EXN.1.AHL.TZ0.15

pestleMathematicsAIHLPaper 1EXN· sl-5-6-stationary-points-local-max-and-minsource ↗

Consider the function $f (x) = \sqrt{- a x^{2} + x + a}, a \in ℝ^{+}$ .

For $a > 0$ the curve $y = f (x)$ has a single local maximum.

Find $f' (x)$ .

[2]

Find in terms of $a$ the value of $x$ at which the maximum occurs.

[2]

Hence find the value of $a$ for which $y$ has the smallest possible maximum value.

[4]

Markscheme / solution

* This sample question was produced by experienced DP mathematics senior examiners to aid teachers in preparing for external assessment in the new MAA course. There may be minor differences in formatting compared to formal exam papers.

$f' (x) = (- 2 a x + 1) \times \frac{1}{2} \times {(- a x^{2} + x + a)}^{- \frac{1}{2}}$

Note: M1 is for use of the chain rule.

$= \frac{- 2 a x + 1}{2 \sqrt{- a x^{2} + x + a}}$ M1A1

[2 marks]

$- 2 a x + 1 = 0$ (M1)

$x = \frac{1}{2 a}$ A1

[2 marks]

Value of local maximum $= \sqrt{- a \times \frac{1}{4 a^{2}} + \frac{1}{2 a} + a}$ M1A1

$= \sqrt{\frac{1}{4 a} + a}$

This has a minimum value when $a = 0.5$ (M1)A1

[4 marks]

Examiners’ report

[N/A]