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

## Question

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

Find $${f^{ – 1}}(x)$$ .

[4]
a.

Let $$g(x) = {{\rm{e}}^x}$$ .

FindÂ $$(g \circ f)(x)$$ , giving your answer in the form $$ax + b$$ , where $$a,b \in \mathbb{Z}$$ .

[3]
b.

## Markscheme

METHOD 1

$$\ln (x + 5) + \ln 2 = \ln (2(x + 5))$$ $$( = \ln (2x + 10))$$Â Â Â  Â (A1)

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

e.g. $$x = \ln (2y + 10)$$

evidence of correct manipulationÂ Â Â Â  (A1)

e.g. $${{\rm{e}}^x} = 2y + 10$$

$${f^{ – 1}}(x) = \frac{{{{\rm{e}}^x} – 10}}{2}$$Â Â Â  A1Â Â Â Â  N2

METHOD 2

$$y = \ln (x + 5) + \ln 2$$

$$y – \ln 2 = ln(x + 5)$$Â Â Â Â Â (A1)

evidence of correct manipulationÂ Â Â Â  (A1)

e.g. $${{\rm{e}}^{y – \ln 2}} = x + 5$$

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

e.g. $${{\rm{e}}^{x – \ln 2}} = y + 5$$

$${f^{ – 1}}(x) = {{\rm{e}}^{x – \ln 2}} – 5$$Â Â Â  Â A1Â Â Â Â  N2

[4 marks]

a.

METHOD 1

evidence of composition in correct orderÂ Â Â Â  (M1)

e.g. $$(g \circ f)(x) = g(\ln (x + 5) + \ln 2)$$

$$= {{\rm{e}}^{\ln (2(x + 5))}} = 2(x + 5)$$

$$(g \circ f)(x) = 2x + 10$$Â Â Â  Â A1A1Â Â Â Â  N2

METHOD 2

evidence of composition in correct orderÂ Â Â Â  (M1)

e.g. $$(g \circ f)(x) = {{\rm{e}}^{\ln (x + 5) + \ln 2}}$$

$$= {{\rm{e}}^{\ln (x + 5)}} \times {{\rm{e}}^{\ln 2}} = (x + 5)2$$

$$(g \circ f)(x) = 2x + 10$$Â Â Â  Â A1A1 Â  Â  N2

[3 marks]

b.

## Question

LetÂ $$f(x) = {{\rm{e}}^{x + 3}}$$ .

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

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

[3]
a.

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

[4]
b.

## Markscheme

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

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

correct manipulationÂ Â Â Â  A1

e.g. $$\ln x = y + 3$$ , $$\ln y = x + 3$$

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

(ii) $$x > 0$$Â Â Â Â Â A1 Â  Â  N1Â

[3 marks]

a.

collecting like terms; using laws of logsÂ Â Â Â  (A1)(A1)

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

simplifyÂ Â Â Â  (A1)

e.g.Â $$\ln x = \frac{3}{2}$$ , Â $${x^2} = {{\rm{e}}^3}$$

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

[4 marks]

b.

## Question

LetÂ $$f(x) = {x^2}$$ and $$g(x) = 2x – 3$$Â .

FindÂ $${g^{ – 1}}(x)$$ .

[2]
a.

FindÂ $$(f \circ g)(4)$$ .

[3]
b.

## Markscheme

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

e.g. $$x = 2y – 3$$

$${g^{ – 1}}(x) = \frac{{x + 3}}{2}$$ (accept $$y = \frac{{x + 3}}{2},\frac{{x + 3}}{2}$$ )Â Â Â  Â A1Â Â Â Â  N2

[2 marks]

a.

METHOD 1

$$g(4) = 5$$Â Â Â Â  (A1)

evidence of composition of functionsÂ Â Â Â  (M1)

$$f(5) = 25$$Â Â Â Â  A1 N3

METHOD 2

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

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

= 25Â Â Â Â  A1Â Â Â Â  N3

[3 marks]

b.

## Question

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

(i)Â Â Â Â  Find $$g(0)$$ .

(ii)Â Â Â  Find $$(f \circ g)(0)$$ .

[5]
a.

Find $${f^{ – 1}}(x)$$ .

[3]
b.

## Markscheme

(i) $$g(0) = {{\rm{e}}^0} – 2$$Â Â Â Â Â (A1)

$$= – 1$$Â Â Â Â Â A1Â Â Â Â  N2

(ii) METHOD 1

substituting answer from (i)Â Â Â Â  (M1)

e.g. $$(f \circ g)(0) = f( – 1)$$

correct substitution $$f( – 1) = 2{( – 1)^3} + 3$$Â Â Â Â Â (A1)

$$f( – 1) = 1$$Â Â Â Â  A1Â Â Â Â  N3

METHOD 2

attempt to find $$(f \circ g)(x)$$Â Â Â Â Â (M1)

e.g. $$(f \circ g)(x) = f({{\rm{e}}^{3x}} – 2)$$ $$= 2{({{\rm{e}}^{3x}} – 2)^3} + 3$$

correct expression for $$(f \circ g)(x)$$Â Â Â  Â (A1)

e.g. $$2{({{\rm{e}}^{3x}} – 2)^3} + 3$$

$$(f \circ g)(0) = 1$$Â Â Â Â  A1Â Â Â Â  N3

[5 marks]

a.

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

e.g. $$x = 2{y^3} + 3$$

attempt to solveÂ Â Â Â  (M1)

e.g. $${y^3} = \frac{{x – 3}}{2}$$

$${f^{ – 1}}(x) = \sqrt[3]{{\frac{{x – 3}}{2}}}$$Â Â Â Â  A1Â Â Â Â  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.

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

[4]
b.

## Markscheme

METHOD 1

recognizing that $$f(8) = 1$$Â Â Â Â Â (M1)

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

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

e.g. $$1 = 3k$$

$$k = \frac{1}{3}$$Â Â Â Â  A1Â Â Â Â  N2

METHOD 2

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

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

substituting 1 and 8Â Â Â Â  (M1)

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

$$k = \frac{1}{{{{\log }_2}8}}$$Â $$\left( {k = \frac{1}{3}} \right)$$Â Â Â Â Â A1Â Â Â Â  N2

[3 marks]

a.

METHOD 1

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

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

$${\log _2}x = 2$$Â Â Â Â  (A1)

$${f^{ – 1}}\left( {\frac{2}{3}} \right) = 4$$ (accept $$x = 4$$)Â Â Â Â  A2Â Â Â Â  N3

METHOD 2

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

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

correct inverseÂ Â Â Â  (A1)

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

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

## Markscheme

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

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

evidence of correct manipulationÂ Â Â Â  A1

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

$${f^{ – 1}}(x) = {3^{2x}}$$Â Â Â Â  AG Â  Â  N0Â

[2 marks]

a.

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

[1 mark]

b.

METHOD 1

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

attempt to substituteÂ Â Â Â  (M1)

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

evidence of using log or index ruleÂ Â Â Â  (A1)

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

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

METHOD 2

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

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

evidence of using log or index ruleÂ Â Â Â  (A1)

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

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

$$({f^{ – 1}} \circ g)(2) = 4$$Â Â Â Â  A1Â Â Â Â  N1

[4 marks]

c.

## Question

LetÂ $$f(x) = 7 – 2x$$ and $$g(x) = x + 3$$Â .

FindÂ $$(g \circ f)(x)$$ .

[2]
a.

Write down $${g^{ – 1}}(x)$$ .

[1]
b.

Find $$(f \circ {g^{ – 1}})(5)$$ .

[2]
c.

## Markscheme

attempt to form compositeÂ Â Â Â  (M1)

e.g. $$g(7 – 2x)$$ , $$7 – 2x + 3$$

$$(g \circ f)(x) = 10 – 2x$$Â Â Â Â Â A1Â Â Â Â  N2

[2 marks]

a.

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

[1 mark]

b.

METHOD 1

valid approachÂ Â Â Â  (M1)

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

$$f(2) = 3$$Â Â Â Â Â A1 Â  Â  N2

METHOD 2

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

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

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

[2 marks]

c.

## Question

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

The graph passes through A(6, 2) , has an x-intercept at (âˆ’2, 0) and has an asymptote at $$x =Â – 3$$ .

Find p .

[4]
a.

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

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

(ii) Â Â  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.

## Markscheme

evidence of substituting the point AÂ Â Â Â  (M1)

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

manipulating logsÂ Â Â Â  A1

e.g. $${p^2} = 9$$

$$p = 3$$Â Â Â Â  A2Â Â Â Â  N2

[4 marks]

a.

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

(ii)

Â Â Â Â  A1A1A1A1Â Â Â Â  N4

Note: Award A1 for asymptote at $$y = – 3$$Â , 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]Â

b.

METHOD 1

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

evidence of valid approachÂ Â Â Â  (M1)

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

correct manipulationÂ Â Â Â  (A1)

e.g. $${3^x} = y + 3$$

$$g(x) = {3^x} – 3$$Â Â Â Â  A1Â Â Â Â  N3

METHOD 2

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

identifying vertical translationÂ Â Â Â  (A1)

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

evidence of valid approachÂ Â Â Â  (M1)

e.g. substituting point to identify the base

$$g(x) = {3^x} – 3$$Â Â Â Â  A1 Â  Â  N3

[4 marks]

c.

## Question

LetÂ $$f(x) = 2x – 1$$ and Â $$g(x) = 3{x^2} + 2$$ .

FindÂ $${f^{ – 1}}(x)$$ .Â

[3]
a.

FindÂ $$(f \circ g)(1)$$ .Â

[3]
b.

## Markscheme

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

e.g. $$x = 2y – 1$$

correct manipulationÂ Â Â Â  (A1)

e.g. $$x + 1 = 2y$$

$${f^{ – 1}}(x) = \frac{{x + 1}}{2}$$Â Â Â  Â  A1Â Â Â Â  N2

[3 marks]

a.

METHOD 1

attempt to find orÂ $$g(1)$$ or $$f(1)$$Â Â Â  Â (M1)

$$g(1) = 5$$Â Â Â  Â (A1)

$$f(5) = 9$$Â Â Â  Â A1 Â  Â  N2Â

[3 marks]

METHOD 2

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

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

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

$$(f \circ g)(1) = 9$$Â Â Â  Â A1Â Â Â Â  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 $$(f \circ {g^{ – 1}})(3)$$Â  .

[3]
b.

## Markscheme

METHOD 1

attempt to set up equationÂ Â Â Â  (M1)

eg Â  $$2 = \sqrt {y – 5}$$ , $$2 = \sqrt {x – 5}$$

correct workingÂ Â Â Â  (A1)

eg Â  $$4 = y – 5$$ , $$x = {2^2} + 5$$

$${f^{ – 1}}(2) = 9$$Â Â Â Â  A1Â Â Â Â  N2

METHOD 2

interchanging $$x$$ and $$y$$ (seen anywhere)Â Â Â Â  (M1)

eg Â  $$x = \sqrt {y – 5}$$

correct workingÂ Â Â Â  (A1)

eg Â  $${x^2} = y – 5$$ , $$y = {x^2} + 5$$

$${f^{ – 1}}(2) = 9$$Â Â Â Â  A1 Â  Â  N2

[3 marks]

a.

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

egÂ Â  $$f(30)$$

correct workingÂ Â Â Â  (A1)

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

$$(f \circ {g^{ – 1}})(3) = 5$$Â Â Â Â  A1Â Â Â Â  N2

Note: Award A0 for multiple values, eg $$\pm 5$$Â .

[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.

## Markscheme

interchanging $$x$$ and $$y$$ (seen anywhere)Â Â Â Â  (M1)

eg Â  $$x = 4y – 2$$

evidence of correct manipulationÂ Â Â Â  (A1)

eg Â  $$x + 2 = 4y$$

$${f^{ – 1}}(x) = \frac{{x + 2}}{4}$$ (accept $$y = \frac{{x + 2}}{4}$$ , $$\frac{{x + 2}}{4}$$ , $${f^{ – 1}}(x) = \frac{1}{4}x + \frac{1}{2}$$Â Â Â Â  A1Â Â Â Â  N2

[3 marks]

a.

METHOD 1

attempt to substitute $$1$$ into $$g(x)$$Â Â Â Â Â (M1)

eg Â  $$g(1) =Â – 2 \times {1^2} + 8$$

$$g(1) = 6$$Â Â Â Â  (A1)

$$f(6) = 22$$Â Â Â Â  A1Â Â Â Â  N3

METHOD 2

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

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

correct substitution

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

$$f(6) = 22$$Â Â Â Â  A1 Â  Â  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.

## Markscheme

$$f(2) = 3$$Â Â Â  Â A1Â Â Â Â  N1

[1 mark]

a.i.

$${f^{ – 1}}( – 1) = 0$$Â Â Â Â Â A2Â Â Â Â  N2

[2 marks]

a.ii.

EITHER

attempt to draw $$y = x$$Â on gridÂ Â Â Â  (M1)

OR

attempt to reverse x and y coordinatesÂ Â Â Â  (M1)

eg Â  writing or plotting at least two of the points

$$( – 2, – 1)$$ , $$( – 1,0)$$ , $$(0,1)$$ , $$(3,2)$$

THEN

correct graphÂ Â Â Â  A2Â Â Â Â  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.

## Markscheme

interchanging $$x$$ and $$y$$ Â  Â  (M1)

eg Â  Â  $$x = 3y – 2$$

$${f^{ – 1}}(x) = \frac{{x + 2}}{3}{\text{ Â }}\left( {{\text{accept }}y = \frac{{x + 2}}{3},{\text{ }}\frac{{x + 2}}{3}} \right)$$ Â  Â  A1 Â  Â  N2

[2 marks]

a.

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

eg Â  Â  $$g\left( {\frac{{x + 2}}{3}} \right),{\text{ }}\frac{{\frac{5}{{3x}} + 2}}{3}$$

correct substitution Â  Â  A1

eg Â  Â  $$\frac{5}{{3\left( {\frac{{x + 2}}{3}} \right)}}$$

$$\left( {g \circ {f^{ – 1}}} \right)(x) = \frac{5}{{x + 2}}$$ Â  Â  AG Â  Â  N0

[2 marks]

b.

valid approach Â  Â  (M1)

eg Â  Â  $$h(0),{\text{ }}\frac{5}{{0 + 2}}$$

$$y = \frac{5}{2}{\text{ Â }}\left( {{\text{accept (0, 2.5)}}} \right)$$ Â  Â  A1 Â  Â  N2

[2 marks]

c(i).

Â  Â

A1A2 Â  Â  N3

Notes: Â  Â Â AwardÂ A1Â for approximately correct shape (reciprocal, decreasing, concave up).

Â Â  Â  OnlyÂ if thisÂ A1Â is awarded, awardÂ A2Â for all the following approximately correct features:Â y-intercept at $$(0, 2.5)$$, asymptotic toÂ x-axis, correct domain $$x \geqslant 0$$.

Â Â  Â  If only two of these features are correct, awardÂ A1.

[3 marks]

c(ii).

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

[1 mark]

d(i).

$$x = 0$$ Â  (must be an equation) Â  Â  A1 Â  Â  N1

[1 mark]

d(ii).

METHOD 1

attempt to substitute $$3$$ into $$h$$ (seen anywhere) Â  Â  (M1)

eg Â  Â  $$h(3),{\text{ }}\frac{5}{{3 + 2}}$$

correct equation Â  Â  (A1)

eg Â  Â  $$a = \frac{5}{{3 + 2}},{\text{ }}h(3) = a$$

$$a = 1$$ Â  Â  A1 Â  Â  N2

[3 marks]

METHOD 2

attempt to find inverse (may be seen in (d)) Â  Â  (M1)

eg Â  Â  $$x = \frac{5}{{y + 2}},{\text{ }}{h^{ – 1}} = \frac{5}{x} – 2,{\text{ }}\frac{5}{x} + 2$$

correct equation, $$\frac{5}{x} – 2 = 3$$ Â  Â  (A1)

$$a = 1$$ Â  Â  A1 Â  Â  N2

[3 marks]

e.

## Question

The following diagram shows the graph of $$y = f(x)$$, forÂ $$– 4 \le x \le 5$$.

Write down the value of $$f( – 3)$$.

[1]
a(i).

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

[1]
a(ii).

Find the domain of $${f^{ – 1}}$$.

[2]
b.

On the grid above, sketch the graph of $${f^{ – 1}}$$.

[3]
c.

## Markscheme

$$f( – 3) =Â – 1$$ Â  Â  A1 Â  Â  N1

[1 mark]

a(i).

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

[1 mark]

a(ii).

domain of $${f^{ – 1}}$$ is range of $$f$$ Â  Â  (R1)

eg Â  Â  $${\text{R}}f = {\text{D}}{f^{ – 1}}$$

correct answer Â  Â  A1 Â  Â  N2

eg Â  Â  $$– 3 \leqslant x \leqslant 3,{\text{ }}x \in [ – 3,{\text{ }}3]{\text{ Â (accept }} – 3 < x < 3,{\text{ }} – 3 \leqslant y \leqslant 3)$$

[2 marks]

b.

Â  Â  Â A1A1 Â  Â  N2

Note:Â Graph must be approximately correct reflection in $$y = x$$.

Â Â  Â  Only if the shape is approximately correct, award the following:

Â Â  Â  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.

## Markscheme

valid approach Â  Â  (M1)

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

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

[2 marks]

a.

attempt to find $$f( – 1)$$ Â  Â  (M1)

eg$$\;\;\;$$line on graph

$$f( – 1) = 2$$ Â  Â  (A1)

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

[3 marks]

b.

Â Â  Â  A1A1 Â  Â  N2

Note: Â  Â  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.

## Markscheme

interchanging $$x$$ and $$y$$ (seen anywhere) Â  Â  (M1)

eg$$\;\;\;x = {(y – 5)^3}$$

evidence of correct manipulation Â  Â  (A1)

eg$$\;\;\;y – 5 = \sqrt[3]{x}$$

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

Notes: Â  Â  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) Â  Â  (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 Â  Â  (A1)

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

$$g(x) = 2{x^2} + 5$$ Â  Â  A1 Â  Â  N2

METHOD 2

recognising inverse relationship Â  Â  (M1)

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

correct working

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

$$g(x) = 2{x^2} + 5$$ Â  Â  A1 Â  Â  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.

## Markscheme

$$g(2) = 8$$ Â  Â A1 Â  Â  N1

[1 mark]

a.

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

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

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

[2 marks]

b.

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

eg$$\,\,\,\,\,$$$$x = 8y + 3$$

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

[2 marks]

c.

## Question

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

FindÂ $${f^{ – 1}}(x)$$.

[2]
a.

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

[3]
b.

## Markscheme

interchanging $$x$$ and $$x$$ Â  Â  (M1)

eg$$\,\,\,\,\,$$$$x = 5y$$

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

[2 marks]

a.

METHOD 1

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

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

$$g(7) = 50$$ Â  Â  (A1)

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

METHOD 2

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

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

correct substitution Â  Â  (A1)

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

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

[3 marks]

b.

## Question

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

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.

## Markscheme

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

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

[1 mark]

a.

$$f(2) = 3$$ Â  Â  A1 Â  Â  N1

[1 mark]

b.i.

$${f^{ – 1}}(2) = 0$$ Â  Â  A1 Â  Â  N1

[1 mark]

b.ii.

Â  Â  Â A1A1A1 Â  Â  N3

Notes: Â  Â  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 fâ€‰(x) , for âˆ’2 â‰¤ x â‰¤ 2 . The following diagram shows the graph of f.

Write down the value ofÂ fâ€‰(0).

[1]
a.i.

Write down the value ofÂ fâ€‰âˆ’1â€‰(1).

[1]
a.ii.

Write down theÂ range ofÂ fâ€‰âˆ’1.

[1]
b.

On the grid above, sketch the graph of fâ€‰âˆ’1.

[4]
c.

## Markscheme

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

[1 mark]

a.i.

fâ€‰âˆ’1â€‰(1) = 2Â  Â  Â A1 N1

[1 mark]

a.ii.

âˆ’2Â â‰¤ yÂ â‰¤ 2, yâˆˆ [âˆ’2, 2]Â  (acceptÂ âˆ’2Â â‰¤Â x â‰¤ 2)Â  Â  Â A1 N1

[1 mark]

b.

A1A1A1A1Â  N4

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

(yÂ =Â 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.