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.

(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]

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.