Question
The function f is defined by f(x) = \(\frac{7x + 7}{2x – 4}\) for \(x\epsilon \mathbb{R}\), x ≠ 2.
(a) Find the zero of f(x).
(b) For the graph of y = f(x), write down the equation of
(i) the vertical asymptote;
(ii) the horizontal asymptote.
(c) Find \(f^{-1}\) (x), the inverse function of f(x).
Answer/Explanation
Answer:
(a) recognizing f(x) = 0
x = -1
(b) (i) x = 2 (must be an equation with x)
(ii) y = \(\frac{7}{2}\) (must be an equation with y)
(c) EITHER
interchanging x and y
2xy – 4x = 7y + 7
correct working with y terms on the same side: 2xy – 7y = 4x + 7
OR
2yx – 4y = 7x +7
correct working with x terms on the same side: 2yx – 7x = 4x + 7
interchanging x and y OR making x the subject x = \(\frac{4y + 7}{2y – 7}\)
THEN
\(f^{-1}\)(x) = \(\frac{4x + 7}{2x – 7}\) (or equivalent) (\(x ≠ \frac{7}{2}\))
Question
Let \(f(x) = \ln (x + 5) + \ln 2\) , for \(x > – 5\) .
Find \({f^{ – 1}}(x)\) .[4]
Let \(g(x) = {{\rm{e}}^x}\) .
Find \((g \circ f)(x)\) , giving your answer in the form \(ax + b\) , where \(a,b \in \mathbb{Z}\) .[3]
Answer/Explanation
Markscheme
METHOD 1
\(\ln (x + 5) + \ln 2 = \ln (2(x + 5))\) \(( = \ln (2x + 10))\) (A1)
interchanging x and y (seen anywhere) (M1)
e.g. \(x = \ln (2y + 10)\)
evidence of correct manipulation (A1)
e.g. \({{\rm{e}}^x} = 2y + 10\)
\({f^{ – 1}}(x) = \frac{{{{\rm{e}}^x} – 10}}{2}\) A1 N2
METHOD 2
\(y = \ln (x + 5) + \ln 2\)
\(y – \ln 2 = ln(x + 5)\) (A1)
evidence of correct manipulation (A1)
e.g. \({{\rm{e}}^{y – \ln 2}} = x + 5\)
interchanging x and y (seen anywhere) (M1)
e.g. \({{\rm{e}}^{x – \ln 2}} = y + 5\)
\({f^{ – 1}}(x) = {{\rm{e}}^{x – \ln 2}} – 5\) A1 N2
[4 marks]
METHOD 1
evidence of composition in correct order (M1)
e.g. \((g \circ f)(x) = g(\ln (x + 5) + \ln 2)\)
\( = {{\rm{e}}^{\ln (2(x + 5))}} = 2(x + 5)\)
\((g \circ f)(x) = 2x + 10\) A1A1 N2
METHOD 2
evidence of composition in correct order (M1)
e.g. \((g \circ f)(x) = {{\rm{e}}^{\ln (x + 5) + \ln 2}}\)
\( = {{\rm{e}}^{\ln (x + 5)}} \times {{\rm{e}}^{\ln 2}} = (x + 5)2\)
\((g \circ f)(x) = 2x + 10\) A1A1 N2
[3 marks]
Question
Let \(f(x) = 2{x^3} + 3\) and \(g(x) = {{\rm{e}}^{3x}} – 2\) .
(i) Find \(g(0)\) .
(ii) Find \((f \circ g)(0)\) .[5]
Find \({f^{ – 1}}(x)\) .[3]
Answer/Explanation
Markscheme
(i) \(g(0) = {{\rm{e}}^0} – 2\) (A1)
\( = – 1\) A1 N2
(ii) METHOD 1
substituting answer from (i) (M1)
e.g. \((f \circ g)(0) = f( – 1)\)
correct substitution \(f( – 1) = 2{( – 1)^3} + 3\) (A1)
\(f( – 1) = 1\) A1 N3
METHOD 2
attempt to find \((f \circ g)(x)\) (M1)
e.g. \((f \circ g)(x) = f({{\rm{e}}^{3x}} – 2)\) \( = 2{({{\rm{e}}^{3x}} – 2)^3} + 3\)
correct expression for \((f \circ g)(x)\) (A1)
e.g. \(2{({{\rm{e}}^{3x}} – 2)^3} + 3\)
\((f \circ g)(0) = 1\) A1 N3
[5 marks]
interchanging x and y (seen anywhere) (M1)
e.g. \(x = 2{y^3} + 3\)
attempt to solve (M1)
e.g. \({y^3} = \frac{{x – 3}}{2}\)
\({f^{ – 1}}(x) = \sqrt[3]{{\frac{{x – 3}}{2}}}\) A1 N3
[3 marks]