H2 Mathematics – Functions
Solutions (2007 – 2016)
Inverse and Composite Functions
1. 2007/P1/2
(i) gf(x) =1
(x� 3)2, x 2 R, x 6= 3
Rg = [0,1)
Df = R\{3}
Rg * Df
) fg does not exist
(ii) Let y = f(x) =1
x� 3
=) x =1
y+ 3
) f�1(x) =1
x+ 3
=1 + 3x
x
Df�1 = Rf = R \ {0}
c�infinitymaths.sg 2016 1
2. 2008/P2/4
(i) f(x) = (x� 4)2 + 1, x > 4
(ii) Let y = f(x), x > 4
x = 4±py � 1 = 4 +
py � 1 since x > 4
) f�1(x) = 4 +px� 1
Df�1 = Rf = (1,1)
(iii) f�1(x) = 4 +px� 1, x > 1
(iv) y = x
Solving f(x) = f�1(x) is equivalent to solving f(x) = x.
(x� 4)2 + 1 = x
=) x2 � 9x+ 17 = 0
=) x =9±
p92 � 4(1)(17)
2(1)
) x =9 +
p13
2since x > 4
c�infinitymaths.sg 2016 2
3. 2009/P2/3
(i) Let y = f(x) =ax
bx� a
=) (bx� a)y = ax
bxy � ay � ax = 0
(by � a)x = ay
x =ay
by � a
) f�1(x) =ax
bx� a
f(x) = f�1(x)
↵(x) = ↵�1(x)
f 2(x) = x
Df2 = Df = R \nab
o
) Rf2 = R \nab
o
(ii) Rg = R \ {0} * Df =) fg does not exist
(iii) f�1(x) = x
ax
bx� a= x
ax = (bx� a)x
bx2 � 2ax = 0
x(bx� 2a) = 0
) x = 0 or2a
b
c�infinitymaths.sg 2016 3
4. 2010/P2/4
(i)
(ii) When x � 0, f is a one-one function and f�1 exists. Hence the least value of k is 0.
(iii) fg(x) =1
✓1
x� 3
◆2
� 1
=1
1
(x� 3)2� 1
=(x� 3)2
1� (x� 3)2
=(x� 3)2
�x2 + 6x� 8
=(x� 3)2
(4� x)(x� 2)
(iv) fg(x) > 0 =) (x� 3)2
(4� x)(x� 2)> 0
) 2 < x < 3 or 3 < x < 4
(v) fg(x) =(x� 3)2
(4� x)(x� 2)= �1 +
1
(4� x)(x� 2)
From the graph of y = fg(x), where x 6= 2, x 6= 3, x 6= 4,
Rfg = (�1,�1) [ (0,1).
c�infinitymaths.sg 2016 4
5. 2011/P2/3
(i) Let y = f(x) =) y = ln (2x+ 1) + 3
Rearranging, x = 12 (e
y�3 � 1) =) f�1(x) = 12 (e
x�3 � 1)
Df�1 = Rf = R
Rf�1 = Df =�� 1
2 ,1�
(ii) A(0, 3), B
✓12
�e�3 � 1
�, 0
◆, C
✓0, 1
2
�e�3 � 1
�◆, D(3, 0)
(iii) The points of intersection lie on the line y = x. Hence the x -coordinates satisfy the equation
ln (2x+ 1) + 3 = x =) ln (2x+ 1) = x� 3
From GC, the x -coordinates of the points of intersection are 5.482 and �0.4847.
c�infinitymaths.sg 2016 5
6. 2012/P1/7
(i) Let y = g(x) =x+ k
x� 1
=) xy � y = x+ k
=) x(y � 1) = y + k
=) x =y + k
y � 1
) g�1(x) =x+ k
x� 1
Hence g is self-inverse.
(ii) y = g(x)
(iii) Since g is self-inverse, y = x is a line of symmetry of the curve.
y =x+ k
x� 1= 1 +
k + 1
x� 1
Translate
✓10
◆, stretch factor (k + 1) parallel to y-axis, translate
✓01
◆.
c�infinitymaths.sg 2016 6
7. 2013/P2/1
(i) Rg = R
Df = R \ {1}
) Rg * Df =) fg does not exist
(ii) gf(x) = 1� 2
✓2 + x
1� x
◆
=(1� x)� 2(2 + x)
1� x
=�3x� 3
1� x
=3(x+ 1)
x� 1
Let x = (gf)�1(5) =) gf(x) = 5
3(x+ 1)
x� 1= 5
3(x+ 1) = 5(x� 1)
3x+ 3 = 5x� 5
x = 4
) (gf)�1(5) = 4
8. 2014/P1/1
(i) f 2(x) =1
1� 1
1� x
⇥ 1� x
1� x
=1� x
(1� x)� 1
=1� x
�x
=x� 1
x
Let y = f(x) =1
1� x
=) y � xy = 1
=) x =y � 1
y
) f�1(x) =x� 1
x
Hence f 2(x) = f�1(x).
(ii) f 2(x) = f�1(x) =) ↵ 2(x) = ↵�1(x) =) f 3(x) = x
c�infinitymaths.sg 2016 7
9. 2015/P2/3
(a) (i) Any horizontal line y = k, k 2 R, cuts the graph of y = f(x) at most once.Hence f is one-one and f�1 exists.
(ii) Let y = f(x)
=) y =1
1� x2
1� x2 =1
y
x2 = 1� 1
y
=) x = ±r1� 1
y
Since x > 1, x =
r1� 1
y
) f�1(x) =
r1� 1
x
Df�1 = Rf = (�1, 0)
(b) Let y =2 + x
1� x2
=) y(1� x2) = 2 + x
=) yx2 + x+ (2� y) = 0
Discriminant = 12 � 4y(2� y) = 4y2 � 8y + 1
For values that y can take, the equation has real solutions for x.
=) Discriminant � 0
=) 4y2 � 8y + 1 � 0
=) y � 1 +
p3
2or y 1�
p3
2
) Rg =
�1, 1�
p3
2
#["1 +
p3
2,1!
c�infinitymaths.sg 2016 8
10. 2016/P1/10a
(i) Let y = f(x) =) y = 1 +px
Rearranging, x = (y � 1)2 =) f�1(x) = (x� 1)2, x 2 R, x � 1
(ii) ↵(x) = x =) 1 +q1 +
px = x
1 +px = (x� 1)2
1 +px = x2 � 2x+ 1
x =�x2 � 2x
�2
x4 � 4x3 + 4x2 � x = 0
x(x3 � 4x2 + 4x� 1) = 0
x3 � 4x2 + 4x� 1 = 0 or x = 0
=) x3 � 4x2 + 4x� 1 = 0 or x = 0 (rejected as 1 +p
1 +p0 6= 0)
Solving x3 � 4x2 + 4x� 1 = 0 with calculator, x = 0.382, 2.62 or 1
Only x = 2.62 satisfies 1 +p1 +
px = x
↵(x) = x =) f�1↵(x) = f�1(x) =) f(x) = f�1(x)
Hence any value of x which satisfies ↵(x) = x will also satisfy f(x) = f�1(x).
c�infinitymaths.sg 2016 9
Piecewise and Periodic Functions
1. 2009/P1/4
(i) f(27) = f(3)
= 2(3)� 1
= 5
f(45) = f(1)
= 7� 12
= 6
) f(27) + f(45) = 11
(ii)
(iii)
Z 3
�4f(x) dx = 2
Z 2
07� x2 dx+
Z 4
22x� 1 dx+
Z 3
22x� 1 dx
= 683 + 10 + 4
= 1103
c�infinitymaths.sg 2016 10
2. 2013/P1/5
(i) (ii)
Z p3a2
a2
f(x) dx =
Z p3a2
a2
r1� x2
a2dx
=
Z 13⇡
16⇡
p1� sin2 ✓ (a cos ✓) d✓
= a
Z 13⇡
16⇡
pcos2 ✓ (cos ✓) d✓
= a
Z 13⇡
16⇡
cos2 ✓ d✓
= a
Z 13⇡
16⇡
12 (cos 2✓ + 1) d✓
=a
2
12 sin 2✓ + ✓
� 13⇡
16⇡
=a
2
p3
4+
⇡
3�
p3
4� ⇡
6
!
=a⇡
12
3. 2015/P1/5
(i) Translate
✓30
◆, stretch factor 1
4 parallel to y-axis (either order) or
Stretch factor 2 parallel to x -axis, translate
✓30
◆or
Translate
✓1.50
◆, stretch factor 2 parallel to x -axis.
(ii)
(iii)
c�infinitymaths.sg 2016 11
4. 2016/P1/10b
(i) g(0) = 1
g(1) = 1 + g(0) = 2
g(2) = 2 + g(1) = 4
g(3) = 1 + g(2) = 5
g(4) = 2 + g(2) = 6
g(6) = 2 + g(3) = 7
g(7) = 1 + g(6) = 8
g(12) = 2 + g(6) = 9
(ii) g(5) = 1 + g(4) = 7
g(5) = g(6) = 7
Therefore, g is not a one-one function and hence does not have an inverse.
c�infinitymaths.sg 2016 12
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