Date | May 2010 | Marks available | 8 | Reference code | 10M.1.hl.TZ2.7 |
Level | HL only | Paper | 1 | Time zone | TZ2 |
Command term | Find | Question number | 7 | Adapted from | N/A |
Question
The function f is defined by \(f(x) = {{\text{e}}^{{x^2} - 2x - 1.5}}\).
(a) Find \(f'(x)\).
(b) You are given that \(y = \frac{{f(x)}}{{x - 1}}\) has a local minimum at x = a, a > 1. Find the
value of a.
Markscheme
(a) \(\left( {u = {x^2} - 2x - 1.5;\frac{{{\text{d}}u}}{{{\text{d}}x}} = 2x - 2} \right)\)
\(\frac{{{\text{d}}f}}{{{\text{d}}x}} = \frac{{{\text{d}}f}}{{{\text{d}}u}}\frac{{{\text{d}}u}}{{{\text{d}}x}} = {{\text{e}}^u}(2x - 2)\) (M1)
\( = 2(x - 1){{\text{e}}^{{x^2} - 2x - 1.5}}\) A1
(b) \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{(x - 1) \times 2(x - 1){{\text{e}}^{{x^2} - 2x - 1.5}} - 1 \times {{\text{e}}^{{x^2} - 2x - 1.5}}}}{{{{(x - 1)}^2}}}\) M1A1
\( = \frac{{2{x^2} - 4x + 1}}{{{{(x - 1)}^2}}}{{\text{e}}^{{x^2} - 2x - 1.5}}\) (A1)
minimum occurs when \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0\) (M1)
\(x = 1 \pm \sqrt {\frac{1}{2}} \,\,\,\,\,\left( {{\text{accept }}x = \frac{{4 \pm \sqrt 8 }}{4}} \right)\) A1
\(a = 1 + \sqrt {\frac{1}{2}} \,\,\,\,\,\left( {{\text{accept }}a = \frac{{4 + \sqrt 8 }}{4}} \right)\) R1
[8 marks]
Examiners report
Part (a) was successfully answered by most candidates. Most candidates were able to make significant progress with part (b) but were then let down by being unable to simplify the expression or by not understanding the significance of being told that a > 1.