2016 H2 Maths Paper 1 Q3 — Worked Solution

Solution

6 marks

Approach

The form $k(x - l)^{4} + m$ is $x^{4}$ translated so that its turning point $(0,0)$ moves to $(l,m)$ , then stretched vertically by factor $k$ . Since the turning point of $x^{4}$ corresponds to $(a,b)$ on $y = f(x)$ , matching turning points gives $l$ and $m$ directly; $k$ then follows from the second given point $(0,c)$ .

The turning point of $y = x^{4}$ is $(0,0)$ , so under the transformation to $y = k(x - l)^{4} + m$ this point becomes $(l,m)$ . Since this corresponds to $(a,b)$ ,

l = a,\ \ m = b

Since $y = f(x)$ also passes through $(0,c)$ ,

f(0) = k(0 - a)^{4} + b = c

ka^{4} = c - b

k = \frac{c - b}{a^{4}}

$\therefore$ $k = \frac{c - b}{a^{4}}$ , $l = a$ , $m = b$ .

Approach

Since $a,b,c$ are positive with $c > b$ , we have $k > 0$ , so $y = f(x)$ is an upward-opening quartic with minimum value $b > 0$ at $x = a$ — it lies entirely above the $x$ -axis. Reciprocating a curve that stays strictly positive and tends to $\infty$ turns its minimum into a maximum of $\frac{1}{f(x)}$ , sends the large values of $\left| f(x) \right|$ to values near $0$ , and never crosses the $x$ -axis (since $f(x)$ is never $0$ ).

We first sketch $y = f(x)$ : a quartic with minimum turning point $(a,b)$ and $y$ -intercept $(0,c)$ , where $c > b > 0$ so the curve never meets the $x$ -axis.

Since $f(x) \geq b > 0$ for all $x$ , the curve $y = \frac{1}{f(x)}$ is defined for all $x$ and is always positive. As $f(x) \rightarrow \infty$ when $x \rightarrow \pm \infty$ , we get $\frac{1}{f(x)} \rightarrow 0^{\text{+}}$ , so $y = 0$ is a horizontal asymptote. The minimum of $f(x)$ at $x = a$ becomes a maximum of $\frac{1}{f(x)}$ there, of value $\frac{1}{b}$ . The $y$ -intercept becomes $\frac{1}{f(0)} = \frac{1}{c}$ . Since $f(x)$ is never $0$ , $y = \frac{1}{f(x)}$ never crosses the $x$ -axis.

$\therefore$ $y = \frac{1}{f(x)}$ crosses the $y$ -axis at $\left( 0,\ \frac{1}{c} \right)$ , does not cross the $x$ -axis, and has a maximum turning point at $\left( a,\ \frac{1}{b} \right)$ .

2016 H2 Maths Paper 1, Question 3

Solution

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