Practice Paper 2 (Cambridge International IGCSE Further Maths)

On the axes, sketch the graph of $y = 5 ( x + 1 ) ( 3 x - 2 ) ( x - 2 )$, stating the intercepts with the coordinate axes.

Extract:

On the axes, sketch the graph of y = 5(x+1)(3x-2)(x-2), stating the intercepts with the coordinate axes.

From the mark scheme: The graph is cubic. Solve 5(x+1)(3x-2)(x-2) = 0 to find where the curve will cross the x axis. x = -1, x = 2/3, x = 2 Substitute x = 0 into y = 5(x+1)(3x-2)(x-2) to find where the curve will cross the y axis. y = 5(1)(-2)(-2) y = 20

Hence find the values of $x$ for which $5 ( x + 1 ) ( 3 x - 2 ) ( x - 2 ) > 0$.

$p(x)=ax^{3}+3x^{2}+bx-12$ has a factor of $2x+1$. When $p(x)$ is divided by $x-3$ the remainder is 105.

Find the value of $a$ and of $b$.

Extract:

p(x) = ax³ + 3x² + bx - 12 has a factor of 2x + 1.

a = 6, b = -24

Using your values of $a$ and $b$, write $p(x)$ as a product of $2x+1$ and a quadratic factor.

Extract:

Using your values of a and b, write p(x) as a product of 2x + 1 and a quadratic factor.

a = 6, b = -24 so p(x) = 6x³ + 3x² - 24x - 12

Rearrange p(x) so that the first 2 terms have a common factor of 2x and the second 2 terms have a common factor of 1

p(x) = 6x³ - 24x + 3x² - 12

Take out the common factor of the first 2 terms and the second 2 terms

p(x) = 2x(3x² - 12) + 1(3x² - 12)

Factorise

p(x) = (2x + 1)(3x² - 12)

Hence solve$p(x)$= 0.

$f(x)=x^{2}+2x-3\mathrm{for}x\geq-1$

Given that the minimum value of $x^{2}+2x-3$ occurs when $x=-1$, explain why $f(x)$ has an inverse.

Extract:

straight f left parenthesis x right parenthesis space equals space x squared space plus 2 x minus 3 space for space x space greater or equal than negative 1

Given that the minimum value of x squared space plus 2 x minus 3 occurs when x space equals negative 1

On the axes below, sketch the graph of $y=f(x)$ and the graph of $y=f^{-1}(x)$. Label each graph and state the intercepts on the coordinate axes.

For variables $x$ and $y$, plotting $\mathrm{ln}y$ against $\mathrm{ln}x$ gives a straight-line graph passing through the points $(6,5)$ and $(8,9)$. Show that $y$ = $e^{p}x^{q}$ where $p$ and $q$ are integers to be found.

The curve $y=2x^{2}+k+4$ intersects the straight line $y=(k+4)x$ at two distinct points. Find the possible values of $k$.

Solve the simultaneous equations.

$x^{2}+3xy=4$

$2x+5y=4$

At time $ts$, a particle travelling in a straight line has acceleration $( 2 t + 1 )^{- 1 \frac{1}{2}} ms^{- 2}$. When $t=0$, the particle is $4m$ from a fixed point $O$ and is travelling with velocity $8\mathrm{ms}^{-1}$ away from $O$.

Find the velocity of the particle at time $ts$.

Extract:

At time t s, a particle travelling in a straight line has acceleration (2t+1)^(-1/2) ms^(-2). When t = 0, the particle is 4 m from a fixed point O and is travelling with velocity 8 ms^(-1) away from O.

Find the displacement of the particle from $O$ at time $ts$.

The radius, $r\mathrm{cm}$, of a circle is increasing at the rate of $5\mathrm{cms}^{--1}$. Find, in terms of $π$, the rate at which the area of the circle is increasing when $r=3$.

The diagram shows part of the graphs of $y=4x^{\frac{2}{3}}$ and $y=(x-3)^{2}$ .The graph of $y=(x-3)^{2}$ meets the $x$-axis at the point $A(a,0)$ and the two graphs intersect at the point $B(b,4)$.

Find the value of $a$ and of $b$.

Extract:

The diagram shows part of the graphs of y = 4x^(2/3) and y = (x - 3)^2. The graph of y = (x - 3)^2 meets the x-axis at the point A(a, 0) and the two graphs intersect at the point B(b, 4).

From part a: a = 3 and b = 1

Find the area of the shaded region.

The point $P ( 8 , 8 )$ lies on the circle $( x - 2 )^{2} + y^{2} = 100$ with centre $C$.

The circle and the tangent to the circle at $P$ are shown below.

Find the equation of the tangent to the circle at $P$, giving your answer in the form $y=mx+c$.

The tangent intersects the $y$-axis at the point $Q$.

Find the area of triangle $PCQ$.

Solutions to this question by accurate drawing will not be accepted. The points $A$ and $B$ are (4, 3) and (12, −7) respectively.

Find the equation of the line $L$, the perpendicular bisector of the line $AB$.

Extract:

The points A and B are (4, 3) and (12, −7) respectively.

Find the equation of the line L, the perpendicular bisector of the line AB.

From the mark scheme: Calculate the gradient between points A and B: gradient_AB = (-7 - 3)/(12 - 4) gradient_AB = -5/4

Find the midpoint, M, of AB: M = ((4 + 12)/2, (3 + (-7))/2) M = (8, -2)

Find the gradient of the perpendicular line, L: gradient_L = 4/5

Find the equation of line L through the midpoint M = (8, -2) with gradient 4/5: y - (-2) = 4/5(x - 8) or y + 2 = 4/5(x - 8)

The line parallel to $AB$ which passes through the point (5, 12) intersects $L$ at the point $C$. Find the coordinates of $C$.

Find the coefficient of $x^{2}$ in the expansion of $( x - \frac{3}{x} ) ( x + \frac{2}{x} )^{5}$.

A committee of 5 people is to be formed from 6 doctors, 4 dentists and 3 nurses. Find the number of different committees that could be formed if

(i) there are no restrictions,

(ii) the committee contains at least one doctor,

(iii) the committee contains all the nurses.

The circles with centres $C_{1}$ and $C_{2}$ have equal radii of length $r\mathrm{cm}$. The line $C_{1}C_{2}$ is a radius of both circles. The two circles intersect at $A$ and $B$.

Given that the perimeter of the shaded region is $4π\mathrm{cm}$, find the value of $r$. [4]

Extract:

The circles with centres C₁ and C₂ have equal radii of length r cm. The line C₁C₂ is a radius of both circles. The two circles intersect at A and B.

Given that the perimeter of the shaded region is 4π cm, find the value of r.

From the solution to part (a): r = 3 and angle ACB = 2π/3

Find the exact area of the shaded region.