1.6.4 Function, SPM Practice (Short Question)

Posted on May 18, 2020 by Myhometuition

Question 10 (4 marks):
Given the function g : x → 2x – 8, find

\begin{array}{l} (a) g^{- 1} (x), \\ (b) the value of p such that g^{2} (\frac{3 p}{2}) = 30. \end{array}

Solution:
(a)

\begin{array}{l} Let y = g (x) \\ = 2 x - 8 \\ 2 x - 8 = y \\ 2 x = y + 8 \\ x = \frac{y + 8}{2} \\ Thus, g^{- 1} (x) = \frac{x + 8}{2} \end{array}

(b)

\begin{array}{l} g (x) = 2 x - 8 \\ g^{2} (x) = g [g (x)] \\ = g (2 x - 8) \\ = 2 (2 x - 8) - 8 \\ = 4 x - 16 - 8 \\ = 4 x - 24 \\ g^{2} (\frac{3 p}{2}) = 30 \\ 4 (\frac{3 p}{2}) - 24 = 30 \\ 6 p = 54 \\ p = 9 \end{array}

Question 11 (4 marks):
Diagram 9 shows the relation between set A, set B and set C.

Diagram 9

It is given that set A maps to set B by the function

\frac{x + 1}{2}

and maps to set C by fg : x → x² + 2x + 4.
(a) Write the function which maps set A to set B by using the function notation.
(b) Find the function which maps set B to set C.

Solution:

(a)

g : x \to \frac{x + 1}{2}

(b)

\begin{array}{l} g (x) = \frac{x + 1}{2} \\ f g (x) = x^{2} + 2 x + 4 \\ f [g (x)] = x^{2} + 2 x + 4 \\ f (\frac{x + 1}{2}) = x^{2} + 2 x + 4 \\ Let \frac{x + 1}{2} = y \\ x + 1 = 2 y \\ x = 2 y - 1 \\ ∴ f (y) = {(2 y - 1)}^{2} + 2 (2 y - 1) + 4 \\ f (y) = 4 y^{2} - 4 y + 1 + 4 y - 2 + 4 \\ f (y) = 4 y^{2} + 3 \\ f (x) = 4 x^{2} + 3 \\ Thus, function which maps set B to set C \\ is f (x) = 4 x^{2} + 3 \end{array}

SPM Practice (Long Question)

Posted on May 17, 2020 by Myhometuition

Question 4:

\begin{array}{l} The function f is denoted by f : x \to \frac{1 + x}{1 - x}, x \neq 1. \\ Find f^{2}, f^{3}, f^{4} and hence write down the functions f^{51} and f^{52} . \end{array}

Solution:

\begin{array}{l} f (x) = \frac{1 + x}{1 - x}, x \neq 1 \\ f^{2} (x) = f [f (x)] = f (\frac{1 + x}{1 - x}) \\ = \frac{1 + (\frac{1 + x}{1 - x})}{1 - (\frac{1 + x}{1 - x})} = \frac{\frac{1 - x + 1 + x}{1 - x}}{\frac{1 - x - 1 - x}{1 - x}} \\ = \frac{2}{- 2 x} = - \frac{1}{x} \\ f^{3} (x) = f [f^{2} (x)] = f (- \frac{1}{x}) \\ = \frac{1 + (- \frac{1}{x})}{1 - (- \frac{1}{x})} = \frac{\frac{x - 1}{x}}{\frac{x + 1}{x}} \\ = \frac{x - 1}{x + 1} \\ f^{4} (x) = f [f^{3} (x)] = f (\frac{x - 1}{x + 1}) \\ = \frac{1 + (\frac{x - 1}{x + 1})}{1 - (\frac{x - 1}{x + 1})} = \frac{\frac{x + 1 + x - 1}{x + 1}}{\frac{x + 1 - x + 1}{x + 1}} \\ = \frac{2 x}{2} = x \\ f^{5} (x) = f [f^{4} (x)] = f (x) = \frac{1 + x}{1 - x} (recurring) \\ ∴ f^{51} (x) = f^{3} [f^{48} (x)] = f^{3} (x) \\ = \frac{x - 1}{x + 1} \\ f^{52} (x) = f^{4} [f^{48} (x)] = f^{4} (x) = x \end{array}

SPM Practice (Long Question)

Posted on May 17, 2020 by Myhometuition

Question 2:
The function f and g is defined by

\begin{array}{l} f (x) = 3 x - 2 \\ g (x) = \frac{3}{x}, x \neq 0 \\ Find \\ (a) f^{- 1} (2), \\ (b) g f (- 3), \\ (c) function h if h f (x) = 3 x + 2, \\ (d) function k if f k (x) = 4 x - 7. \end{array}

Solution:
(a)

\begin{array}{l} Let f^{- 1} (2) = x, \\ thus f (x) = 2 \\ 3 x - 2 = 2 \\ 3 x = 4 \\ x = \frac{4}{3} \\ f^{- 1} (2) = \frac{4}{3} \end{array}

(b)

\begin{array}{l} g f (- 3) = g [3 (- 3) - 2] \\ = g (- 11) \\ = - \frac{3}{11} \end{array}

(c)

\begin{array}{l} h [f (x)] = 3 x + 2 \\ h (3 x - 2) = 3 x + 2 \\ Let y = 3 x - 2 \\ thus x = \frac{y + 2}{3} \\ h (y) = 3 (\frac{y + 2}{3}) + 2 \\ = y + 2 + 2 \\ = y + 4 \\ ∴ h (x) = x + 4 \end{array}

(d)

\begin{array}{l} f [k (x)] = 4 x - 7 \\ 3 k (x) - 2 = 4 x - 7 \\ 3 k (x) = 4 x - 5 \\ k (x) = \frac{4 x - 5}{3} \end{array}

SPM Practice (Long Question)

Posted on May 17, 2020 by Myhometuition

Question 6 (8 marks):
It is given that g : x → 2x – 3 and h : x → 1 – 3x.
(a) Find
(i) h (5)
(ii) the value of k if

g (k + 2) = \frac{1}{7} h (5),

(iii) hg(x).

(b) Hence, sketch the graph of y = | hg(x) | for –1 ≤ x ≤ 3.
State the range of y.

Solution:
(a)(i)

\begin{array}{l} h (x) = 1 - 3 x \\ h (5) = 1 - 3 (5) \\ = - 14 \end{array}

(a)(ii)

\begin{array}{l} g (x) = 2 x - 3 \\ g (k + 2) = \frac{1}{7} h (5) \\ 2 (k + 2) - 3 = \frac{1}{7} (- 14) \\ 2 k + 4 - 3 = - 2 \\ 2 k = - 3 \\ k = - \frac{3}{2} \end{array}

(a)(iii)

\begin{array}{l} g (x) = 2 x - 3, h (x) = 1 - 3 x \\ h g (x) = h (2 x - 3) \\ = 1 - 3 (2 x - 3) \\ = 1 - 6 x + 9 \\ = 10 - 6 x \end{array}

(b)
y = |hg(x)|,
y = |10 – 6x|
Range of y : 0 ≤ y ≤ 16

1.6.3 Function, SPM Practice (Short Question)

Posted on May 15, 2020 by Myhometuition

Question 7:
Diagram below shows the function g : x → x – 2k, where k is a constant.

Find the value of k.

Solution:

\begin{array}{l} Given g (6) = 12 \\ g (6) = 6 - 2 k \\ 12 = 6 - 2 k \\ 2 k = 6 - 12 \\ k = - 3 \end{array}

Question 8:
Diagram below shows the relation between set M and set N in the arrow diagram.

(a) Represent the relation in the form of ordered pairs.
(b) State the domain of the relation.

Solution:
(a) Relation in the form of ordered pairs = {(–4, 8), (3, 3), (4, 8)}.
(b) Domain of the relation = {–4, 3, 4}.

Question 9 (3 marks):
Diagram shows the graph of the function f : x → |1 – 2x| for the domain –2 ≤ x ≤ 4.

Diagram

State
(a) the object of 7,
(b) the image of 3,
(c) the domain of 0 ≤ f(x) ≤ 5.

Solution:
(a)
The object of 7 is 4.

(b)
f (x) = |1 – 2x|
f (3) = |1 – 2(3)|
= |1 – 6|
= |–5|
= 5

The image of 3 is 5.

(c)
|1 – 2x| = 5
1 – 2x = ±5
Given when f(x) = 5, x = –2.

When f(x) = –5
1 – 2x = –5
2x = 6
x = 3

Domain: –2 ≤ x ≤ 3.

1.6.2 Function, SPM Practice (Short Question)

Posted on May 15, 2020 by Myhometuition

Question 4:
It is given the functions g(x) = 3x and h(x) = m – nx, where m and n are constants.
Express m in terms of n such that hg(1) = 4.

Solution:

\begin{array}{l} h g (x) = h (3 x) \\ = m - n (3 x) \\ = m - 3 n x \\ h g (1) = 4 \\ m - 3 n (1) = 4 \\ m - 3 n = 4 \\ m = 4 + 3 n \end{array}

Question 5:
Diagram below shows the relation between set M and set N in the graph form.

State
(a) the range of the relation,
(b) the type of the relation between set M and set N.

Solution:
(a) Range of the relation = {p, r, s}.
(b) Type of the relation between set M and set N is many to one relation.

Question 6:
Diagram below shows the relation between set P and set Q.

State
(a) the object of 3,
(b) the range of the relation.

Solution:
(a) The object of 3 is 7.
(b) The range of the relation is {–3, –1, 1, 3}.

SPM Practice (Long Question)

Posted on April 18, 2020 by user

Question 3:
Given that f : x → hx + k and f² : x → 4x + 15.
(a) Find the value of h and of k.
(b) Take the value of h > 0, find the values of x for which f (x² ) = 7x

Solution:
(a)
Given f (x) = hx + k
f² (x) = ff (x) = f (hx + k)
= h (hx + k) + k
= h² x + hk + k

f² (x) = 4x + 15
h² x + hk + k = 4x + 15
h² = 4
h = ± 2
when, h = 2
hk + k = 15
2k + k = 15
k = 5

When, h = –2
hk + k = 15
–2k + k = 15
k = –15

(b)
h > 0, h = 2, k = 5
Given f (x) = hx + k
f (x) = 2x + 5 f (x² ) = 7x
2 (x² ) + 5 = 7x
2x² – 7x + 5 = 0
(2x – 5)(x –1) = 0
2x – 5 = 0 or x –1= 0
x = 5/2 x = 1

SPM Practice (Long Question)

Posted on April 18, 2020 by user

Question 1:
The function f and g is defined by

\begin{array}{l} f : x \mapsto 2 x - 3 \\ g : x \mapsto \frac{2}{x}; x \neq 0 \end{array}

Find the expression for each of the following functions
(a) ff,
(b) gf,
(c) f^-1 ,
Calculate the value of x such that ff(x) = gf(x).

Solution:
(a)

\begin{array}{l} f f (x) = f [f (x)] \\ = f (2 x - 3) \\ = 2 (2 x - 3) - 3 \\ = 4 x - 9 \\ f f : x \mapsto 4 x - 9 \end{array}

(b)

\begin{array}{l} g f (x) = g [f (x)] \\ = g (2 x - 3) \\ = \frac{2}{2 x - 3} \\ g f : x \mapsto \frac{2}{2 x - 3} \end{array}

(c)

\begin{array}{l} Let f^{- 1} (x) = y, \\ thus f (y) = x \\ 2 y - 3 = x \\ y = \frac{x + 3}{2} \\ ∴ f^{- 1} (x) = \frac{x + 3}{2} \\ f^{- 1} : x \mapsto \frac{x + 3}{2} \\ When f f (x) = g f (x), \\ 4 x - 9 = \frac{2}{2 x - 3} \\ (4 x - 9) (2 x - 3) = 2 \\ 8 x^{2} - 30 x + 27 = 2 \\ 8 x^{2} - 30 x + 25 = 0 \\ (4 x - 5) (2 x - 5) = 0 \\ 4 x - 5 = 0 or 2 x - 5 = 0 \\ x = \frac{5}{4} or x = \frac{5}{2} \end{array}

SPM Practice (Short Question)

Posted on April 18, 2020 by user

Question 1:
Given the function g : x → 3x – 2, find
(a) the value of x when g(x) maps onto itself,
(b) the value of k such that g(2 – k) = 4k.

Solution:
(a)

\begin{array}{l} g (x) = x \\ 3 x - 2 = x \\ 3 x - x = 2 \\ 2 x = 2 \\ x = 1 \end{array}

(b)

\begin{array}{l} g (x) = 3 x - 2 \\ g (2 - k) = 4 k \\ 3 (2 - k) - 2 = 4 k \\ 6 - 3 k - 2 = 4 k \\ - 7 k = - 4 \\ k = \frac{4}{7} \end{array}

Question 2:
Given the functions f : x → px + 1, g : x → 3x – 5 and fg(x) = 3px + q.
Express p in terms of q.

Solution:

\begin{array}{l} f (x) = p x + 1, g (x) = 3 x - 5 \\ f g (x) = p (3 x - 5) + 1 \\ = 3 p x - 5 p + 1 \\ Given f g (x) = 3 p x + q \\ 3 p x - 5 p + 1 = 3 p x + q \\ - 5 p + 1 = q \\ - 5 p = q - 1 \\ 5 p = 1 - q \\ p = \frac{1 - q}{5} \end{array}

Question 3:
Given the functions h : x → 3x + 1, and gh : x → 9x² + 6x – 4, find
(a) h^-1 (x),
(b) g(x).

Solution:
(a)

\begin{array}{l} Let h^{- 1} (x) = y, \\ thus h (y) = x \\ 3 y + 1 = x \\ 3 y = x - 1 \\ y = \frac{x - 1}{3} \\ ∴ h^{- 1} (x) = \frac{x - 1}{3} \\ h^{- 1} : x \mapsto \frac{x - 1}{3} \end{array}

(b)

\begin{array}{l} g [h (x)] = 9 x^{2} + 6 x - 4 \\ g (3 x + 1) = 9 x^{2} + 6 x - 4 \\ Let y = 3 x + 1 \\ thus x = \frac{y - 1}{3} \\ g (y) = 9 {(\frac{y - 1}{3})}^{2} + 6 (\frac{y - 1}{3}) - 4 \\ = \frac{9 {(y - 1)}^{2}}{9} + 2 (y - 1) - 4 \\ = y^{2} - 2 y + 1 + 2 y - 2 - 4 \\ = y^{2} - 5 \\ ∴ g (x) = x^{2} - 5 \end{array}

Inverse Function Example 4

Posted on April 18, 2020 by user

Example 4:

(a) If f : x \to x - 2, find f^{- 1} (5),

(b) if f : x \to \frac{x + 9}{x - 5}, x \neq 5, find f^{- 1} (3) .

Solution:

(a)

f (x) = x– 2

Let y = f ^-1(5)

f (y) = 5

y – 2 = 5

y = 7

therefore, f ^-1(5) = 7

(b)

\begin{array}{l} f (x) = \frac{x + 9}{x - 5} \\ Let y = f^{- 1} (3) \\ f (y) = 3 \\ \frac{y + 9}{y - 5} = 3 \\ y + 9 = 3 y - 15 \\ 2 y = 24 \\ y = 12 \\ ∴ f^{- 1} (3) = 7 \end{array}