(a)     Find in terms of $a$

  (i)     the zeros of $f$ ;

  (ii)     the values of $x$ for which $f$ is decreasing;

  (iii)     the values of $x$ for which $f$ is increasing;

  (iv)     the range of $f$ .

(b)     State the concavity of the graph of $f$ .

(a)

(i)     $x - a\sqrt x$    M1

 $\sqrt x \sqrt x  - a = 0$     (A1)

  2 $x = 0$, $x = {a^2}$     A1     N2

(ii)     $f'(x) = 1 - \frac{a}{{2\sqrt x }}$     A1

  $f$ is decreasing when $f' < 0$     (M1)

  $1 - \frac{a}{{2\sqrt x }} < 0 \Rightarrow \frac{{2\sqrt x  - a}}{{2\sqrt x }} < 0 \Rightarrow x > \frac{{{a^2}}}{4}$     A1

(iii)     $f$ is increasing when $f' > 0$

   $1 - \frac{a}{{2\sqrt x }} > 0 \Rightarrow \frac{{2\sqrt x  - a}}{{2\sqrt x }} > 0 \Rightarrow x > \frac{{{a^2}}}{4}$     A1

  Note: Award the M1 mark for either (ii) or (iii).

(iv)     minimum occurs at $x = \frac{{{a^2}}}{4}$

   minimum value is $y = - \frac{{{a^2}}}{4}$     (M1)A1

   hence $y \geqslant  - \frac{{{a^2}}}{4}$     A1

[10 marks]

(b)     concave up for all values of $x$     R1

[1 mark]

Total [11 marks]

This was generally a well answered question.