User interface language: English | Español

Date May 2011 Marks available 4 Reference code 11M.1.hl.TZ1.8
Level HL only Paper 1 Time zone TZ1
Command term Find Question number 8 Adapted from N/A

Question

Consider the functions given below.

\[f(x) = 2x + 3\]\[g(x) = \frac{1}{x},x \ne 0\]

(i)     Find \(\left( {g \circ f} \right)\left( x \right)\) and write down the domain of the function.

(ii)     Find \(\left( {f \circ g} \right)\left( x \right)\) and write down the domain of the function.

[2]
a.

Find the coordinates of the point where the graph of \(y = f(x)\) and the graph of \(y = \left( {{g^{ - 1}} \circ f \circ g} \right)(x)\) intersect.

[4]
b.

Markscheme

(i)     \(\left( {g \circ f} \right)\left( x \right) = \frac{1}{{2x + 3}}\), \(x \ne - \frac{3}{2}\) (or equivalent)     A1

 

(ii)     \(\left( {f \circ g} \right)\left( x \right) = \frac{2}{x} + 3\), \(x \ne 0\) (or equivalent)     A1

[2 marks]

a.

EITHER

\(f(x) = \left( {{g^{ - 1}} \circ f \circ g} \right)(x) \Rightarrow \left( {f \circ g} \right)\left( x \right)\)     (M1)

\(\frac{1}{{2x + 3}} = \frac{2}{x} + 3\)     A1

OR

\(\left( {{g^{ - 1}} \circ f \circ g} \right)(x) = \frac{1}{{\frac{2}{x} + 3}}\)     A1

\(2x + 3 = \frac{1}{{\frac{2}{x} + 3}}\)     M1

THEN

\(6{x^2} + 12x + 6 = 0\) (or equivalent)     A1

\(x = - 1\), \(y = 1\) (coordinates are (−1, 1) )     A1

[4 marks]

b.

Examiners report

Part (a) was in general well answered and part (b) well attempted. Some candidates had difficulties with the order of composition and in using correct notation to represent the domains of the functions.

a.

Part (a) was in general well answered and part (b) well attempted. Some candidates had difficulties with the order of composition and in using correct notation to represent the domains of the functions.

b.

Syllabus sections

Topic 2 - Core: Functions and equations » 2.1 » Concept of function \(f:x \mapsto f\left( x \right)\) : domain, range; image (value)

View options