2017 H2 Mathematics Paper 1 Question 5

Differentiation I: Tangents and Normals, Parametric Curves

Answers

{a=- \frac{3}{2}, \;}

{b=\frac{3}{2}, \;}

{c=7}

{x=-1.33}

{x=-0.145, \;}

{x=1.15}

By the remainder theorem,

\begin{align*} 1^3 + a(1)^2 + b(1) + c &= 8 \\ 2^3 + a(2)^2 + b(2) + c &= 12 \\ 3^3 + a(3)^2 + b(3) + c &= 25 \\ \end{align*}

\begin{align} && \quad a + b + c &= 7 \\ && \quad 4a + 2b + c &= 4 \\ && \quad 9a + 3b + c &= -2 \\ \end{align}

Solving with a GC,

\begin{align*} a &= - \frac{3}{2} \; \blacksquare \\ b &= \frac{3}{2} \; \blacksquare \\ c &= 7 \; \blacksquare \\ \end{align*}

\begin{align*} y &= x^3 - \frac{3}{2} x^2 + \frac{3}{2} x + 7 \\ \frac{\mathrm{d}y}{\mathrm{d}x} &= 3 x^2 - 3 x + \frac{3}{2} \\ &= 3 \left( x^2 - x \right) + \frac{3}{2} \\ &= 3 \Bigg( \left(x- \frac{1}{2} \right)^2 - \left( \frac{1}{2} \right)^2 \Bigg) + \frac{3}{2} \\ &= 3 \left(x- \frac{1}{2} \right)^2 + \frac{3}{4} \\ \end{align*}

Since

{\left(x-\frac{1}{2}\right)^2 \geq 0}

for all real values of

{x,}

{3 \left(x- \frac{1}{2} \right)^2 + \frac{3}{4} > 0}

for all real values of

{x}

so the gradient of the curve is always positive

{\blacksquare}

Hence the curve is increasing and

{f(x)=0}

has only one real root

Using a GC, root to the equation

{f(x)=0:}

x=-1.33 \textrm{ (3 sf)} \; \blacksquare

When the tangent is parallel to the line

{y=2x-3,}

\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} &= 2 \\ 3 x^2 - 3 x + \frac{3}{2} &= 2 \\ 3 x^2 - 3 x - \frac{1}{2} &= 0 \end{align*}

Solving with a GC,

\begin{align*} x &= -0.145 \textrm{ (3 sf)} \; \blacksquare \\ \textrm{or} \quad x &= 1.15 \textrm{ (3 sf)} \; \blacksquare \end{align*}