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Date May 2011 Marks available 4 Reference code 11M.2.sl.TZ1.9
Level SL only Paper 2 Time zone TZ1
Command term Find Question number 9 Adapted from N/A

Question

The following diagram shows the graph of \(f(x) = {{\rm{e}}^{ - {x^2}}}\) .


The points A, B, C, D and E lie on the graph of f . Two of these are points of inflexion.

Identify the two points of inflexion.

[2]
a.

(i)     Find \(f'(x)\) .

(ii)    Show that \(f''(x) = (4{x^2} - 2){{\rm{e}}^{ - {x^2}}}\) .

[5]
b(i) and (ii).

Find the x-coordinate of each point of inflexion.

[4]
c.

Use the second derivative to show that one of these points is a point of inflexion.

[4]
d.

Markscheme

B, D     A1A1     N2

[2 marks]

a.

(i) \(f'(x) = - 2x{{\rm{e}}^{ - {x^2}}}\)     A1A1     N2

Note: Award A1 for \({{\rm{e}}^{ - {x^2}}}\) and A1 for \( - 2x\) .

(ii) finding the derivative of \( - 2x\) , i.e. \( - 2\)     (A1)

evidence of choosing the product rule     (M1)

e.g. \( - 2{{\rm{e}}^{ - {x^2}}}\) \( - 2x \times - 2x{{\rm{e}}^{ - {x^2}}}\)

\( - 2{{\rm{e}}^{ - {x^2}}} + 4{x^2}{{\rm{e}}^{ - {x^2}}}\)     A1

\(f''(x) = (4{x^2} - 2){{\rm{e}}^{ - {x^2}}}\)     AG     N0

[5 marks]

b(i) and (ii).

valid reasoning     R1

e.g. \(f''(x) = 0\)

attempting to solve the equation     (M1)

e.g. \((4{x^2} - 2) = 0\) , sketch of \(f''(x)\)

\(p = 0.707\) \(\left( { = \frac{1}{{\sqrt 2 }}} \right)\) , \(q = - 0.707\) \(\left( { = - \frac{1}{{\sqrt 2 }}} \right)\)     A1A1     N3

[4 marks]

c.

evidence of using second derivative to test values on either side of POI     M1

e.g. finding values, reference to graph of \(f''\) , sign table

correct working     A1A1

e.g. finding any two correct values either side of POI,

checking sign of \(f''\) on either side of POI

reference to sign change of \(f''(x)\)     R1     N0

[4 marks]

d.

Examiners report

Most candidates were able to recognize the points of inflexion in part (a).

a.

Most candidates were able to recognize the points of inflexion in part (a) and had little difficulty with the first and second derivatives in part (b). A few did not recognize the application of the product rule in part (b).

b(i) and (ii).

Obtaining the x-coordinates of the inflexion points in (c) usually did not cause many problems.

c.

Only the better-prepared candidates understood how to set up a second derivative test in part (d). Many of those did not show, or clearly indicate, the values of x used to test for a point of inflexion, but merely gave an indication of the sign. Some candidates simply resorted to showing that \(f''\left( { \pm \frac{1}{{\sqrt 2 }}} \right) = 0\) , completely missing the point of the question. The necessary condition for a point of inflexion, i.e. \(f''(x) = 0\) and the change of sign for \(f''(x)\) , seemed not to be known by the vast majority of candidates.

d.

Syllabus sections

Topic 2 - Functions and equations » 2.7 » Use of technology to solve a variety of equations, including those where there is no appropriate analytic approach.

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