2012 H2 Mathematics Paper 2 Question 3

Graphs and Transformations

Answers

{x=2}

{x=-1}

{x^3 + x^2 - 2 x - 4=4}

{x^3 + x^2 - 2 x - 4=-4}

{x=0, }

{x=- 2, }

{x=1, }

{x=1}

\begin{align*} & f(2) \\ & = 2^3 + 2^2 - 2(2) - 4 \\ & = 8 + 4 - 4 - 4 \\ & = 4 \end{align*}

Hence the integer solution to

{f(x)=4}

x=2 \; \blacksquare

x^3 + x^2 - 2 x - 8=(x - 2)(ax^2+bx+c)

Comparing coefficients,

\begin{align*} &x^3:& a &= 1 \\ &x^0:& -2c &= -8 \\ && c &= 4 \\ &x^2:& -2a+b &= 1 \\ && b &= 3 \end{align*}

f(x)-4=(x - 2)(x^2 + 3 x + 4)

Discriminant for

{x^2 + 3 x + 4:}

\begin{align*} & b^2 - 4ac \\ & = 3^2 - 4(1)(4) \\ & = -7 \end{align*}

Since

{b^2-4ac<0,}

{x^2 + 3 x + 4=0}

has no real roots. Hence there are no other real solutions to the equation

{f(x)=4 \; \blacksquare}

Replacing

{x}

with

{x+3}

in part (ii),

\begin{align*} x+3 &= 2 \\ x &= -1 \; \blacksquare \end{align*}

Two cubic equations that give the roots of the equation

{|f(x)|=4:}

\begin{align} && \quad x^3 + x^2 - 2 x - 4 = 4 \; \blacksquare \\ && \quad x^3 + x^2 - 2 x - 4 = -4 \; \blacksquare \\ \end{align}

\begin{align*} x^3 + x^2 - 2 x - 4 &= -4 \\ x^3 + x^2 - 2 x - 8 &= 0 \\ x(x^2 + x - 2) &= 0 \\ x(x + 2)(x - 1) &= 0 \\ \end{align*}

Combining with the solution

{x=2}

from part (ii),

\begin{align*} && x &= 0 \; \blacksquare \\ &\textrm{or}& x &= - 2 \; \blacksquare \\ &\textrm{or}& x &= 1 \; \blacksquare \\ &\textrm{or}& x &= 1 \; \blacksquare \\ \end{align*}