal is to keep the liquid at a constant height of 10 cm, at what rate should we pour the liquid into the container?
• You may need this: The surface area of a cone is πrl, where r is the radius and l is the slant height(斜高).
Page 15
Chapter 03-Problem Plus
Solution:
Let
Vpour(t) be the volume of the liquid pours into the container at time t Vooze(t) be the volume of the liquid oozes through the container at time t h(t) be the height of the liquid in the container at time t,as h in the figure r(t) be the radius of the liquid in the container at time t,as r in the figure V(t) be the volume of liquid in the container at time t
by similarity of triangles : r(t) 5 = h(t)
16 ⇒ r(t) = 5 16h(t) we also have :
V0 (initial volume of the liquid) + Vpour(t) − Vooze(t) = V (t)
=1
3πr2(t)h(t) = 25
768πh3(t)...(4%)
⇒ Vpour0 (t) − Vooze0 (t) = V0(t) = 25
256πh2(t)h0(t)...(4%) notice that : h0(t0) = −0.3 at h(t0) = 10, Vpour0 (t0) = 2
⇒ 2 − Vooze0 (t0) = 25
256π(102)(−0.3) = −375π 128 ...(2%)
⇒ Vooze0 (t0) = 2 +375π 128
so if Vpour0 (t0) = Vooze0 (t0) = 2 +375π
128 (cm3/min) ...(2%)
then V0(t0) = Vpour0 (t0) − Vooze0 (t0) = 0,hence h0(t0) = 0 which is what we want
(If you finish the second part(state the differential equation),then the first 4 points will be included.There is 0 points for the third part if the first two parts are not both correct.If the relation of functions are wrong,there is 0 points for this problem!)
5. (12%) If xy + ey= e, (a) (3%) find dy
dx.
(b) (9%) find the values of y, y0 and y00 at the point where x = 0.
Solution:
(a) y + xdy
dx+ eydy
dx = 0 ⇒ dy
dx(x + ey) = −y ⇒ dy
dx = −y x + ey
(b) x = 0 ⇒ ey = 1 ⇒ y = 1 , and y0+ y0+ xy00+ e(y0)2+ eyy00= 0 , then y0= −1 0 + e =−1
e when x = 0 , then take y = 0, y0 =−1
e , we get −2 e + e(1
e)2+ ey00= 0 ⇒ ey00= 1
e ⇒ y00= 1 e2
6. (10%) Define f (x) =
x2sin 1
x if x 6= 0,
0 if x = 0.
(a) (3%) Show that f (x) is continuous at x = 0.
(b) (4%) Calculate f0(x) when x 6= 0.
(c) (3%) Find f0(0) if it exists.
Solution:
(a) We need to check that lim
x→0f (x) = f (0) = 0. (2pts) lim
x→0f (x) = lim
x→0x2sin(1 x) = 0.
(since ∀x 6= 0, −x2≤ x2sin(1
x) ≤ x2 & lim
x→0x2= lim
x→0−x2= 0.
Then by squeezing theorem, we have lim
x→0x2sin(1
x) = 0.) (1pt) (b) When x 6= 0, f0(x) = 2x sin1
x+ x2(cos1 x)(−1
x2) = 2x sin1
x− cos1 x (4pts) (c) f0(0) = lim
x→0
f (x) − f (0) x − 0 = lim
x→0
x2sin1x− 0 x − 0 = lim
x→0x sin1
x = 0. (2pts) (since ∀x 6= 0, −|x| ≤ x sin(1
x) ≤ |x| & lim
x→0x = lim
x→0−x = 0.
Then by squeezing theorem, we have lim
x→0x sin1
x = 0.) (1pt)
7. (24%) Let f (x) = x4
(1 + x)3. Answer the following questions by filling each blank below. Show your work (computa- tions and reasoning) in the space following. Put None in the blank if the item asked does not exist.
(a) The function is increasing on the interval(s) and decreasing on the
interval(s) (6% total). The local maximal point(s) (x, y) = (2%),
The local minimal point(s) (x, y) = (2%).
(b) The function is concave upward on the interval(s) and concave downward
on the interval(s) (6% total).
The inflection point(s) (x, y) = (2%).
(c) The vertical asymptote line(s) of the function is(are) . The horizontal asymptote line(s) is(are) (2%)
(d) Sketch the graph of the function. Indicate, if any, where it is increasing/decreasing, where it concaves up- ward/downward, all relative maxima/minima, inflection points and asymptotic line(s) (if any).(4%)
Solution:
(a) f0(x) = 4x3(1 + x)3− 3(1 + x)2x4
(1 + x)6 =4x3(1 + x) − 3x4
(1 + x)4 = x3(x + 4)
(1 + x)4 (4 points) f (x) = 0 ⇒ x3(x + 4) = 0 ⇒ x = 0 or 4
f0(x) > 0 on (−∞, −4) ∪ (0, −∞)
⇒ f is increasing on (−∞, −4) ∪ (0, −∞). (1 point) f0(x) < 0 on (−4, −1) ∪ (−1, 0)
⇒ f is decreasing on (−4, −1) ∪ (−1, 0). (1 point)
⇒ f has a local minimum at (0, 0) (2 points) and a local maximun at (−4, −256
27 ). (2 points)
(b) f00(x) =(3x2(x + 4) + x3)(1 + x)4− 4(1 + x)3(x3(x + 4))
(1 + x)8 =(1 + x)(4x3+ 12x2) − (4x4+ 16x3) (1 + x)5
= 12x2
(1 + x)5 (4 points) f00(x) > 0 ⇒ 12x2
(1 + x)5 > 0, x 6= 0⇒ x > −1 , x 6= 0.
f00(x) < 0 ⇒ 12x2
(1 + x)5 < 0⇒ x < −1 . f is concave upward on (−1, ∞), (1 point) and concave downward on (−∞, −1). (1 point)
(0, 0) is not an inflection point, since f00(x) > 0 on (−1, 0) and (0, ∞).
8. (10%) Consider any point (x0, y0), where x0> 0 and y0> 0, on the hyperbola xy = k, where k > 0 is a constant.
(a) (6%) Find the equation of the tangent line at (x0, y0).
(b) (3%) Let A and B denote respectively the x-intercept and the y-intercept of the tangent line at (x0, y0). Find the area of the triangle enclosed by the origin O and A, B.
(c) (1%) Is the area of 4OAB a constant? That is, is the area of 4OAB independent of x0and y0?
Solution:
Method 1 (a) To find the tangent line of xy = k at (x0, y0) in the first quadrant, we differentiate xy = k implicitly with respect to x to get
y + xdy
dx = 0 ⇒ dy dx = −y
x (3%) ⇒ dy dx (x
0,y0)
= −y0 x0
. (1%)
So the tangent line of xy = k at (x0, y0) is y − y0= −y0
x0(x − x0). (2%) (b) We get x-intercept and y-intercept of the tangent line as follows:
• Let y = 0, then x = 2x0, so the coordinates of A is (2x0, 0). (1%)
• Let x = 0, then y = 2y0, so the coordinates of B is (0, 2y0). (1%) Hence the area of 4OAB is 1
2· 2x0· 2y0= 2x0y0. (1%)
(c) Yes, the area of 4OAB is 2x0y0= 2k, which is a constant. (1%) The hyperbola xy = k can be written as y = k
x. Method 2 (a) Since y = k
x, we have dy dx= − k
x2 (3%) ⇒ dy dx = −k
x2 x=x
0
= − k x20. (1%)
So the tangent line of xy = k at (x0, y0) is y − y0= −k
x20(x − x0). (2%) (b) We get x-intercept and y-intercept of the tangent line as follows:
• Let y = 0, then x = x0+x20y0
k , so the coordinates of A is (x0+x20y0
k , 0). (1%)
• Let x = 0, then y = y0+ k
x0, so the coordinates of B is (0, y0+ k
x0). (1%) Hence the area of 4OAB is 1
2
x0+x20y0 k
y0+ k
x0
. (1%)
(c) Yes, the area of 4OAB is 1 2
x0y0+ k +x20y02 k + x0y0
= 2k, which is a constant. (1%) 微分計算錯誤，代點得到錯誤的值，但是有寫出直線方程式(斜率當然有誤)，得 2 分。
因為 (a) 計算錯誤，但有確實找直線和兩軸的交點以及面積，得 2 分。
(c)寫了 Yes，但是在 (b) 或 (c) 的過程中並未寫出算得之面積為 2k(常數)，不給分。
