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Date November 2010 Marks available 9 Reference code 10N.1.sl.TZ0.10
Level SL only Paper 1 Time zone TZ0
Command term Show that Question number 10 Adapted from N/A

Question

Let \(f(x) = {x^3}\). The following diagram shows part of the graph of f .


The point \({\rm{P}}(a,f(a))\) , where \(a > 0\) , lies on the graph of f . The tangent at P crosses the x-axis at the point \({\rm{Q}}\left( {\frac{2}{3},0} \right)\) . This tangent intersects the graph of f at the point R(−2, −8) .

 

The equation of the tangent at P is \(y = 3x - 2\) . Let T be the region enclosed by the graph of f , the tangent [PR] and the line \(x = k\) , between \(x = - 2\) and \(x = k\) where \( - 2 < k < 1\) . This is shown in the diagram below.


(i)     Show that the gradient of [PQ] is \(\frac{{{a^3}}}{{a - \frac{2}{3}}}\) .

(ii)    Find \(f'(a)\) .

(iii)   Hence show that \(a = 1\) .

[7]
a(i), (ii) and (iii).

Given that the area of T is \(2k + 4\) , show that k satisfies the equation \({k^4} - 6{k^2} + 8 = 0\) .

[9]
b.

Markscheme

(i) substitute into gradient \( = \frac{{{y_1} - {y_2}}}{{{x_1} - {x_2}}}\)     (M1)

e.g. \(\frac{{f(a) - 0}}{{a - \frac{2}{3}}}\)

substituting \(f(a) = {a^3}\)

e.g. \(\frac{{{a^3} - 0}}{{a - \frac{2}{3}}}\)     A1

gradient \(\frac{{{a^3}}}{{a - \frac{2}{3}}}\)     AG     N0

(ii) correct answer     A1     N1

e.g. \(3{a^2}\) , \(f'(a) = 3\) , \(f'(a) = \frac{{{a^3}}}{{a - \frac{2}{3}}}\)

(iii) METHOD 1

evidence of approach     (M1)

e.g. \(f'(a) = {\rm{gradient}}\) , \(3{a^2} = \frac{{{a^3}}}{{a - \frac{2}{3}}}\)

simplify     A1

e.g. \(3{a^2}\left( {a - \frac{2}{3}} \right) = {a^3}\)

rearrange     A1

e.g. \(3{a^3} - 2{a^2} = {a^3}\)

evidence of solving     A1

e.g. \(2{a^3} - 2{a^2} = 2{a^2}(a - 1) = 0\)

\(a = 1\)     AG     N0

METHOD 2

gradient RQ \( = \frac{{ - 8}}{{ - 2 - \frac{2}{3}}}\)     A1

simplify     A1

e.g. \(\frac{{ - 8}}{{ - \frac{8}{3}}},3\)

evidence of approach     (M1)

e.g. \(f'(a) = {\rm{gradient}}\) , \(3{a^2} = \frac{{ - 8}}{{ - 2 - \frac{2}{3}}}\) , \(\frac{{{a^3}}}{{a - \frac{2}{3}}} = 3\)

simplify     A1

e.g. \(3{a^2} = 3\) , \({a^2} = 1\)

\(a = 1\)     AG     N0

[7 marks]

a(i), (ii) and (iii).

approach to find area of T involving subtraction and integrals    (M1)

e.g. \(\int {f - (3x - 2){\rm{d}}x} \) , \(\int_{ - 2}^k {(3x - 2) - \int_{ - 2}^k {{x^3}} } \) , \(\int {({x^3} - 3x + 2)} \)

correct integration with correct signs     A1A1A1

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

correct limits \( - 2\) and k (seen anywhere)     A1

e.g. \(\int_{ - 2}^k {({x^3} - 3x + 2){\rm{d}}x} \) , \(\left[ {\frac{1}{4}{x^4} - \frac{3}{2}{x^2} + 2x} \right]_{ - 2}^k\)

attempt to substitute k and \( - 2\)     (M1)

correct substitution into their integral if 2 or more terms     A1

e.g. \(\left( {\frac{1}{4}{k^4} - \frac{3}{2}{k^2} + 2k} \right) - (4 - 6 - 4)\)

setting their integral expression equal to \(2k + 4\) (seen anywhere)     (M1)

simplifying     A1

e.g. \(\frac{1}{4}{k^4} - \frac{3}{2}{k^2} + 2 = 0\)

\({k^4} - 6{k^2} + 8 = 0\)     AG     N0

[9 marks]

b.

Examiners report

Part (a) seemed to be well-understood by many candidates, and most were able to earn at least partial marks here. Part (ai) was a "show that" question, and unfortunately there were some candidates who did not show how they arrived at the given expression.

a(i), (ii) and (iii).

In part (b), the concept seemed to be well-understood. Most candidates saw the necessity of using definite integrals and subtracting the two functions, and the integration was generally done correctly. However, there were a number of algebraic and arithmetic errors which prevented candidates from correctly showing the desired final result.

b.

Syllabus sections

Topic 6 - Calculus » 6.2 » Derivative of \({x^n}\left( {n \in \mathbb{Q}} \right)\) , \(\sin x\) , \(\cos x\) , \(\tan x\) , \({{\text{e}}^x}\) and \(\ln x\) .
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