Question 7 2015 STEP 1

Submitted by Mathssteeple on Wed, 03/13/2019 - 09:41

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Please can someone explain why it isn’t a^3/9 for all a greater than or equal to two. In the mark scheme it says it’s always bigger than 3a-6 but then let’s M(a) be 3a-6 ? Mark scheme at: https://correspondence.maths.org/sites/correspondence.maths.org/files/fe...

2015 S1 Q7

Permalink Submitted by cg213 on Fri, 03/15/2019 - 09:43

If $\frac a 3>1$ then the local maximum doesn't lie in the range $-\frac 1 3 \le x \le 1$ . The maximal value will be at one of the end points.

So if $\frac a 3 > 1 \implies a>3$ then the maximum value will be at $f(1)=3a-6$ . $\frac{a^3}9$ will be a bigger value but $x=\frac a 3$ will not lie in the range of $x$ when $a>3$ .