1011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. (12 points) Evaluate the following limits.
(a) lim
x→2−
x −√ 2x
|x2− 4|. (b) lim
x→0
1 − sin2(ax) − cos(ax)
1 + sin2(bx) − cos(bx), where a and b are non-zero real numbers.
Solution:
(1)
lim
x→2−
x −√ 2x
|x2− 4| = lim
x→2−
x −√ 2x
4 − x2 (1.pt)
= lim
x→2−
x2− 2x (4 − x2)(x +√
2x) (2.pts)
= lim
x→2−
−x (2 + x)(x +√
2x) (1pt.)
= −1
8 (1.pt)
(2)
x→0lim
1 − sin2(ax) − cos (ax) 1 + sin2(bx) − cos (bx) = lim
x→0
2 sin2 a2x − sin2(ax)
2 sin2 b2x + sin2(bx) (2.pts)
= lim
x→0
2 ·a24x2 · (sinaa2x
2x )2− (a2x2) · (sin (ax)ax )2 2 ·b24x2 · (sinbb2x
2x )2+ (b2x2) · (sin (bx)bx )2
(2.pts)
= lim
x→0
a2 b2
·12· (sinaa2x
2x )2− (sin (ax)ax )2
1
2· (sinbb2x
2x )2+ (sin (bx)bx )2
(1.pt)
= a2 b2
1 2− 1
1
2+ 1 = −a2
3b2 (Since lim
x→0
sin x
x = 1) (2.pts)
One who uses L’H¨opital rule without proof or giving a wrong proof will only get at most (3.pts) for each problem. The following is the answer of using L’H¨opital rule to calculate the problems without proof and the credit distribution:
(1)
lim
x→2−
x −√ 2x
|x2− 4| = lim
x→2−
x −√ 2x 4 − x2
= lim
x→2−
1 −√1
2x
−2x = − lim
x→2−
√2x − 1
(2x)32 (2.pts)
= −1
8 (1.pt)
(2)
lim
x→0
1 − sin2(ax) − cos ax 1 + sin2bx − cos bx = lim
x→0
−2a sin (ax) cos (ax) + a sin (ax)
2b sin (ax) cos (bx) + b sin (bx) (1.pt)
= lim
x→0
−2a2 cos2(ax) − sin2(ax) + a2cos (ax)
2b2 cos2(ax) − sin2(bx) + b2cos (bx) (1.pt)
= −2a2+ a2 2b2+ b2 = −a2
3b2 (1.pt)
2. (12 points) Suppose that f (x) is differentiable on (−1, 1) with lim f (x)
x2 = L, where L is a constant.
Define g(x) = (
f (x) sin(1
x), for 0 < |x| < 1;
A, for x = 0.
(a) Find f (0) and f0(0).
(b) If g is continuous at x = 0, find the value of A and compute g0(0).
(c) Write down a formula of g0(x) in terms of f (x) and f0(x) for 0 < |x| < 1.
(d) Suppose that f0(x) and g0(x) are both continuous at 0. Find the value of L.
Solution:
(a) Since f is differentiable, f is continuous. Hence f (0) = lim
x→0f (x) (1)
= lim
x→0
f (x) x2 · x2
= lim
x→0
f (x) x2 · lim
x→0x2 since both limits exist
= L · 0 = 0. (1%) (2)
By definition of derivative,
f0(0) = lim
h→0
f (h) − f (0) h − 0 (1%)
= lim
h→0
f (h)
h by (2) (3)
= lim
h→0
f (h) h2 · h
= lim
h→0
f (h) h2 · lim
h→0h since both limits exist
= L · 0 = 0. (1%) (4)
(b) Since g is continuous at x = 0,
A = g(0) = lim
x→0g(x)
= lim
x→0f (x) sin(1
x) since x 6= 0.
Since −1 ≤ sin(1/x) ≤ 1, we have
−|f (x)| ≤ f (x) sin(1
x) ≤ |f (x)|.
Then by Pinching Theorem, we have by (1), 0 = lim
x→0−|f (x)| ≤ lim
x→0f (x) sin(1
x) ≤ lim
x→0|f (x)| = 0, (1%) which means
A = g(0) = 0. (5)
By definition of derivative,
g0(0) = lim
h→0
g(h) − g(0) h − 0 (1%)
= lim
h→0
g(h)
h by (5)
= lim
h→0
f (h) h sin1
x
= 0 by (3) and Pinching Theorem. (1%) (6)
(c) For x 6= 0, by product rule and chain rule, g0(x) = d
dxf (x) sin(1/x) = f0(x) sin(1/x) + f (x) cos(1/x)(−x−2). (3%) (7)
(d) Since g0(x) is continuous at x = 0, by (7), g0(0) = lim
x→0g0(x)
= lim
x→0{f0(x) sin(1/x) + f (x) cos(1/x)(−x−2)} (1%). (8) Since f0(x) is continuous at x = 0, by (4),
x→0limf0(x) = f0(0) = 0.
Hence by Pinching Theorem,
x→0limf0(x) sin(1/x) = 0. (1%) Therefore, by (6), (8) becomes
0 = g0(0) = lim
x→0−f (x)
x2 cos(1/x). (9)
If L 6= 0,
x→0lim−f (x)
x2 cos(1/x) = −L lim
x→0cos(1/x), which doesn’t exist and contradict to (9). (1%)
If L(= lim
x→0
f (x)
x2 ) = 0, by Pinching Theorem,
x→0lim f (x)
x2 cos(1/x) = 0, which coincide with (9). Hence
L = 0. (10)
3. (12 points) Let a < b. A function f in said to be a contraction on [a, b] if there exists K, 0 < K < 1, such that for all x1, x2∈ [a, b] we have |f (x1) − f (x2)| ≤ K |x1− x2|.
(a) Show by the − δ definition that if f is a contraction on [a, b], then f is continuous on [a, b].
(b) Suppose that f is continuous on [a, b] and differentiable on (a, b) with |f0(x)| ≤ q, 0 < q < 1, for all x ∈ (a, b).
Show that f is a contraction on [a, b] and has at most one fixed point on [a, b]. (A point c ∈ [a, b] is called a fixed point of f if f (c) = c.)
Solution:
(a)Take c ∈ [a, b]. Let > 0 be given, take δ =
K. Then for all x ∈ [a, b] and |x − c| < δ, we have
|f (x) − f (c)| ≤ K|x − c| < K ·
K = . So f is continuous at c. As c is arbitrary, f is continuous on [a, b].(4points) (b)(1)By Mean-value theorem, for any x1 < x2 and x1, x2 ∈ [a, b], we have f0(x3) = f (x2) − f (x1)
x2− x1
, where x3∈ [x1, x2]. Hence we have |f (x2)−f (x1)| = |f0(x3)||x2−x1| ≤ q|x2−x1|. So we have |f (x2)−f (x1)| ≤ q|x2−x1|, where 0 < q < 1. Then we can conclude f is a contraction.(Note that if x1= x2, the result follows directly.(4 points)
(2)If not, ∃x4 < x5 and f (x4) = x4, f (x5) = x5. By mean-value theorem, we have f0(x6) = f (x5) − f (x4) x5− x4
= x5− x4
x5− x4
= 1,. f0(x6) = 1 contradicates the assumption, f0(x) < 1, for all x ∈ [a, b].Hence f has at most one fixed point on [a, b].(4points)
4. (12 points) Suppose ABC is a triangle with vertices A = (−5, 0), B = (0, 10) and C = (5, 0). Let P be a point on the line segment that join B to the origin. Find the position of P that minimizes the sum of distances between P and the three vertices of the triangle ABC.
Solution:
Let P = (0, y) and set
S(y) = (10 − y) + 2p
25 + y2 (2.pts) One who wrote the wrong function will not get any point !
Then, we solve S0(y) = 0, we have
S0(y) = −1 + 2y
py2+ 25 = 0 ⇒ y = 5
√3
[(2.pts) for Calculate S0(y), (2.pts) for solving y = 5
√3 and (1.pt) for choosing the right sign of y]
Next, we exam whether S(y) takes minimum at y = 5
√
3, by Second derivative test, we have S00(y)
y=√5
3
= 50 − y2 (y2+ 25)32
y=√5
3
> 0
[((3.pts) for calculate S00(y)), (1.pt) for S00( 5
√3) > 0)]
One who didn’t check the minimality will not get this (4.pts), other methods to check are 1. First derivative test, 2. using the Extreme value theorem to check critical point and end points.
Hence S(y) has a minimum at y =5 3, and
P = (0, 5
√3) (1.pt)
5. (12 points) Suppose that f (x) is continuous and increasing on [−1, 2] with f (x) > 0. Let F (x) = Z x
0
t
Z t 1
f (u)du
dt.
(a) Classify all critical points of F (x) in (−1, 2).
(b) Show that F00(x) is increasing on (0, 1) and there is a point of inflection of F (x) on (1 2, 1).
Solution:
(a) F (x) =
Z x 0
t
Z t 1
f (u)du
dt
⇒ F0(x) = x Z x
1
f (u)du (1 point)
⇒ F00(x) = Z x
1
f (u)du + x · f (x) F0(x) = 0 when x = 0 or
Z x 1
f (u)du = 0 i.e. x = 0 (1 point) or x = 1 (1 point)
Observe that
F0(x) > 0 when x ∈ (−1, 0) ∪ (1, 2), (1 point) F0(x) < 0 when x ∈ (0, 1). (1 point)
( Or, F00(0) = Z 0
1
f (u)du < 0 and F00(1) = f (1) > 0 ) Thus x = 0 is a local maximum, (1 point)
x = 1 is a local minimum. (1 point)
(b)(1) F00(x) =
Z x 1
f (u)du + x · f (x) (1 point)
For x > 0,
∵ d dx
Z x 1
f (u)du
= f (x) > 0
∴ Z x
1
f (u)du is increasing.
Since x, f (x), Z x
1
f (u)du are all increasing, F00(x) is increasing on (0, 1). (1 point)
(2)
∵ F00(1) = f (1) > 0, (1 point)
F00(1 2) =
Z 12
1
f (x)dx +1 2 · f (1
2)
= − Z 1
1 2
f (x)dx + Z 1
1 2
f (1 2)dx
= Z 1
1 2
f (1
2) − f (x)
dx
< 0,
∵ f (x) is increasing on [1
2, 1] ⇒ f (1
2) < f (x), ∀x ∈ (1 2, 1]
(1 point)
And also, F00(x) is continuous on [1 2, 1]
∴ By Intermediate Value Theorem, F00(c) = 0 for some c ∈ (1
2, 1). (1 point) Since F00(x) is increasing on (0, 1),
⇒ F00(x) < 0 for x ∈ (1
2, c) and F00(x) > 0 for x ∈ (c, 1)
⇒ x = c is a point of inflection.
6. (12 points) From the equation p 1 + y −
Z x2−1 0
dt
1 + t2 + tan(xy) = 1 a differentiable function y = y(x) can be determined around (x, y) = (1, 0).
(a) Evaluate y0 at (x, y) = (1, 0).
(b) Evaluate y00 at (x, y) = (1, 0) and determine the concavity of y = y(x) around this point.
Solution:
(i) Apply d
dx on both sides of the equation.
(Calculation of derivatives 5%)
1 2√
1 + yy0− 1
1 + (x2− 1)2(2x) + (y + xy0) sec2(xy) = 0 1st term: 1%
2nd term: FTC 1%, chain rule 1%
3rd term: d tan x
dx 1%, chain rule 1%
Put in (x, y) = (1, 0)
(Evaluation 1%. Your answer should be matched.)
1
2y0|(1,0)− 2 + y0|(1,0)= 0 y0|(1,0)=4
3
(ii) Apply d
dx on both sides of the above equation again.
y00 2√
1 + y − y0
4p(1 + y)3y0−2(1 + (x2− 1)2) − 2x · 2(x2− 1)2x (1 + (x2− 1)2)2
+ (y0+ y0+ xy00) sec2(xy) + (y + xy0)2 sec(xy)[sec(xy) tan(xy)](y + xy0) = 0
(If the calculation of derivatives in (i) is correct, there is total 4% in this part; otherwise, there is total 3% in this part. You get -1% for one mistake with respect to your result in (i) until you get 0% in this part.)
Put in (x, y) = (1, 0)
(Evaluation 1%. Your answer should be matched.)
y00|(1,0)
2 −1
4(4
3)2− 2 + (2 · 4
3+ y00|(1,0)) + 0 = 0 y00|(1,0)= −4
27
(Determine your concavity with respect to your evaluation. 1%) Since y00< 0 at (1, 0), the graph is concave down around (1, 0).
7. (12 points) Express the following limit as a definite integral of certain function and then evaluate the integral:
n→∞lim
√1 n
n
X
k=1
1 pn +√
nk .
Solution:
n→∞lim
√1 n
n
X
k=1
1 pn +√
nk
= lim
n→∞
1 n
n
X
k=1
1 r
1 + qk
n
(3 points)
= Z 1
0
1 p1 +√
xdx (3 points) (11)
Using the following change of variable
u = 1 +√ x x = (u − 1)2
dx = 2(u − 1)du (1 point) the integral (11) is transformed into
Z 2 1
2(u − 1)
√u du = Z 2
1
(2u12 − 2u−12)du (2 points)
= 4
3u32 − 4u12
2 1
= (4 32√
2 − 4√ 2) − (4
3− 4) = 8 3 −4
3
√
2 (3 points)
8. (16 points) Let f (x) = 3 x23 x − 1. (a) Find all critical points.
(b) Find the intervals of increasing and intervals of decreasing.
(c) Find the intervals on which f is concave up and intervals on which f is concave down.
(d) Find the points of inflection.
(e) Determine whether f has any vertical tangent or vertical cusps.
(f) Find all vertical or horizontal asymptotes.
(g) Draw the graph of f (x).
Solution:
(a) (3 pt.)
f0(x) = −(x + 2)
x13(x − 1)2 Critical Points x = −2 and x = 0 f (x) does not exist at x = 1
(b) (2 pt.)
increasing [−2, 0]
decreasing (−∞, −2] [0, 1), (1, ∞) (c) (3 pt.)
f00(x) =(−23x−13 +23x−43 )(x − 1) + 2(x23 + 2x−13 ) (x − 1)3
Concave up [−4 − 3√ 2
2 , 0), (0,−4 + 3√ 2
2 , ] (1, ∞) Concave down (−∞,−4 − 3√
2
2 ], [−4 + 3√ 2 2 , 1) (d) (2 pt.)
x = −4 ± 3√ 2 2 (e) (2 pt.)
cusp at x = 0 no vertical tangent
f (x) is not continuous at x = 1 (f) (2 pt.)
vertical asymptotes: x = 1 horizontal asymptotes: y = 0 (g) (2 pt.)
