IB Math Analysis & Approaches Questionbank-Topic: SL 2.5 Identity function SL Paper 1

 

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Question

Let \(f(x) = \ln (x + 5) + \ln 2\) , for \(x > – 5\) .

Find \({f^{ – 1}}(x)\) .

[4]
a.

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]
b.
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]

a.

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]

b.

Question

Let \(f(x) = {{\rm{e}}^{x + 3}}\) .

(i)     Show that \({f^{ – 1}}(x) = \ln x – 3\) .

(ii)    Write down the domain of \({f^{ – 1}}\) .

[3]
a.

Solve the equation \({f^{ – 1}}(x) = \ln \frac{1}{x}\) .

[4]
b.
Answer/Explanation

Markscheme

(i) interchanging x and y (seen anywhere)     M1

e.g. \(x = {{\rm{e}}^{y + 3}}\)

correct manipulation     A1

e.g. \(\ln x = y + 3\) , \(\ln y = x + 3\)

\({f^{ – 1}}(x) = \ln x – 3\)     AG     N0

(ii) \(x > 0\)     A1     N1 

[3 marks]

a.

collecting like terms; using laws of logs     (A1)(A1)

e.g. \(\ln x – \ln \left( {\frac{1}{x}} \right) = 3\) , \(\ln x + \ln x = 3\) , \(\ln \left( {\frac{x}{{\frac{1}{x}}}} \right) = 3\) , \(\ln {x^2} = 3\)

simplify     (A1)

e.g. \(\ln x = \frac{3}{2}\) ,  \({x^2} = {{\rm{e}}^3}\)

\(x = {{\rm{e}}^{\frac{3}{2}}}\left( { = \sqrt {{{\rm{e}}^3}} } \right)\)     A1     N2

[4 marks]

b.

Question

Let \(f(x) = {x^2}\) and \(g(x) = 2x – 3\) .

Find \({g^{ – 1}}(x)\) .

[2]
a.

Find \((f \circ g)(4)\) .

[3]
b.
Answer/Explanation

Markscheme

for interchanging x and y (may be done later)     (M1)

e.g. \(x = 2y – 3\)

\({g^{ – 1}}(x) = \frac{{x + 3}}{2}\) (accept \(y = \frac{{x + 3}}{2},\frac{{x + 3}}{2}\) )     A1     N2

[2 marks]

a.

METHOD 1

\(g(4) = 5\)     (A1)

evidence of composition of functions     (M1)

\(f(5) = 25\)     A1 N3

METHOD 2

\(f \circ g(x) = {(2x – 3)^2}\)     (M1)

\(f \circ g(4) = {(2 \times 4 – 3)^2}\)     (A1)

= 25     A1     N3

[3 marks]

b.

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]
a.

Find \({f^{ – 1}}(x)\) .

[3]
b.
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]

a.

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]

b.

Question

Let \(f(x) = k{\log _2}x\) .

Given that \({f^{ – 1}}(1) = 8\) , find the value of \(k\) .

[3]
a.

Find \({f^{ – 1}}\left( {\frac{2}{3}} \right)\) .

[4]
b.
Answer/Explanation

Markscheme

METHOD 1

recognizing that \(f(8) = 1\)     (M1)

e.g. \(1 = k{\log _2}8\)

recognizing that \({\log _2}8 = 3\)     (A1)

e.g. \(1 = 3k\)

\(k = \frac{1}{3}\)     A1     N2

METHOD 2

attempt to find the inverse of \(f(x) = k{\log _2}x\)     (M1)

e.g. \(x = k{\log _2}y\) , \(y = {2^{\frac{x}{k}}}\)

substituting 1 and 8     (M1)

e.g. \(1 = k{\log _2}8\) , \({2^{\frac{1}{k}}} = 8\)

\(k = \frac{1}{{{{\log }_2}8}}\) \(\left( {k = \frac{1}{3}} \right)\)     A1     N2

[3 marks]

a.

METHOD 1

recognizing that \(f(x) = \frac{2}{3}\)     (M1)

e.g. \(\frac{2}{3} = \frac{1}{3}{\log _2}x\)

\({\log _2}x = 2\)     (A1)

\({f^{ – 1}}\left( {\frac{2}{3}} \right) = 4\) (accept \(x = 4\))     A2     N3

METHOD 2

attempt to find inverse of \(f(x) = \frac{1}{3}{\log _2}x\)     (M1)

e.g. interchanging x and y , substituting \(k = \frac{1}{3}\) into \(y = {2^{\frac{x}{k}}}\)

correct inverse     (A1)

e.g. \({f^{ – 1}}(x) = {2^{3x}}\) , \({2^{3x}}\)

\({f^{ – 1}}\left( {\frac{2}{3}} \right) = 4\)     A2    N3

[4 marks]

b.

Question

Let \(f(x) = lo{g_3}\sqrt x \) , for \(x > 0\) .

Show that \({f^{ – 1}}(x) = {3^{2x}}\) .

[2]
a.

Write down the range of \({f^{ – 1}}\) .

[1]
b.

Let \(g(x) = {\log _3}x\) , for \(x > 0\) .

Find the value of \(({f^{ – 1}} \circ g)(2)\) , giving your answer as an integer.

[4]
c.
Answer/Explanation

Markscheme

interchanging x and y (seen anywhere)     (M1)

e.g. \(x = \log \sqrt y \) (accept any base)

evidence of correct manipulation     A1

e.g. \(3^x = \sqrt y \) , \({3^y} = {x^{\frac{1}{2}}}\) , \(x = \frac{1}{2}{\log _3}y\) , \(2y = {\log _3}x\)

\({f^{ – 1}}(x) = {3^{2x}}\)     AG     N0 

[2 marks]

a.

\(y > 0\) , \({f^{ – 1}}(x) > 0\)     A1     N1

[1 mark]

b.

METHOD 1

finding \(g(2) = lo{g_3}2\) (seen anywhere)     A1

attempt to substitute     (M1)

e.g. \(({f^{ – 1}} \circ g)(2) = {3^{2\log {_3}2}}\)

evidence of using log or index rule     (A1)

e.g. \(({f^{ – 1}} \circ g)(2) = {3^{\log {_3}4}}\) , \({3^{{{\log }_3}2^2}}\)

\(({f^{ – 1}} \circ g)(2) = 4\)     A1     N1

METHOD 2

attempt to form composite (in any order)     (M1)

e.g. \(({f^{ – 1}} \circ g)(x) = {3^{2{{\log }_3}x}}\)

evidence of using log or index rule     (A1)

e.g. \(({f^{ – 1}} \circ g)(x) = {3^{{{\log }_3}{x^2}}}\) , \({3^{{{\log }_3}{x^{}}}}^2\)

\(({f^{ – 1}} \circ g)(x) = {x^2}\)     A1

\(({f^{ – 1}} \circ g)(2) = 4\)     A1     N1

[4 marks]

c.

Question

Let \(f(x) = 7 – 2x\) and \(g(x) = x + 3\) .

Find \((g \circ f)(x)\) .

[2]
a.

Write down \({g^{ – 1}}(x)\) .

[1]
b.

Find \((f \circ {g^{ – 1}})(5)\) .

[2]
c.
Answer/Explanation

Markscheme

attempt to form composite     (M1)

e.g. \(g(7 – 2x)\) , \(7 – 2x + 3\)

\((g \circ f)(x) = 10 – 2x\)     A1     N2

[2 marks]

a.

\({g^{ – 1}}(x) = x – 3\)     A1     N1

[1 mark]

b.

METHOD 1

valid approach     (M1)

e.g. \({g^{ – 1}}(5)\) , \(2\) , \(f(5)\)

\(f(2) = 3\)     A1     N2

METHOD 2

attempt to form composite of f and \({g^{ – 1}}\)     (M1)

e.g. \((f \circ {g^{ – 1}})(x) = 7 – 2(x – 3)\) , \(13 – 2x\)

\((f \circ {g^{ – 1}})(5) = 3\)     A1     N2

[2 marks]

c.

Question

Let \(f(x) = {\log _p}(x + 3)\) for \(x >  – 3\) . Part of the graph of f is shown below.


The graph passes through A(6, 2) , has an x-intercept at (−2, 0) and has an asymptote at \(x =  – 3\) .

Find p .

[4]
a.

The graph of f is reflected in the line \(y = x\) to give the graph of g .

(i)     Write down the y-intercept of the graph of g .

(ii)    Sketch the graph of g , noting clearly any asymptotes and the image of A.

[5]
b.

The graph of \(f\) is reflected in the line \(y = x\) to give the graph of \(g\) .

Find \(g(x)\) .

[4]
c.
Answer/Explanation

Markscheme

evidence of substituting the point A     (M1)

e.g. \(2 = {\log _p}(6 + 3)\)

manipulating logs     A1

e.g. \({p^2} = 9\)

\(p = 3\)     A2     N2

[4 marks]

a.

(i) \(y = – 2\) (accept \((0{\text{, }} – 2))\)     A1     N1

(ii)


     A1A1A1A1     N4

Note: Award A1 for asymptote at \(y = – 3\) , A1 for an increasing function that is concave up, A1 for a positive x-intercept and a negative y-intercept, A1 for passing through the point \((2{\text{, }}6)\) .

[5 marks] 

b.

METHOD 1

recognizing that \(g = {f^{ – 1}}\)     (R1)

evidence of valid approach     (M1)

e.g. switching x and y (seen anywhere), solving for x

correct manipulation     (A1)

e.g. \({3^x} = y + 3\)

\(g(x) = {3^x} – 3\)     A1     N3

METHOD 2

recognizing that \(g(x) = {a^x} + b\)     (R1) 

identifying vertical translation     (A1)

e.g. graph shifted down 3 units, \(f(x) – 3\)

evidence of valid approach     (M1)

e.g. substituting point to identify the base

\(g(x) = {3^x} – 3\)     A1     N3

[4 marks]

c.

Question

Let \(f(x) = 2x – 1\) and  \(g(x) = 3{x^2} + 2\) .

Find \({f^{ – 1}}(x)\) . 

[3]
a.

Find \((f \circ g)(1)\) . 

[3]
b.
Answer/Explanation

Markscheme

interchanging x and y (seen anywhere)     (M1)

e.g. \(x = 2y – 1\)

correct manipulation     (A1)

e.g. \(x + 1 = 2y\)

\({f^{ – 1}}(x) = \frac{{x + 1}}{2}\)      A1     N2

[3 marks]

a.

METHOD 1

attempt to find or \(g(1)\) or \(f(1)\)     (M1)

\(g(1) = 5\)     (A1)

\(f(5) = 9\)     A1     N2 

[3 marks]

METHOD 2

attempt to form composite (in any order)     (M1)

e.g. \(2(3{x^2} + 2) – 1\) , \(3{(2x – 1)^2} + 2\)

\((f \circ g)(1) = 2(3 \times {1^2} + 2) – 1\) \(( = 6 \times {1^2} + 3)\)     (A1)

\((f \circ g)(1) = 9\)     A1     N2

[3 marks]

b

Question

Let \(f(x) = \sqrt {x – 5} \) , for \(x \ge 5\) .

Find \({f^{ – 1}}(2)\) .

[3]
a.

Let \(g\) be a function such that \({g^{ – 1}}\) exists for all real numbers. Given that \(g(30) = 3\) , find \((f \circ {g^{ – 1}})(3)\)  .

[3]
b.
Answer/Explanation

Markscheme

METHOD 1

attempt to set up equation     (M1)

eg   \(2 = \sqrt {y – 5} \) , \(2 = \sqrt {x – 5} \)

correct working     (A1)

eg   \(4 = y – 5\) , \(x = {2^2} + 5\)

\({f^{ – 1}}(2) = 9\)     A1     N2

METHOD 2

interchanging \(x\) and \(y\) (seen anywhere)     (M1)

eg   \(x = \sqrt {y – 5} \)

correct working     (A1)

eg   \({x^2} = y – 5\) , \(y = {x^2} + 5\)

\({f^{ – 1}}(2) = 9\)     A1     N2

[3 marks]

a.

recognizing \({g^{ – 1}}(3) = 30\)     (M1)

eg   \(f(30)\)

correct working     (A1)

eg   \((f \circ {g^{ – 1}})(3) = \sqrt {30 – 5} \) , \(\sqrt {25} \)

\((f \circ {g^{ – 1}})(3) = 5\)     A1     N2

Note: Award A0 for multiple values, eg \( \pm 5\) .

[3 marks]

b.

Question

Let \(f(x) = 4x – 2\) and \(g(x) = – 2{x^2} + 8\) .

Find \({f^{ – 1}}(x)\) .

[3]
a.

Find \((f \circ g)(1)\) .

[3]
b.
Answer/Explanation

Markscheme

interchanging \(x\) and \(y\) (seen anywhere)     (M1)

eg   \(x = 4y – 2\)

evidence of correct manipulation     (A1)

eg   \(x + 2 = 4y\)

\({f^{ – 1}}(x) = \frac{{x + 2}}{4}\) (accept \(y = \frac{{x + 2}}{4}\) , \(\frac{{x + 2}}{4}\) , \({f^{ – 1}}(x) = \frac{1}{4}x + \frac{1}{2}\)     A1     N2

[3 marks]

a.

METHOD 1

attempt to substitute \(1\) into \(g(x)\)     (M1)

eg   \(g(1) =  – 2 \times {1^2} + 8\)

\(g(1) = 6\)     (A1)

\(f(6) = 22\)     A1     N3

METHOD 2

attempt to form composite function (in any order)     (M1)

eg   \((f \circ g)(x) = 4( – 2{x^2} + 8) – 2\) \(( =  – 8{x^2} + 30)\)

correct substitution

eg   \((f \circ g)(1) = 4( – 2 \times {1^2} + 8) – 2\) , \( – 8 + 30\)

\(f(6) = 22\)     A1     N3

[3 marks]

b.

Question

The diagram below shows the graph of a function \(f\) , for \( – 1 \le x \le 2\) .


Write down the value of \(f(2)\).

[1]
a.i.

Write down the value of \({f^{ – 1}}( – 1)\) .

[2]
a.ii.

Sketch the graph of \({f^{ – 1}}\) on the grid below.


[3]
b.
Answer/Explanation

Markscheme

\(f(2) = 3\)     A1     N1

[1 mark]

a.i.

\({f^{ – 1}}( – 1) = 0\)     A2     N2

[2 marks]

a.ii.

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]

b.

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 \({f^{ – 1}}(x)\).

[2]
a.

Show that \(\left( {g \circ {f^{ – 1}}} \right)(x) = \frac{5}{{x + 2}}\).

[2]
b.

Find the \(y\)-intercept of the graph of \(h\).

[2]
c(i).

Hence, sketch the graph of \(h\).

[3]
c(ii).

For the graph of \({h^{ – 1}}\), write down the \(x\)-intercept;

[1]
d(i).

For the graph of \({h^{ – 1}}\), write down the equation of the vertical asymptote.

[1]
d(ii).

Given that \({h^{ – 1}}(a) = 3\), find the value of \(a\).

[3]
e.
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]

a.

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]

b.

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]

c(i).

   

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]

c(ii).

\(x = \frac{5}{2}{\text{   }}\left( {{\text{accept (2.5, 0)}}} \right)\)     A1     N1

[1 mark]

d(i).

\(x = 0\)   (must be an equation)     A1     N1

[1 mark]

d(ii).

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]

e.

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]
a(i).

Write down the value of  \({f^{ – 1}}(1)\).

[1]
a(ii).

Find the domain of \({f^{ – 1}}\).

[2]
b.

On the grid above, sketch the graph of \({f^{ – 1}}\).

[3]
c.

Markscheme

\(f( – 3) =  – 1\)     A1     N1

[1 mark]

a(i).

\({f^{ – 1}}(1) = 0\)   (accept \(y = 0\))     A1     N1

[1 mark]

a(ii).

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]

b.


     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]

c.

Question

The following diagram shows the graph of a function \(f\).

Find \({f^{ – 1}}( – 1)\).

[2]
a.

Find \((f \circ f)( – 1)\).

[3]
b.

On the same diagram, sketch the graph of \(y = f( – x)\).

[2]
c.
Answer/Explanation

Markscheme

valid approach     (M1)

eg\(\;\;\;\)horizontal line on graph at \( – 1,{\text{ }}f(a) =  – 1,{\text{ }}( – 1,5)\)

\({f^{ – 1}}( – 1) = 5\)     A1     N2

[2 marks]

a.

attempt to find \(f( – 1)\)     (M1)

eg\(\;\;\;\)line on graph

\(f( – 1) = 2\)     (A1)

\((f \circ f)( – 1) = 1\)     A1     N3

[3 marks]

b.

     A1A1     N2

Note:     The shape must be an approximately correct shape (concave down and increasing). Only if the shape is approximately correct, award the following for points in circles:

A1 for the \(y\)-intercept,

A1 for any two of these points \(( – 5,{\text{ }} – 1),{\text{ }}( – 2,{\text{ }}1),{\text{ }}(1,{\text{ }}2)\).

[2 marks]

Total [7 marks]

c.

Question

Let \(f(x) = {(x – 5)^3}\), for \(x \in \mathbb{R}\).

Find \({f^{ – 1}}(x)\).

[3]
a.

Let \(g\) be a function so that \((f \circ g)(x) = 8{x^6}\). Find \(g(x)\).

[3]
b.
Answer/Explanation

Markscheme

interchanging \(x\) and \(y\) (seen anywhere)     (M1)

eg\(\;\;\;x = {(y – 5)^3}\)

evidence of correct manipulation     (A1)

eg\(\;\;\;y – 5 = \sqrt[3]{x}\)

\({f^{ – 1}}(x) = \sqrt[3]{x} + 5\;\;\;({\text{accept }}5 + {x^{\frac{1}{3}}},{\text{ }}y = 5 + \sqrt[3]{x})\)     A1     N2

Notes:     If working shown, and they do not interchange \(x\) and \(y\), award A1A1M0 for \(\sqrt[3]{y} + 5\).

If no working shown, award N1 for \(\sqrt[3]{y} + 5\).

a.

METHOD 1

attempt to form composite (in any order)     (M1)

eg\(\;\;\;g\left( {{{(x – 5)}^3}} \right),{\text{ }}{\left( {g(x) – 5} \right)^3} = 8{x^6},{\text{ }}f(2{x^2} + 5)\)

correct working     (A1)

eg\(\;\;\;g – 5 = 2{x^2},{\text{ }}{\left( {(2{x^2} + 5) – 5} \right)^3}\)

\(g(x) = 2{x^2} + 5\)     A1     N2

METHOD 2

recognising inverse relationship     (M1)

eg\(\;\;\;{f^{ – 1}}(8{x^6}) = g(x),{\text{ }}{f^{ – 1}}(f \circ g)(x) = {f^{ – 1}}(8{x^6})\)

correct working

eg\(\;\;\;g(x) = \sqrt[3]{{(8{x^6})}} + 5\)     (A1)

\(g(x) = 2{x^2} + 5\)     A1     N2

b.

Question

Let \(f(x) = 8x + 3\) and \(g(x) = 4x\), for \(x \in \mathbb{R}\).

Write down \(g(2)\).

[1]
a.

Find \((f \circ g)(x)\).

[2]
b.

Find \({f^{ – 1}}(x)\).

[2]
c.
Answer/Explanation

Markscheme

\(g(2) = 8\)    A1     N1

[1 mark]

a.

attempt to form composite (in any order)     (M1)

eg\(\,\,\,\,\,\)\(f(4x),{\text{ }}4 \times (8x + 3)\)

\((f \circ g)(x) = 32x + 3\)     A1     N2

[2 marks]

b.

interchanging \(x\) and \(y\) (may be seen at any time)     (M1)

eg\(\,\,\,\,\,\)\(x = 8y + 3\)

\({f^{ – 1}}(x) = \frac{{x – 3}}{8}\,\,\,\,\,\left( {{\text{accept }}\frac{{x – 3}}{8},{\text{ }}y = \frac{{x – 3}}{8}} \right)\)     A1     N2

[2 marks]

c.

Question

Let \(f(x) = 5x\) and \(g(x) = {x^2} + 1\), for \(x \in \mathbb{R}\).

Find \({f^{ – 1}}(x)\).

[2]
a.

Find \((f \circ g)(7)\).

[3]
b.
Answer/Explanation

Markscheme

interchanging \(x\) and \(x\)     (M1)

eg\(\,\,\,\,\,\)\(x = 5y\)

\({f^{ – 1}}\left( x \right) = \frac{x}{5}\)     A1     N2

[2 marks]

a.

METHOD 1

attempt to substitute 7 into \(g(x)\) or \(f(x)\)     (M1)

eg\(\,\,\,\,\,\)\({7^2} + 1,{\text{ }}5 \times 7\)

\(g(7) = 50\)     (A1)

\(f\left( {50} \right) = 250\)     A1     N2

METHOD 2

attempt to form composite function (in any order)     (M1)

eg\(\,\,\,\,\,\)\(5({x^2} + 1),{\text{ }}{(5x)^2} + 1\)

correct substitution     (A1)

eg\(\,\,\,\,\,\)\(5 \times ({7^2} + 1)\)

\((f \circ g)(7) = 250\)     A1     N2

[3 marks]

b.

Question

The following diagram shows the graph of a function \(f\), with domain \( – 2 \leqslant x \leqslant 4\).

N17/5/MATME/SP1/ENG/TZ0/03

The points \(( – 2,{\text{ }}0)\) and \((4,{\text{ }}7)\) lie on the graph of \(f\).

Write down the range of \(f\).

[1]
a.

Write down \(f(2)\);

[1]
b.i.

Write down \({f^{ – 1}}(2)\).

[1]
b.ii.

On the grid, sketch the graph of \({f^{ – 1}}\).

[3]
c.
Answer/Explanation

Markscheme

correct range (do not accept \(0 \leqslant x \leqslant 7\))     A1     N1

eg\(\,\,\,\,\,\)\([0,{\text{ }}7],{\text{ }}0 \leqslant y \leqslant 7\)

[1 mark]

a.

\(f(2) = 3\)     A1     N1

[1 mark]

b.i.

\({f^{ – 1}}(2) = 0\)     A1     N1

[1 mark]

b.ii.

N17/5/MATME/SP1/ENG/TZ0/03.c/M     A1A1A1     N3

Notes:     Award A1 for both end points within circles,

A1 for images of \((2,{\text{ }}3)\) and \((0,{\text{ }}2)\) within circles,

A1 for approximately correct reflection in \(y = x\), concave up then concave down shape (do not accept line segments).

[3 marks]

c.

Question

Consider a function f (x) , for −2 ≤ x ≤ 2 . The following diagram shows the graph of f.

Write down the value of f (0).

[1]
a.i.

Write down the value of f −1 (1).

[1]
a.ii.

Write down the range of f −1.

[1]
b.

On the grid above, sketch the graph of f −1.

[4]
c.
Answer/Explanation

Markscheme

\(f\left( 0 \right) =  – \frac{1}{2}\)     A1 N1

[1 mark]

a.i.

f −1 (1) = 2     A1 N1

[1 mark]

a.ii.

−2 ≤ y ≤ 2, y∈ [−2, 2]  (accept −2 ≤ x ≤ 2)     A1 N1

[1 mark]

b.

A1A1A1A1  N4

Note: Award A1 for evidence of approximately correct reflection in y = x with correct curvature.

(y = x does not need to be explicitly seen)

Only if this mark is awarded, award marks as follows:

A1 for both correct invariant points in circles,

A1 for the three other points in circles,

A1 for correct domain.

[4 marks]

c.

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