MATH1141 Test 2 past paper solutions
c UNSW MATHSOC 2012.
V2.0
These solutions were written by Johann Blanco and typed up by Georgia Tsambos.
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summarised some sections of the solutions (especially when I used L’Hopital’s rule) or
skipped explanation altogether.
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of solution should have. There are sometimes multiple methods of solving the same
question - and I offer one. Remember in the real class test, you will be expected to
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Version History
V1.0: Solutions
V2.0: Minor corrections, usage of split function theorem (4.5.6) for brevity, and
improved wording (VN)
1
2008 V8a
1. f is continuous on [0, 1] and [2, 3]. Note that f (0) = 3, f (1) = −1. Since
−1 < 0 < 3, then by the intermediate value theorem there exists a c ∈ (0, 1) such
that f (c) = 0. Hence, f has a zero on the interval [0, 1]. Similarly, for f on [2, 3],
we note that f (2) = −3 and f (3) = 3. 2. We are given f (x) = −2x2 + x. Then by the definition of the derivative,
f 0 (x) = lim
h→0
f (x + h) − f (x)
h
(−2(x + h)2 + (x + h)) − (−2x2 + x)
= lim
h→0
h
−2(x2 + 2xh + h2 ) + x + h + 2x2 − x
h→0
h
= lim
−2x2 − 4xh − 2h2 + h + 2x2
h→0
h
= lim
h(−4x − 2h + 1)
h→0
h
= lim
= lim −4x − 2h + 1
h→0
= −4x + 1. 3.
A=L×W ⇒
⇒
dA
dL
dW
=W
+L
(by the product rule)
dt
dt
dt
dA
= (10 × −2) + (13 × 4) = 32cm2 /s. dt
4. (Mean value theorem.) If f is continuous on [a, b] and differentiable on (a, b), then
there exists at least one c ∈ (a, b) such that
f 0 (c) =
f (b) − f (a)
.
b−a
5−5
⇒ c(3c − 4) = 0. Therefore
0
since c ∈ (0, 2). Now f (c) = 103
. The point is ( 43 , 103
). 27
27
Now f 0 (x) = 3x2 − 4x, so (1) becomes 3c2 − 4c =
c=
4
3
(1)
2
5.
lim
x→0
x sin x
sin x + x cos x
=L’H lim
(product rule)
x→0
1 − cos x
sin x
x
)
= lim (1 +
x→0
tan x
x
= 1 + lim
x→0 tan x
1
= L’H 1 + lim
x→0 sec x2
=1+1
= 2. 3
2008 V4b
1. (Mean value
theorem) f is continuous on [1, 3] and differentiable on (1, 3). For
√
f (x) = x − 1 on [1, 3] we must find a c ∈ (1, 3) such that
f 0 (c) =
Now f 0 (x) =
f (3) − f (1)
.
3−1
(2)
1
−1
(x − 1) 2 , so (2) becomes
2
1
√
=
2 c−1
⇒
√
2−0
2
1
=2
c−1
⇒c=
3
∈ (1, 3).
2
1
3 1
Now f (c) = √ . A point is ( , √ ). 2 2
2
2.
lim
x→0
3 sin 3x
1 − cos 3x L’H
=
lim
2
x→0
x
2x
9 cos 3x
=L’H lim
x→0
2
9
= .
2
3. f is continuous on [−2, −1] and [1, 2], where f (x) = x3 − 3x2 − 2x + 5. Note that
f (−2) = −11 and f (−1) = 3. Since f (−2) < 0 < f (−1), by the intermediate
value theorem, there exists a c ∈ (−2, −1) such that f (c) = 0. Hence, f has a
zero on [−2, −1]. Similarly for f on [1, 2] where f (1) = 1 and f (2) = −3. Hence,
f has a zero on [1, 2]. 4. Since f (x) = −x3 , we have
f 0 (x) = lim
h→0
f (x + h) − f (x)
h
= lim
−(x + h)3 − (−x3 )
h
= lim
x3 − x3 − 3x2 h − 3xh2 − h3
(binomial theorem)
h
h→0
h→0
4
= lim
h→0
−h(3x2 + 3xh + h2 )
h
= −3x2 . 5. We want the line tangent to x2 +y 3 −x2 y = 1 at (1, 1). By differentiating implicitly
with respect to x, we see that
2x + 3y 2
⇒
dy
dy
− (2xy + x2 ) = 0 (product rule on x2 y)
dx
dx
dy
−2x + 2xy
=
.
dx
3y 2 − x2
−(2 · 1) + (2 · 1 · 1)
dy
=
= 0.
dx
(3 · 12 ) − (12 )
Since the gradient of the line is 0 and passes through a point with a y co-ordinate
of 1, the line is given by y = 1.
At (1, 1),
5
2009 V8b
1. (Mean value theorem) f is continuous on [0, 1] and differentiable on (0, 1), where
f (x) = x3 − x2 + 3 . Thus by the mean value theorem, there exists a c ∈ (0, 1)
such that
f 0 (c) =
f (1) − f (0)
.
1−0
(3)
Now f 0 (x) = 3x2 − 2x, so (3) becomes
3c2 − 2c =
3−3
1
c(3c − 2) = 0
which has solutions c = 0, c = 23 . Now
77
point is ( 23 , 27
). 2
3
∈ (0, 1), and f ( 23 ) =
77
.
27
So a suitable
2.
lim
x→1
3x3 − 5x2 + x + 1 L’H
9x2 − 10x + 1
=
lim
x→1
x2 − 2x + 1
2x − 2
18x − 10
x→1
2
18 − 10
=
2
= L’H lim
= 4. 3.
f (x) =
x3
(x + 1)3 + 2
for x < 1
for x ≥ 1
Clearly, f is continuous away from x = 1. If f is continuous at x = 1, then
lim f (x) = f (1) = lim+ f (x).
x→1−
x→1
But f (1) = (1 − 1)3 + 2 = 2, and
lim f (x) = lim− x3 = 1.
x→1−
x→1
As f (1) 6= lim− f (x), f is not continuous at x = 1. Thus f is continuous for all
x 6= 1. x→1
4. See 2008 V4b Q4
6
2009 V1a
x−2
.
− 3x + 2
f is continuous at x = 2 if lim f (x) = f (2). Now,
1. f (x) =
x2
x→2
lim
x→2 x2
x−2
1
= L’H lim
= 1.
x→2 2x − 3
− 3x + 2
Hence, we require f (2) = 1 for f to be continuous. (
p(x) x ≥ α
2. From Theorem 4.5.6, for f =
, for continuous p and q in an interval
q(x) x < α
containing α, if f is continuous at α and p0 (α) = q 0 (α), then
f is differentiable
and α = 1. p is a
at x = α. In this case p(x) = ax + b and q(x) = tan πx
4
polynomial and hence continuous everywhere and q is continuous in (−2, 2) which
contains x = 1. For f to be continuous at x = 1,
lim f (x) = f (1) = lim+ f (x).
x→1−
x→1
So
f (1) = a · 1 + b = a + b,
lim f (x) = lim− ax + b = a + b and
x→1−
x→1
lim+ f (x) = lim− tan
x→1
x→1
π
πx
= tan = 1.
4
4
Hence, a + b = 1.
Note that p0 (x) = a and q 0 (x) = π4 sec2
x = 1, we require p0 (1) = q 0 (1), i.e.:
πx
4
. Now for f to be differentiable at
π π
sec2
4
4
π
= .2
4
π
=
2
a=
Hence for f to be differentiable at x = 1, a =
π
2
and b = 1 − π2 . 3. Let f (x) = |x2 − 2x|.
i. Critical points on [0, 5] are the endpoints of the interval, where f is not
differentiable or when f 0 (x) = 0. In this case, the critical points are:
- (0, 0) and (5, 15), the endpoints of f on [0, 5]
7
- (1, 1), where f has a stationary point
- (2, 0), as f is not differentiable there
b
The stationary point is at x = − 2a
=
differentiable at x = 2.
−2
2(1)
= 1, and clearly f is not
ii. The absolute minimum value on [0, 5] is y = 0 when x = 0 or x = 2. The
absolute maximum on [0, 5] is y = 15, when x = 5. 4.
lim
x→1
8x3 − 9x2 + 1
2x4 − 3x3 + x L’H
=
lim
x→1
(x − 1)2
2x − 2
= L’H lim
x→1
=
24 − 18
2
= 3. 8
24x2 − 18x
2
2009 V3a
1. Using implicit differentiation with respect to x, we see that
1+
1
dy 2 dy
=
+
x
dx y dx
1 + x1
dy
.
⇒
= 2
dx
+1
y
At (1, 1),
dy
2
= . Use the point-gradient formula with x1 = 1, y1 = 1.
dx
3
y − y1 = m(x − x1 )
2
y − 1 = (x − 1)
3
2
1
y = x+ .
3
3
2. By drawing a diagram, we can see that
b2 + h2 = 9.
(4)
Using implicit differentiation with respect to t gives
db
dh
+ 2h
= 0.
(5)
dt
dt
√
If b = 1, then by substituting into (4) we get h = 8 as h > 0. Then (5) becomes
2b
2 · 1 · 0.5 + 2 ·
√ dh
dh
1
8
=0⇒
= − √ m/s. dt
dt
2 8
3. (Mean value theorem.) Clearly f is continuous on [1, 3] and differentiable on (1, 3).
So we know by the mean value theorem that there exists a c ∈ (1, 3) such that
f (3) − f (1)
3−1
f 0 (c) =
If f (x) =
√
1
x − 1, then f 0 (x) = √
. So (6) becomes
2 x−1
√
1
2−0
√
=
2
2 c−1
⇒
√
1
c−1= √
2
c=
9
3
∈ (1, 3).
2
(6)
r
Since f (c) =
4.
1
3 1
, the point is ( , √ ). 2
2 2
i. Here’s a quick sketch of f (x).
The critical points are:
- (0, 1) and (2, 1), the endpoints of the interval
- (1, 0), as f is not differentiable there
1
Note that f 0 (x) = 23 (x − 1)− 3 which is never zero on [0, 2].
ii. There is an absolute minimum of 0 at x = 1. There is an absolute minimum of 1
at x = 2. 5.
lim
x→0
tan x L’H
sec2 x
1
= 1. =
lim
=
x→0
x
1
cos2 0
10
2010 V2a
(
p(x) x ≥ α
1. From Theorem 4.5.6, for f =
, for continuous p and q in an interval
q(x) x < α
containing α, if f is continuous at α and p0 (α) = q 0 (α), then f is differentiable at
x = α. In this case, p(x) = aex + b and q(x) = sin(x) and x = 0. Since p is an
exponential function, it is continuous everywhere and since q is a sine wave, it is
also continuous everywhere. Now for f to be continuous at x = 0
lim f (x) = f (0) = lim+ f (x).
x→0−
x→0
Now
lim f (x) = lim− aex + b = ae0 + b = a + b,
x→0−
x→0
f (0) = sin 0 = 0,
lim f (x) = lim− sin x = 0
x→0+
x→0
so a + b = 0.
Note that p0 (x) = aex and q 0 (x) = cos(x). Now for f to be differentiable at x = 0,
we require p0 (0) = q 0 (0), i.e.:
ae0 = cos(0)
a=1
Therefore, f to be differentiable at x = 0, a = 1 and b = −1. 2.
lim
x→0
sec2 x
tan x
L’H
=
lim
x→0 3ex
e3x − 1
= lim
x→0 3ex
1
cos2 x
1
= .
3
3. We use implicit differentiation to obtain
3x2 + 3y 2
dy
dy
− 1 − 2y
=0
dx
dx
⇒ (3x2 − 1) +
dy
(3y 2 − 2y) = 0
dx
dy
1 − 3x2
⇒
= 2
.
dx
3y − 2y
At x = 1 and y = 1,
dy
dx
= −2, so the tangent line is y − 1 = −2(x − 1). 11
4. Let f (x) = x3 − 6x2 + 1 and f 0 (x) = 3x2 − 12x = 3x(x − 4). Hence we have
stationary points at x = 0 and x = 4. Note that f is continuous everywhere as it
is a polynomial. Consider:
f (−1) = −1 − 6 + 1 = −6
f (0) = 1
f (4) = 64 − 6(16) + 1 = −31
f (6) = 63 − 63 + 1 = 1
Now for the the region [−1, 0], we have f (−1) < 0 < f (0). By the Intermediate
Value Theorem, f has a root in this region, and since for x < 0, f 0 > 0 and hence
can have no more than 1 root in this region.
Simliarly for the the region [0, 4], we have f (4) < 0 < f (0). By the Intermediate
Value Theorem, f has a root in this region, and since for 0 < x < 4, f 0 < 0 and
hence can have no more than 1 root in this region.
And finally, for the the region [4, 6], we have f (4) < 0 < f (6). By the Intermediate
Value Theorem, f has a root in this region, and since for x < 0, f 0 > 0 and hence
can have no more than 1 root in this region.
Hence f (x) has 3 real roots. 5. We know that
1
d
tan−1 x =
.
dx
1 + x2
Use the chain rule to obtain
d
4
tan−1 (4x + 1) =
dx
1 + (4x + 1)2
as
d
(4x + 1) = 4. dx
12