# 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.

## 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

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 ofÂ $${f^{ – 1}}$$ .

[3]
a.

Write down the range of $${f^{ – 1}}$$ .

[1]
b.

FindÂ $${f^{ – 1}}(x)$$ .

[3]
c.

## Markscheme

Â Â

Â Â  A1A1A1Â Â Â Â  N3

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

a.

$$0 \le y \le 3.5$$Â Â Â  Â A1 Â  Â  N1

[1 mark]

b.

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

e.g. $$x = {e^{0.5y}}$$

evidence of changing to log formÂ Â Â Â  A1

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

$${f^{ – 1}}(x) = 2\ln x$$Â Â Â  Â A1Â Â Â Â  N1

[3 marks]

c.

## 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.

## 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 $$x \ge 0$$Â ,Â $$x > 3$$ and “all real numbers” as responses.

a.

Where students attempted to solve the equation in (b), most treated $$\ln x – 3$$Â as $$\ln (x – 3)$$Â 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.

## 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

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

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)Â Â Â Â  Find $$f”(1)$$ .

(ii)Â Â Â Â  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.

## Markscheme

substitute $$0$$ into $$f$$Â Â Â Â  (M1)

eg Â  $$\ln (0 + 1)$$Â , $$\ln 1$$

$$f(0) = 0$$Â Â Â Â  A1 N2

[2 marks]

a.

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

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

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

eg Â  $$f'(x) > 0$$Â , diagram of signs

attempt to solve $$f'(x) > 0$$Â Â Â Â Â (M1)

eg Â  $$4{x^3} = 0$$Â , $${x^3} > 0$$

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

[5 marks]

b.

(i)Â Â Â Â  substituting $$x = 1$$Â into $$f”$$Â Â Â Â Â (A1)

egÂ Â  $$\frac{{4(3 – 1)}}{{{{(1 + 1)}^2}}}$$Â , $$\frac{{4 \times 2}}{4}$$

$$f”(1) = 2$$Â Â Â Â  A1Â Â Â Â  N2

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

egÂ Â  no change of sign in $$f”(x)$$Â , no change in concavity,

$$f’$$ increasing both sides of zero

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

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

correct working leading to positive valueÂ Â Â Â  A1

egÂ Â  $$f”( – 1) = 2$$Â , discussing signs of numerator and denominator

there is no point of inflexion at $$x = 0$$Â Â Â Â Â AGÂ Â Â Â  N0

[5 marks]

c.

Â Â Â Â  A1A1A1Â Â Â Â  N3

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

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

Â Â Â  A1 for curve through ($$0$$, $$0$$)Â , A1 for increasing throughout.

Â Â Â  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.

## 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

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.

## Markscheme

$$g(3) = – 18$$ , $$f'(3) = 1$$ , $$h”(2) = – 6$$Â Â Â Â Â A1A1A1Â Â Â Â  N3

[3 marks]

a.

$$h”(3) = 0$$Â Â Â Â  (A1)

valid reasoningÂ Â Â  Â R1

eg Â  $${h”}$$Â changes sign at $$x = 3$$Â , change in concavity of $$h$$ at $$x = 3$$

so P is a point of inflexionÂ Â Â Â  AGÂ Â Â Â  N0

[2 marks]

b.

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

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

$$h(3) = – 54$$Â Â Â Â  A1 N1

[2 marks]

c.

recognizing need to find derivative of $$h$$Â Â Â Â  (R1)

eg Â  $${h’}$$Â , $$h'(3)$$

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

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

correct substitutionÂ Â Â Â  (A1)

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

$$h'(3) = – 27$$Â Â Â  A1

attempt to find the gradient of the normalÂ Â Â Â  (M1)

eg Â  $$– \frac{1}{m}$$ , $$– \frac{1}{{27}}x$$Â

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

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

correct equation in any formÂ Â Â Â  A1Â Â Â Â  N4

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

[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.

## 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  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

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

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

LetÂ $$f\left( x \right) = \sqrt {x + 2}$$ for xÂ â‰¥ 2 and g(x) = 3xÂ âˆ’ 7 forÂ $$x \in \mathbb{R}$$.

Write down fâ€‰(14).

[1]
a.

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

[2]
b.

FindÂ gâˆ’1(x).

[3]
c.

## Markscheme

fâ€‰(14) = 4Â  Â  Â A1 N1

[1 mark]

a.

attempt to substituteÂ  Â  Â (M1)

egÂ  Â gâ€‰(4), 3 Ã— 4 âˆ’ 7

5Â  Â  Â A1 N2

[2 marks]

b.

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

egÂ  Â x = 3yÂ âˆ’ 7

evidence of correct manipulationÂ  Â  Â (A1)

egÂ  Â x + 7 = 3y

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

[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.