Edexcel IGCSE Math B 4 Functions

[Maximum number: 9]

Figure 3

Information about the function f is shown in Figure 3.
Given that f is the mapping $\mathrm{f}: x \mapsto a x^{3}+b$ where a and b are constants,

(a)

use the information in Figure 3 to write down and simplify two equations in a and b.

[ 2 ]

(b)

Using your values of a and b, express the inverse function $f^{-1}$ in the form $f^{-1}:x\mapsto\ldots$ .

[ 2 ]

$y=2x^3+3$ $x=\sqrt[3]{\frac{y-3}{2}}$

Answer:
$f^{-1}:x\mapsto\sqrt[3]{\frac{x-3}{2}}$

Marks: M1, A1.

(c)

Find g(2 p), simplifying your answer.

[ 2 ]

$g(2p)=\frac{p(2p)}{2p-p}=2p$

Answer: 2p.
Marks: M1, A1.

(d)

Show algebraically that $\operatorname{gg}(x)=x$

Show clear algebraic working.

[ 2 ]

$g(g(x))=\frac{p\left(\frac{px}{x-p}\right)}{\frac{px}{x-p}-p}$ $=\frac{\frac{p^2x}{x-p}}{\frac{px-p(x-p)}{x-p}} =\frac{p^2x}{x-p}\times\frac{x-p}{p^2}=x$

Marks: M1, A1.

(e)

Hence write down the inverse function $\mathrm{g}^{-1}$ in the form $\mathrm{g}^{-1}(x)=\ldots$

[ 1 ]

Answer:
$g^{-1}(x)=\frac{px}{x-p}$

Marks: B1.

[Maximum number: 13]

The equation of the straight line L is
y=2-3x.

(a)

Find the exact coordinates of the points where L crosses the coordinate axes.

[ 2 ]

Coordinate-axis intersections:
$(0,2),\qquad \left(\frac23,0\right).$

Marking notes:
Equivalent unambiguous y=2 and $x=\frac23$ statements are allowed;
the x-intercept must be exact.

(b)

The functions f and g are defined as
$f:x\mapsto2-3x, \qquad g:x\mapsto\frac{3+x}{1-2x},\quad x\ne\frac12.$
Find g(-2).

[ 1 ]

(c)

Find the values of x for which ff(x)+g(x)=0

Show your working clearly and give your answer in the form

\frac{a \pm \sqrt{b}}{c}

where a, b and c are integers.

[ 5 ]

ff(x)=f(2-3x)=2-3(2-3x)=9x-4.
Set
ff(x)+g(x)=0: $9x-4+\frac{3+x}{1-2x}=0.$
This gives
$18x^{2}-18x+1=0.$
Using the quadratic formula:
$x=\frac{18\pm\sqrt{(-18)^2-4(18)(1)}}{36}.$
Final answer:
$x=\frac{3\pm\sqrt7}{6}.$

(d)

Express the inverse of the composite function fg in the form $(\mathrm{fg})^{-1}: x \mapsto \ldots$

\left[\text { Solutions of } a x^{2}+b x+c=0 \text { are } x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\right]

[ 5 ]

$fg(x)=f\left(\frac{3+x}{1-2x}\right) = 2-3\left(\frac{3+x}{1-2x}\right) = \frac{7x+7}{2x-1}.$
Let y=fg(x):
y(2x-1)=7x+7.2xy-7x=7+y.x(2y-7)=7+y.
Therefore
$(fg)^{-1}:x\mapsto\frac{7+x}{2x-7}.$

Marking notes:
Equivalent form $\frac{-x-7}{7-2x}$ is allowed.

[Maximum number: 4]

Here are the functions f, g and h.
$\begin{aligned}f:x&\mapsto 2x-5,\quad x>1,\\g:x&\mapsto 7-\frac{8x}{2x+1},\\h:x&\mapsto 2x^2+4x-5,\quad x>-1.\end{aligned}$

(a)

Find the value of g(-1).

[ 1 ]

(b)

Write down the value of x for which g is not defined.

[ 1 ]

Final answer:
$x=-\frac12$

B1 oe. Allow x=-0.5 or $x\neq-0.5$ . Do not allow x<-0.5 or x>-0.5;
must be a single value.

(c)

Find $h^{-1}(x)$ .

[ 2 ]

$y=2(x+1)^{2}-7$ $\frac{y+7}{2}=(x+1)^{2}$

Final answer:
$h^{-1}(x)=\sqrt{\frac{x+7}{2}}-1$

M1 for rearranging the completed-square form.
A1 ft from part (e);
must be in terms of x. If $\pm$ is given, award M1 only.

[Maximum number: 2]

Figure 3

Information about the function f is shown in Figure 3.
Given that f is the mapping $\mathrm{f}: x \mapsto a x^{3}+b$ where a and b are constants,

use the information in Figure 3 to write down and simplify two equations in a and b.

[Maximum number: 1]

The equation of the straight line L is
y=2-3x.

The functions f and g are defined as
$f:x\mapsto2-3x, \qquad g:x\mapsto\frac{3+x}{1-2x},\quad x\ne\frac12.$
Find g(-2).

[Maximum number: 1]

Here are the functions f, g and h.
$\begin{aligned}f:x&\mapsto 2x-5,\quad x>1,\\g:x&\mapsto 7-\frac{8x}{2x+1},\\h:x&\mapsto 2x^2+4x-5,\quad x>-1.\end{aligned}$

Write down the value of x for which g is not defined.

Final answer:
$x=-\frac12$

B1 oe. Allow x=-0.5 or $x\neq-0.5$ . Do not allow x<-0.5 or x>-0.5;
must be a single value.

[Maximum number: 2]

Figure 3

Information about the function f is shown in Figure 3.
Given that f is the mapping $\mathrm{f}: x \mapsto a x^{3}+b$ where a and b are constants,

Show algebraically that $\operatorname{gg}(x)=x$

Show clear algebraic working.

$g(g(x))=\frac{p\left(\frac{px}{x-p}\right)}{\frac{px}{x-p}-p}$ $=\frac{\frac{p^2x}{x-p}}{\frac{px-p(x-p)}{x-p}} =\frac{p^2x}{x-p}\times\frac{x-p}{p^2}=x$

Marks: M1, A1.

[Maximum number: 1]

3 y varies inversely as the square of x.

Write down a formula for y in terms of x and a constant k.

Answer:
$y=\frac{k}{x^2}$

Marks: B1.

[Maximum number: 1]

Figure for Question 9 — Edexcel IGCSE Math B

The vertices of triangle A are the points with coordinates (2,2),(3,2) and (2,4).

On the grid, draw and label triangle A.

[Maximum number: 2]

The equation of the straight line L is
y=2-3x.

Find the exact coordinates of the points where L crosses the coordinate axes.

Coordinate-axis intersections:
$(0,2),\qquad \left(\frac23,0\right).$

Marking notes:
Equivalent unambiguous y=2 and $x=\frac23$ statements are allowed;
the x-intercept must be exact.

[Maximum number: 5]

Figure for Question 11 — Edexcel IGCSE Math B

Diagrams NOT accurately drawn.

Figure 3 shows a rectangle with dimensions 20 cm by 10 cm from which a square with sides of length x cm is removed from each of the corners.
The shape in Figure 3 is then folded along the dotted lines to form a box, without a lid, in the shape of a cuboid, shown in Figure 4.
The volume of the box is $V\,\mathrm{cm}^3$ .

(a)

Figure for Question (a) — Edexcel IGCSE Math B

On the grid opposite, plot the points from your completed table and, using your answer to part (b), join them to form a smooth curve.

[ 2 ]

Fully correct graph.

Marking notes:
Award B2 for a fully correct graph, allowing the maximum at $1.5<x\le2.5$ .
Award B1 for all points plotted correctly, or 4 points plotted correctly with a curve joining all plotted points.

(b)

Starting with a square of side 15 cm and removing a square with sides of length x cm from each corner, a second box without a lid is formed by folding as in part (a).
The volume of this box is $B\,\mathrm{cm}^3$ , where
$B=4x^3-60x^2+225x.$
Given that B=200, find, by drawing a suitable straight line on the grid, estimates, to one decimal place, of the possible values of x.

[ 3 ]

[Maximum number: 2]

Figure for Question 12 — Edexcel IGCSE Math B

Diagrams NOT accurately drawn.

Figure 3 shows a rectangle with dimensions 20 cm by 10 cm from which a square with sides of length x cm is removed from each of the corners.
The shape in Figure 3 is then folded along the dotted lines to form a box, without a lid, in the shape of a cuboid, shown in Figure 4.
The volume of the box is $V\,\mathrm{cm}^3$ .

By drawing on the grid a tangent to the curve, find an estimate of the gradient of the curve at the point where x=1.5.

[Maximum number: 2]

Given
$y=\frac{20x^4+26x^3-6x^2}{5x^2-x}, \quad x\ne0, \quad x\ne\frac15.$
Use your answer to part (a) to find $\frac{dy}{dx}$ .

From part (a):
$y=2x(2x+3)=4x^2+6x$ $\frac{dy}{dx}=8x+6$

Answer: 8x+6.

Marks: M1, A1.

[Maximum number: 4]

Figure for Question 14 — Edexcel IGCSE Math B

Diagrams NOT accurately drawn.

Figure 3 shows a rectangle with dimensions 20 cm by 10 cm from which a square with sides of length x cm is removed from each of the corners.
The shape in Figure 3 is then folded along the dotted lines to form a box, without a lid, in the shape of a cuboid, shown in Figure 4.
The volume of the box is $V\,\mathrm{cm}^3$ .

Find, to 3 significant figures, the value of x such that $\frac{dV}{dx}=0$ .

$\text{Solutions of } ax^2+bx+c=0 \text{ are } x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.$

$\frac{dV}{dx}=12x^{2}-120x+200$ $12x^{2}-120x+200=0$ $x=\frac{120\pm\sqrt{120^{2}-4\times12\times200}}{2\times12}$
Final answer:
x=2.11

Marking notes:
Allow $\frac{15-5\sqrt3}{3}$ . Do not accept 7.89.

[Maximum number: 2]

A particle P is moving along a straight line through the fixed point O.
The speed, v m/s, of P at time t seconds is given by
$v=\frac{t^2}{27}+2+\frac{3}{t^2}$
for $1\leqslant t\leqslant5$ .

The acceleration of P at time t seconds, where $1\leqslant t\leqslant5$ , is a m/s $^2$ .
Find an expression for a in terms of t.

Differentiating:
$a=\frac{2t}{27}-6t^{-3}.$
Final answer:
$a=\frac{2t}{27}-6t^{-3}$

Marking notes:
M1 for attempting to differentiate with one term in t correct. Penalise extra terms.

Edexcel IGCSE Math B 4 Functions Questions

Covered syllabus points

Covered syllabus points

Question 1

Question (a)

Question (b)

Question (c)

Question (d)

Question (e)

Question 2

Question (a)

Question (b)

Question (c)

Question (d)

Question 3

Question (a)

Question (b)

Question (c)

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question (a)

Question (b)

Question 12

Question 13

Question 14

Question 15