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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.

Syllabus sections

Topic 6 - Core: Calculus » 6.3 » Local maximum and minimum values.
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