Transcript Additional Mathematics

Additional
Mathematics
Functions
Questions
1.Given function f : x  mx + 4 , x=n.
x–n
If f(2) = 10 and f(8) = 4, find
a) the values of m and n
b) the image of 3
c) the object of 11
Answer
2.Given that f(x) = 6 x = -b ,where
ax + b
a
4.The graph shows the absolute values
a and b are constants, f(1) = 2 and
function f(x) = I hx – k I
f(5) = 6/7, find the value of a and b
a) Find the possible values of h and k
Answer
b) If the value of f(m) = 6, find the
value of m
Answer
3. Given that f:x  1 , x= k
x–k
and f(3) = 1,
Find
a) the value of k
b) the values of a such that
f(a²) = f(3a)
5.Sketch the graph of the function g:x
I3x – 5I for the domain 0≤x≤4 and
state the range of g(x) corresponding
to the domain.
Answer
Answer
Questions
6.Given the functions f:x  ax + b and g : 9.Given that f(x) = 3x – 8 and
x  5 – x. if fg(x) = 19 – 3x, find
fg(x) = 3x² - 2, find the function g.
a) the values of a and b
Answer
b) the value of x such that
fg(-x) = f(7)
Answer
7.Given that f:x  2x + 1 and g:x  3
x
x = 0, find the composite function
a) fg
b) gf
c) f²
d) g²
Answer
10.Given that g(x) = 2 – x, fg(x) =
2x² + x + 8, find
a) the function f
b) the value of g(x) if fg(x) = 9
8.Given that h:x2x + 1 and k:x 2x +3
3
Find
a) the value of x if hk(x) = x
b) the value of x if kh(x) = -3
c) the value of x when hk(x) = 3 kh(x)
Answer
11. f(x) = 3x + 2 and
gf(x)= 9x+12x²,
find
a) g(x)
b)fg(2)
Answer
Answer
Questions
12.Given that h(x) = 3 + x and
hk(x) = 1 , x ≠3,find
3–x
a) k(x)
b) the value of k(x) such that x = 4
Answer
13) Given that f:x  2 +x, g(x) ax + b,
where a and b are constants, and gf(x) = 6
+ 10x, find
a) the value of a and b
b) the value of x if g(x) = gf(x)
Answer
14) Given that f(x) = 2x + x² and
g(x) = 4 – 3x,
Find
a) the values of x such that f(x) = 8
b) the composite function fg
c) the values of x if f(x) = 3g(x)
Answer
15. Given that f:x  3x – 1 and gf:x
 2x² - 6x + 4, find
a) the function g
b) the values of x such that
f(x) = 9g(x)
Answer
Solution for Question 1(a)
Given f(x) = mx + 4
Subtracting (1) and (2)
x–n
Where
2m + 10n = 16
f(2) = 10 and f(8) = 4
- 2m + n = 7
f(2) = 10 (substitute x = 2, f(x) = 10)
9n = 9
2m + 4 = 10
9n = 9
2–n
n=1
2m + 4 = 20 – 10n
Substitute n = 1 to (2)
2m + 1 = 7
2m + 10 = 16 ……. (1)
2m = 6
f(8) = 4
m=3
8m + 4 = 4(Substitute x = 8, f(x)=4)
8 – n (bring 8 – n to the right)
8m + 4 = 32 – 4n
8m + 4n = 28(Simplify by dividing
with 4)
2m + n = 7 ……(2)
Therefore, f(x) = 3x + 4
x–1
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Solutions for Question 1(b)
Solution for Question 1(c)
to find the image of 3,
x=3
f(3) = 3(3) + 4
3–1
= 13
2 (answer)
To find the object of 11
f(x) = 11
3x + 4 = 11
x–1
3x + 4 = 11x – 11
8x = 15
x = 15
8 (answer)
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Solution for Question 2
Given f(x) = 6 .
ax + b
f(1) = 2 and f(5) = 6/7
f(1) = 2 ( substitute x = 1, f(x) = 2)
6 =2
ax + b
 (bring a + b to the right)
6 = 2a + 2b
a + b = 3 …… (1)
Then f(5) = 6/7( substitute x = 5, f(x)
= 6/7)
6 =6
ax + b 7 (eliminate 6 on both
sides)
1 =1
5a + b 7
after we eliminate 6, we
cross multiply
therefore, 7 = 5a + b
5a + b = 7 …… (2)
Subtracting (1) from (2)
5a + b = 7
- a+b=3
4a
=4
a=1
Substitute a=1 into (1)
1+b=3
b =2
Therefore, a = 1, b = 2
(answer)
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Solution for Question 3(a)
a) find the value of k
f(x) = 1
x–k
f(3) = 1 (substitute x = 3, f(x) = 1)
1= 1
3 – k ( bring 3 – k over the right)
3–k=1
k=2
Therefore f(x) =
1
x–2
(answer)
Solution for question 3(b)
b) find the values of a such that
f(a²) = f(3a)
1 =
1
a² - 2 3a – 2 (cross multiply for
both sides)
3a – 2 = a² - 2
a² = 3a(both sides cancel out one a)
a = 3 (answer)
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Solution for Question 4(a)
a) Find the possible values of
h and k
f(x) = I hx – k I
Given
f(0) = 8 , f(4) = 0
8 = I h(0) – k I
I-kI = 8
-k = 8 or -k = -8
when modulus sign, I I is
removed, the resulting
value can be negative or
positive.
therefore, k = -8 or 8
When f(4) = 0 x = 4 and
f(x) = 0
If k = -8, 0 = I 4h + 8 I
-4h = 8
h = -2
If k = 8, 0 = I 4h – 8 I
4h = 8
h=2
Therefore, h = -2 or 2
Then f(x) = I -2x + 8 I or
f(x) = I2x – 8I (answer)
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Solution for Question 4(b)
b) If the value of f(m) 6, find
the value of m
Take either one of the
function,
Let
f(x) = I 2x – 8 I
Given f(m) = 6
6 = I 2x – 8 I
-6 = 2x – 8 or 6 = 2x – 8
when modulus sign, I I is
removed, the resulting
value can be negative or
positive
Obtain answers separately
-6 = 2x – 8
2x = 2
x=1
or
6 = 2x – 8
2x = 14
x=7
From the graph, m>4
so, x = 7 (answer)
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Solution for Question 5
Steps to sketch the graph
Given g(x) = I 3x – 5 I
-First, find the value of g(x),
when x = 0
-Second, find the value of g(x),
when x = 4. This is to check
the maximum value of g(x)
-Finally, find the value of x,
when g(x) = 0. This is to check
the point where the turning
point exists.
When x = 0,
g(0) = I 3(0) – 5 I
= I -5 I
=5
When x = 4,
g(4) = I 3(4) – 5 I
=7
When g(x) = 0
0 = I 3x – 5 I
3x – 5 = 0
3x = 5
x = 5/3
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Solution for Question 6(a)
a)the values of a and b
Given
f(x) = ax + b
g(x) = 5 – x
fg(x) = 19 – 3x ……(1)
fg(x) = f[g(x)]
= a(5 – x) + b
= 5a – ax + b (factorize a)
= (5a + b) – ax……(2)
Compare the x-coefficient of
(1) and (2)
-a = -3
a=3
Compare the constant of (1)
and (2)
5a + b = 19
Substitute a = 3
5(3) + b = 19
15 + b = 19
b=4
Therefore, a = 3, b = 4
f(x) = 3x + 4 (answer)
Solution for Question 6(b)
b) Find the value of x
such that
fg(-x) = f(4)
19 + 3x = 3(4) + 4
19 + 3x = 16
3x = 3
x = -1( answer)
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Solution for Question 7(a)
Solution for Question 7(c)
a)fg
fg(x) = f[g(x)] substitute g(x) = 3
c) f²
f²(x) = ff(x)
= f[f(x)](substitute f(x) = 2x + 1)
x
= f(3/x)
= 2(3/x) + 1
=6 + 1 (answer)
x
Solution for Question 7(b)
b) gf
gf(x) = g[f(x)](substitute f(x) = 2x + 1)
= g(2x + 1)
= 3 (answer)
2x + 1
= f(2x + 1)
= 2(2x + 1) + 1
= 4x + 3 (answer)
Solution for Question 7(d)
d) g²
g²(x) = g[g(x)](substitute g(x)
= 3/x)
= g(3/x)
= 3
3/x
= x (answer)
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Solution for Question 8(a)
a)Find the value of x, if hk(x) =
x
First, find hk(x)
hk(x)
= h[k(x)]
= h(2x + 3)
= 2(2x + 3) + 1
3
= 4x + 6 + 1
3
= 4x + 7
3
From hk(x) = x, so
4x + 7 = x
3
4x + 7 = 3x
x = -7 (answer)
Solution for Question 8(b)
b) Find the value of x if
kh(x) = -3
First, find kh(x),
kh(x) = k[h(x)]
= k ( 2x + 1)
3
= 2 ( 2x + 1 ) + 3
3
= 4x + 2 + 3
3
kh(x) = -3
4x + 2 + 3 = -3
3
4x + 2 + 9 = -9
4x = -20
x = -5 (answer)
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Solution for Question 8(c)
c) find the value of x when
hk(x) = 3kh(x)
From (a) From (b)
hk(x)
= 3kh(x)
4x + 7 = (4x + 2 + 3) 3
3
3
4x + 7 = 3(4x + 2 + 9)
4x + 7 = 12x + 6 + 27
-26 = 8x
x = - 13/4 (answer)
Solution for Question 9
fg(x) = 3x² - 2
f[g(x)] = 3x² - 2
Substitute f(x) = 3x – 8
3g(x) – 8 = 3x² - 2
3g(x) = 3x² + 6
g(x) = x² + 2 (answer)
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Solution for Question 10(a)
a)fg(x) = 2x² + x + 8
f(2-x) = 2x² + x + 8
Let y = 2 – x
Hence x = 2 – y
f(y) = 2(2 – y)² + (2 – y) + 8
= 2(4–4y+y²)+ 2– y + 8
= 8 – 8y+2y² + 2 – y + 8
= 2y² - 9y + 18
f(x) = 2x² - 9x + 18(answer)
Solution for Question 10(b)
b) fg(x) = 9
2x² + x + 8 = 9
2x² + x – 1 = 0
(2x – 1)(x + 1) = 0
2x – 1 = 0 or x + 1 = 0
x=½
x = -1
When x = ½
g( ½ ) = 2 – ½
= 3/2
When x = -1
g(-1) = 2 – (-1)
=3
So, g(x) = 3/2 or 3
(answer)
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Solution for Question 11(a)
Solution for Question 11(b)
a)Find g(x)
b) fg(x) = f[g(x)]
gf(x) = 9x + 12x²
= 3(4x² - 7x – 2 ) + 2
g[f(x)] = 9x + 12x²(substitute f(x) = 3x + 2)
3
g(3x + 2) = 9x + 12x²
= 4x² - 7x – 2 + 2
Let y = 3x + 2
= 4x² - 7x (answer)
3x = y – 2
Hence x = y – 2
3
Let u = y – 2
3
g(u) = 9 (y – 2) + 12 (y – 2)²
3
3
=3y – 6 + 12 (y²- 4y +4)
9
= 4y² - 7y -2
3
so,
g(x) = 4x² - 7x – 2
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3
(answer)
Solution for Question 12 (a)
a) Find k(x),
hk(x) = 1 (substitute h(x) = 3 + x)
3–x
h[k(x)] =
1
3–x
3 + k(x) = 1
3–x
[3 + k(x)][3 – x] = 1
9 – 3x + 3k(x) – k(x)x = 1
So,
8 – 3x = k(x)x – 3k(x)
8 – 3x = k(x) (x – 3)
k(x) = 8 – 3x
x – 3 (answer)
Solution for Question 12 (b)
b) Find the value of k(4)
k(4) = 8 – 3(4)
4–3
= 8 – 12
= -4
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Solution for Question 13(a)
a)Find the values of a and b
gf(x) = 6 + 10x
g[f(x)] = 6 + 10x (substitute f(x) = 2
+ 5x)
9(2 + 5x) + b = 6 + 10x
2a + 5ax + b = 6 + 10x
5ax + 2a + b = 6 + 10x
Compare the x-coefficient
5a = 10
a=2
Compare the constant
2a + b = 6 (replace a = 2)
2(2) + b = 6
4+b=6
b = 2 ( answer)
Solution for Question 13(b)
b) the value of x if g(x) = gf(x)
g(x) = gf(x)
2x + 2 = 6 + 10x
8x = -4 (Bring the 8 to the right)
x=-4=-1
8
2
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Solution for Question 14(a)
a)Find the values of x such
that f(x) = 8
f(x) = 8
2x + x² = 8
x² + 2x – 8 = 0
(x + 4)(x – 2) = 0
So,
x+4=0
x = -4
OR
x–2=0
x=2
Solution for Question 14(b)
b) Find the composite fg
fg(x) = f[g(x)] ( substitute g(x) = 4–x)
= f[ 4 – 3x]
= 2(4 – 3x) + (4 – 3x)²
= 8 – 6x + 16 + 9x² - 24x
= 9x² - 30x + 24 (answer)
Solution for Question 14(c)
c)Find the values of x if f(x) =
3g(x)
f(x) = 3g(x)
2x + x² = 3(4 – 3x)
x² + 2x = 12 – 9x
x² + 11x – 12 = 0
So,
(x + 12)(x – 1) = 0
(x + 12) = 0 OR (x – 1) = 0
x = -12
x=1
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Solution for Question 15(a)
g[3x – 1] = 2x² - 6x + 4
Let y = 3x – 1
3x = y + 1
Hence x = y + 1
3
Let u = y + 1
3
g(u) = 2(y + 1)²-6(y + 1)+4
3
3
=2(y²+2y+1)-2y–2+ 4
9
Solution for Question 15(b)
b) f(x) = 9g(x)
3x – 1 = 9(2x²-14x+20)
9
3x – 1 = 2x²-14x+20
2x²-17x+21=0
(2x-3)(x-7) =0
2x – 3 = 0 or x – 7 = 0
x=3
x=7
2
= 2y²+4y+2–2y–2+ 4
9
= 2y²+4y+2–18y + 18
9
= 2y² - 14y + 20
9
g(x) = 2x² - 14x + 20
9
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End of Questions