IB Math Analysis & Approaches Questionbank-Topic: SL 2.2 Inverse function SL Paper 1

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.
Answer/Explanation

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

Let f be the function given by \(f(x) = {{\rm{e}}^{0.5x}}\) , \(0 \le x \le 3.5\) . The diagram shows the graph of f .


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

[3]
a.

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

[1]
b.

Find \({f^{ – 1}}(x)\) .

[3]
c.
Answer/Explanation

Markscheme


  

   A1A1A1     N3

Note: Award A1 for approximately correct (reflected) shape, A1 for right end point in circle, A1 for through \((1{\text{, }}0)\) .

a.

\(0 \le y \le 3.5\)     A1     N1

[1 mark]

b.

interchanging x and y (seen anywhere)     M1

e.g. \(x = {e^{0.5y}}\)

evidence of changing to log form     A1

e.g. \(\ln x = 0.5y\) , \(\ln x = \ln {{\rm{e}}^{0.5y}}\) (any base), \(\ln x = 0.5y\ln {\rm{e}}\) (any base)

\({f^{ – 1}}(x) = 2\ln x\)     A1     N1

[3 marks]

c.

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.

Examiners report

Many candidates interchanged the \(x\) and \(y\) to find the inverse function, but very few could write down the correct domain of the inverse, often giving \(x \ge 0\) , \(x > 3\) and “all real numbers” as responses.

a.

Where students attempted to solve the equation in (b), most treated \(\ln x – 3\) as \(\ln (x – 3)\) and created an incorrect equation from the outset. The few who applied laws of logarithms often carried the algebra through to completion.

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 \(f(x) = \sqrt {x – 5} \) , 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

Consider \(f(x) = \ln ({x^4} + 1)\) .

The second derivative is given by \(f”(x) = \frac{{4{x^2}(3 – {x^4})}}{{{{({x^4} + 1)}^2}}}\) .

The equation \(f”(x) = 0\) has only three solutions, when \(x = 0\) , \( \pm \sqrt[4]{3}\) \(( \pm 1.316 \ldots )\) .

Find the value of \(f(0)\) .

[2]
a.

Find the set of values of \(x\) for which \(f\) is increasing.

[5]
b.

(i)     Find \(f”(1)\) .

(ii)     Hence, show that there is no point of inflexion on the graph of \(f\) at \(x = 0\) .

[5]
c.

There is a point of inflexion on the graph of \(f\) at \(x = \sqrt[4]{3}\) \((x = 1.316 \ldots )\) .

Sketch the graph of \(f\) , for \(x \ge 0\) .

[3]
d.
Answer/Explanation

Markscheme

substitute \(0\) into \(f\)     (M1)

eg   \(\ln (0 + 1)\) , \(\ln 1\)

\(f(0) = 0\)     A1 N2

[2 marks]

a.

\(f'(x) = \frac{1}{{{x^4} + 1}} \times 4{x^3}\) (seen anywhere)     A1A1

Note: Award A1 for \(\frac{1}{{{x^4} + 1}}\) and A1 for \(4{x^3}\) .

recognizing \(f\) increasing where \(f'(x) > 0\) (seen anywhere)     R1

eg   \(f'(x) > 0\) , diagram of signs

attempt to solve \(f'(x) > 0\)     (M1)

eg   \(4{x^3} = 0\) , \({x^3} > 0\)

\(f\) increasing for \(x > 0\) (accept \(x \ge 0\) )     A1     N1

[5 marks]

b.

(i)     substituting \(x = 1\) into \(f”\)     (A1)

eg   \(\frac{{4(3 – 1)}}{{{{(1 + 1)}^2}}}\) , \(\frac{{4 \times 2}}{4}\)

\(f”(1) = 2\)     A1     N2

(ii)     valid interpretation of point of inflexion (seen anywhere)     R1

eg   no change of sign in \(f”(x)\) , no change in concavity,

\(f’\) increasing both sides of zero

attempt to find \(f”(x)\) for \(x < 0\)     (M1)

eg   \(f”( – 1)\) , \(\frac{{4{{( – 1)}^2}(3 – {{( – 1)}^4})}}{{{{({{( – 1)}^4} + 1)}^2}}}\) , diagram of signs

correct working leading to positive value     A1

eg   \(f”( – 1) = 2\) , discussing signs of numerator and denominator

there is no point of inflexion at \(x = 0\)     AG     N0

[5 marks]

c.

     A1A1A1     N3

Notes: Award A1 for shape concave up left of POI and concave down right of POI.

    Only if this A1 is awarded, then award the following:

    A1 for curve through (\(0\), \(0\)) , A1 for increasing throughout.

    Sketch need not be drawn to scale. Only essential features need to be clear.

[3 marks]

d.

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

Consider the functions \(f(x)\) , \(g(x)\) and \(h(x)\) . The following table gives some values associated with these functions.


The following diagram shows parts of the graphs of \(h\) and \(h”\) .


There is a point of inflexion on the graph of \(h\) at P, when \(x = 3\) .

Given that \(h(x) = f(x) \times g(x)\) ,

Write down the value of \(g(3)\) , of \(f'(3)\) , and of \(h”(2)\) .

[3]
a.

Explain why P is a point of inflexion.

[2]
b.

find the \(y\)-coordinate of P.

[2]
c.

find the equation of the normal to the graph of \(h\) at P.

[7]
d.
Answer/Explanation

Markscheme

\(g(3) = – 18\) , \(f'(3) = 1\) , \(h”(2) = – 6\)     A1A1A1     N3

[3 marks]

a.

\(h”(3) = 0\)     (A1)

valid reasoning     R1

eg   \({h”}\) changes sign at \(x = 3\) , change in concavity of \(h\) at \(x = 3\)

so P is a point of inflexion     AG     N0

[2 marks]

b.

writing \(h(3)\) as a product of \(f(3)\) and \(g(3)\)     A1

eg   \(f(3) \times g(3)\) , \(3 \times ( – 18)\)

\(h(3) = – 54\)     A1 N1

[2 marks]

c.

recognizing need to find derivative of \(h\)     (R1)

eg   \({h’}\) , \(h'(3)\)

attempt to use the product rule (do not accept \(h’ = f’ \times g’\) )     (M1)

eg   \(h’ = fg’ + gf’\) ,  \(h'(3) = f(3) \times g'(3) + g(3) \times f'(3)\)

correct substitution     (A1)

eg   \(h'(3) = 3( – 3) + ( – 18) \times 1\)

\(h'(3) = – 27\)    A1

attempt to find the gradient of the normal     (M1)

eg   \( – \frac{1}{m}\) , \( – \frac{1}{{27}}x\) 

attempt to substitute their coordinates and their normal gradient into the equation of a line     (M1)

eg   \( – 54 = \frac{1}{{27}}(3) + b\) , \(0 = \frac{1}{{27}}(3) + b\) , \(y + 54 = 27(x – 3)\) , \(y – 54 = \frac{1}{{27}}(x + 3)\)

correct equation in any form     A1     N4

eg   \(y + 54 = \frac{1}{{27}}(x – 3)\) , \(y = \frac{1}{{27}}x – 54\frac{1}{9}\)

[7 marks]

d.

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.
Answer/Explanation

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

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

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

Let \(f\left( x \right) = \sqrt {x + 2} \) for x ≥ 2 and g(x) = 3x − 7 for \(x \in \mathbb{R}\).

Write down f (14).

[1]
a.

Find \(\left( {g \circ f} \right)\) (14).

[2]
b.

Find g−1(x).

[3]
c.
Answer/Explanation

Markscheme

f (14) = 4     A1 N1

[1 mark]

a.

attempt to substitute     (M1)

eg   g (4), 3 × 4 − 7

5     A1 N2

[2 marks]

b.

interchanging x and y (seen anywhere)     (M1)

eg   x = 3y − 7

evidence of correct manipulation     (A1)

eg   x + 7 = 3y

\({g^{ – 1}}\left( x \right) = \frac{{x + 7}}{3}\)     A1 N3

[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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