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Date May 2009 Marks available 2 Reference code 09M.2.sl.TZ1.10
Level SL only Paper 2 Time zone TZ1
Command term Write down Question number 10 Adapted from N/A

Question

Let \(f(x) = {x^3} - 4x + 1\) .

Expand \({(x + h)^3}\) .

[2]
a.

Use the formula \(f'(x) = \mathop {\lim }\limits_{h \to 0} \frac{{f(x + h) - f(x)}}{h}\) to show that the derivative of \(f(x)\) is \(3{x^2} - 4\) .

[4]
b.

The tangent to the curve of f at the point \({\text{P}}(1{\text{, }} - 2)\) is parallel to the tangent at a point Q. Find the coordinates of Q.

[4]
c.

The graph of f is decreasing for \(p < x < q\) . Find the value of p and of q.

[3]
d.

Write down the range of values for the gradient of \(f\) .

[2]
e.

Markscheme

attempt to expand     (M1)

\({(x + h)^3} = {x^3} + 3{x^2}h + 3x{h^2} + {h^3}\)     A1     N2

[2 marks]

a.

evidence of substituting \(x + h\)     (M1)

correct substitution     A1

e.g. \(f'(x) = \mathop {\lim }\limits_{h \to 0} \frac{{{{(x + h)}^3} - 4(x + h) + 1 - ({x^3} - 4x + 1)}}{h}\)

simplifying     A1

e.g. \(\frac{{({x^3} + 3{x^2}h + 3x{h^2} + {h^3} - 4x - 4h + 1 - {x^3} + 4x - 1)}}{h}\)

factoring out h     A1

e.g. \(\frac{{h(3{x^2} + 3xh + {h^2} - 4)}}{h}\)

\(f'(x) = 3{x^2} - 4\)     AG     N0

[4 marks]

b.

\(f'(1) = - 1\)    (A1)

setting up an appropriate equation     M1

e.g. \(3{x^2} - 4 = - 1\)

at Q, \(x = - 1,y = 4\) (Q is \(( - 1{\text{, }}4)\))    A1    A1

[4 marks]

c.

recognizing that f is decreasing when \(f'(x) < 0\)     R1

correct values for p and q (but do not accept \(p = 1.15{\text{, }}q = - 1.15\) )     A1A1     N1N1

e.g. \(p = - 1.15{\text{, }}q = 1.15\) ; \( \pm \frac{2}{{\sqrt 3 }}\) ; an interval such as \( - 1.15 \le x \le 1.15\)

[3 marks]

d.

\(f'(x) \ge - 4\) , \(y \ge - 4\) , \(\left[ { - 4,\infty } \right[\)     A2     N2

[2 marks]

e.

Examiners report

In part (a), the basic expansion was not done well. Rather than use the binomial theorem, many candidates opted to expand by multiplication which resulted in algebraic errors.

a.

In part (b), it was clear that many candidates had difficulty with differentiation from first principles. Those that successfully set the answer up, often got lost in the simplification.

b.

Part (c) was poorly done with many candidates assuming that the tangents were horizontal and then incorrectly estimating the maximum of f as the required point. Many candidates unnecessarily found the equation of the tangent and could not make any further progress.

c.

In part (d) many correct solutions were seen but only a very few earned the reasoning mark.

d.

Part (e) was often not attempted and if it was, candidates were not clear on what was expected.

e.

Syllabus sections

Topic 2 - Functions and equations » 2.4 » The quadratic function \(x \mapsto a{x^2} + bx + c\) : its graph, \(y\)-intercept \((0, c)\). Axis of symmetry.
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