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}

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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

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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

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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).

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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.

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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

◆.

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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

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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!

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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).

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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

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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)

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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.

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