Question
The diagram below shows the graph of a function \(f\) , for \( – 1 \le x \le 2\) .
Write down the value of \(f(2)\).
Write down the value of \({f^{ – 1}}( – 1)\) .
Sketch the graph of \({f^{ – 1}}\) on the grid below.
Answer/Explanation
Markscheme
\(f(2) = 3\) A1 N1
[1 mark]
\({f^{ – 1}}( – 1) = 0\) A2 N2
[2 marks]
EITHER
attempt to draw \(y = x\) on grid (M1)
OR
attempt to reverse x and y coordinates (M1)
eg writing or plotting at least two of the points
\(( – 2, – 1)\) , \(( – 1,0)\) , \((0,1)\) , \((3,2)\)
THEN
correct graph A2 N3
[3 marks]
Question
Let \(f(x) = 3x – 2\) and \(g(x) = \frac{5}{{3x}}\), for \(x \ne 0\).
Let \(h(x) = \frac{5}{{x + 2}}\), for \(x \geqslant 0\). The graph of h has a horizontal asymptote at \(y = 0\).
Find \({f^{ – 1}}(x)\).
Show that \(\left( {g \circ {f^{ – 1}}} \right)(x) = \frac{5}{{x + 2}}\).
Find the \(y\)-intercept of the graph of \(h\).
Hence, sketch the graph of \(h\).
For the graph of \({h^{ – 1}}\), write down the \(x\)-intercept;
For the graph of \({h^{ – 1}}\), write down the equation of the vertical asymptote.
Given that \({h^{ – 1}}(a) = 3\), find the value of \(a\).
Answer/Explanation
Markscheme
interchanging \(x\) and \(y\) (M1)
eg \(x = 3y – 2\)
\({f^{ – 1}}(x) = \frac{{x + 2}}{3}{\text{ }}\left( {{\text{accept }}y = \frac{{x + 2}}{3},{\text{ }}\frac{{x + 2}}{3}} \right)\) A1 N2
[2 marks]
attempt to form composite (in any order) (M1)
eg \(g\left( {\frac{{x + 2}}{3}} \right),{\text{ }}\frac{{\frac{5}{{3x}} + 2}}{3}\)
correct substitution A1
eg \(\frac{5}{{3\left( {\frac{{x + 2}}{3}} \right)}}\)
\(\left( {g \circ {f^{ – 1}}} \right)(x) = \frac{5}{{x + 2}}\) AG N0
[2 marks]
valid approach (M1)
eg \(h(0),{\text{ }}\frac{5}{{0 + 2}}\)
\(y = \frac{5}{2}{\text{ }}\left( {{\text{accept (0, 2.5)}}} \right)\) A1 N2
[2 marks]
A1A2 N3
Notes: Award A1 for approximately correct shape (reciprocal, decreasing, concave up).
Only if this A1 is awarded, award A2 for all the following approximately correct features: y-intercept at \((0, 2.5)\), asymptotic to x-axis, correct domain \(x \geqslant 0\).
If only two of these features are correct, award A1.
[3 marks]
\(x = \frac{5}{2}{\text{ }}\left( {{\text{accept (2.5, 0)}}} \right)\) A1 N1
[1 mark]
\(x = 0\) (must be an equation) A1 N1
[1 mark]
METHOD 1
attempt to substitute \(3\) into \(h\) (seen anywhere) (M1)
eg \(h(3),{\text{ }}\frac{5}{{3 + 2}}\)
correct equation (A1)
eg \(a = \frac{5}{{3 + 2}},{\text{ }}h(3) = a\)
\(a = 1\) A1 N2
[3 marks]
METHOD 2
attempt to find inverse (may be seen in (d)) (M1)
eg \(x = \frac{5}{{y + 2}},{\text{ }}{h^{ – 1}} = \frac{5}{x} – 2,{\text{ }}\frac{5}{x} + 2\)
correct equation, \(\frac{5}{x} – 2 = 3\) (A1)
\(a = 1\) A1 N2
[3 marks]
Question
The following diagram shows the graph of \(y = f(x)\), for \( – 4 \le x \le 5\).
Write down the value of \(f( – 3)\).[1]
Write down the value of \({f^{ – 1}}(1)\).[1]
Find the domain of \({f^{ – 1}}\).[2]
On the grid above, sketch the graph of \({f^{ – 1}}\).[3]
Answer/Explanation
Markscheme
\(f( – 3) = – 1\) A1 N1
[1 mark]
\({f^{ – 1}}(1) = 0\) (accept \(y = 0\)) A1 N1
[1 mark]
domain of \({f^{ – 1}}\) is range of \(f\) (R1)
eg \({\text{R}}f = {\text{D}}{f^{ – 1}}\)
correct answer A1 N2
eg \( – 3 \leqslant x \leqslant 3,{\text{ }}x \in [ – 3,{\text{ }}3]{\text{ (accept }} – 3 < x < 3,{\text{ }} – 3 \leqslant y \leqslant 3)\)
[2 marks]
A1A1 N2
Note: Graph must be approximately correct reflection in \(y = x\).
Only if the shape is approximately correct, award the following:
A1 for x-intercept at \(1\), and A1 for endpoints within circles.
[2 marks]