2008 H2 Mathematics Paper 1 Question 9

Differentiation I: Tangents and Normals, Parametric Curves

Answers

The graph of

{y=f(x)}

is a horizontal line

{f'(x) = \frac{17}{(2x+1)^2} > 0}

Out of syllabus

\begin{align*} f'(x) &= \frac{a(cx+d)-c(ax+b)}{(cx+d)^2} \\ &= \frac{acx+ad-acx-bc}{(cx+d)^2} \\ &= \frac{ad-bc}{(cx+d)^2} \\ \end{align*}

Hence if

{ad-bc \neq 0,}

then

{f'(x)\neq 0}

for all real values of

{x}

so the graph of

{y=f(x)}

has no turning points

{\blacksquare}

{ad-bc=0, }

then

{f'(x)=0}

for all real values of

{x}

so the graph of

{y=f(x)}

is a horizontal line

{\blacksquare}

{a=3,}

{b=-7,}

{c=2,}

and

{d=1,}

\begin{align*} f'(x) &= \frac{3(1)-(-7)(2)}{(2x+1)^2} \\ &= \frac{17}{(2x+1)^2} \\ \end{align*}

{(2x+1)^2}

is always positive for all

{x \in \mathbb{R}, x \neq -1}

{f'(x) = \frac{17}{(2x+1)^2} > 0}

so the graph has a positive gradient at all points of the graph

{\blacksquare}

(iva)

(ivb)

Out of syllabus