Chemistry and Biology
CONTINUITY
Focus on Exam 7
1 (a) x + 3 =
{
x  3,
x < 3,
x + 3,
x ≥ 3.
(x + 1)(x  3)
x+3
= x  1
(x + 1)(x + 3)
For x ≥ 3, f (x) =
x+3
=x+1
Hence, in the nonmodulus form,
x  1, x < 3,
f (x) =
x + 1,
x ≥ 3.
For x < 3, f (x) =
{
(b) The graph of f(x) is as shown below.
y = −x − 3
y
2
y=x+1
1
−3
−2
−1
x
O
−1
−2
(c) lim f (x) = 2
x → 3
lim f (x) = 2
x → 3+
(d) lim f (x) does not exist because lim f (x) ≠ lim f (x). x → 3
2 (a) lim h(x) = 2
x → 1
1 + p = 2
p=3
x → 3
x → 3+
x = 1 is in the range 3 ≤ x < 0, so the
part of the function x + p is used.
© Oxford Fajar Sdn. Bhd. (008974T) 2012
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2
ACE AHEAD Mathematics (T) Second Term
(b) Since lim h(x) exists,
x → 3
lim h(x) = lim h(x)
x → 3
x → 3+
x2  k
(3)2  k = 3 + 3
k=9
Since lim h(x) exists,
x+3
x→0
lim h(x) = lim h(x)
x→0
x→0+
x+3
0+3=e
0q
ex  q
ln 3 = q
q = ln 3
= ln 31
= ln 1
3
(c) The graph of y = h(x) is as shown below.
{
ex
eq
ex
ln
y = 3e x
1
e3
ex
=
1
3
= 3e x
3 (a) f o g = f [g(x)]
1
=f
x3
1
1
+
3
3
y = x2 − 9
x
=
(1, 8.2)
2
=
e xq =
y
y
h(x) =
x < 3,
3 ≤ x < 0,
x ≥ 0.
x 2  9,
x + 3,
3e x,
1
−4 −3 −2 −1 O
x
1
2
2
2
1
1
x3
= 2(3 + x  3)
= 2x
The domain of f o g is the same as the domain of g, i.e. {x : x ∈ R, x ≠ 3}. Because the domain cannot take the value 3, the range of f o g cannot take the value 2x = 2(3)
= 6.
Hence, the range of f o g is {y : y ∈ R, y ≠ 6}.
(b) The graph of
y = f g(x)
= 2x, x ≠ 3
is as shown below.
=2 3+
1
2
© Oxford Fajar Sdn. Bhd. (008974T) 2012
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Fully Worked Solution
3
y
6
x
O
3
y = 2x
(c) lim f g(x) = 2(3)
x→3
=6
and
lim f g(x) = 2(3)

x→3+
=6
Since lim f g(x) = lim f g(x)
x→3+
x→3+
=6
then lim f g(x) = 6
x→3
4 In the nonmodulus form,
f (x) =
{
x 2  1,
x2 + 1,
(x  2)(x  3),
x < 1,
1 ≤ x < 1,
x ≥ 1.
The graph of y = f (x) is as shown below.
y
2
2
y=x −1
y = (x − 2)(x − 3)
1
−1 O
1
2
3
4
x
y = −x 2 + 1
(b) (i) lim f (x) = 12  1
x→1
=0
lim f (x) = 12 + 1
x→1
=0
f (1) = 12 + 1
=0

+
© Oxford Fajar Sdn. Bhd. (008974T) 2012
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4
ACE AHEAD Mathematics (T) Second Term
Since lim f (x) = lim f (x)
x→1
x→1+
= f (1)
= 0, then f (x) is continuous at x = 1.
(ii) lim f (x) = 12 + 1
x→1
=0
lim f (x) = (1  2)(1  3)
x→1+
=2
Since lim f (x) ≠ lim f (x), then lim f (x) does not exist. x→1
x→1+
x→1
Hence, f (x) is not continuous at x = 1.
5 (a) In the nonmodulus form,
f (x) =
{
x2 , x < 0,
x
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