IB Math Analysis & Approaches Question bank-Topic: SL 5.6 Differentiation of a sum and a multiple of these functions SL Paper 1

Question

Consider $$f(x) = \frac{1}{3}{x^3} + 2{x^2} – 5x$$ . Part of the graph of f is shown below. There is a maximum point at M, and a point of inflexion at N.

Find $$f'(x)$$ .

[3]
a.

Find the x-coordinate of M.

[4]
b.

Find the x-coordinate of N.

[3]
c.

The line L is the tangent to the curve of f at $$(3{\text{, }}12)$$. Find the equation of L in the form $$y = ax + b$$ .

[4]
d.

Markscheme

$$f'(x) = {x^2} + 4x – 5$$Â Â Â  Â A1A1A1 Â  Â  N3

[3 marks]

a.

evidence of attempting to solve $$f'(x) = 0$$Â Â Â  Â (M1)

evidence of correct workingÂ Â Â Â  A1

e.g.Â $$(x + 5)(x – 1)$$ , $$\frac{{ – 4 \pm \sqrt {16 + 20} }}{2}$$ , sketch

$$x = – 5$$, $$x = 1$$Â Â Â  Â (A1)

so $$x = – 5$$Â Â Â  Â A1Â Â Â Â  N2

[4 marks]

b.

METHOD 1

$$f”(x) = 2x + 4$$ (may be seen later)Â Â Â Â  A1

evidence of setting second derivative = 0Â Â Â Â  (M1)

e.g. $$2x + 4 = 0$$

$$x = – 2$$Â Â Â  Â A1Â Â Â Â  N2

METHOD 2

evidence of use of symmetryÂ Â Â Â  (M1)

e.g. midpoint of max/min, reference to shape of cubic

correct calculationÂ Â Â Â  A1

e.g. $$\frac{{ – 5 + 1}}{2}$$

$$x = – 2$$Â Â Â  Â A1Â Â Â Â  N2

[3 marks]

c.

attempting to find the value of the derivative when $$x = 3$$Â Â Â  Â (M1)

$$f'(3) = 16$$Â Â Â  Â A1

valid approach to finding the equation of a lineÂ Â Â Â  M1

e.g. $$y – 12 = 16(x – 3)$$ , $$12 = 16 \times 3 + b$$

$$y = 16x – 36$$Â Â Â  Â A1Â Â Â Â  N2

[4 marks]

d.

Question

A rectangle is inscribed in a circle of radius 3 cm and centre O, as shown below.

The point P(x , y) is a vertex of the rectangle and also lies on the circle. The angle between (OP) and the x-axis is $$\theta$$ radians, where $$0 \le \thetaÂ \le \frac{\pi }{2}$$ .

Write down an expression in terms of $$\theta$$ for

(i)Â Â Â Â  $$x$$ ;

(ii) Â Â  $$y$$ .

[2]
a.

Let the area of the rectangle be A.

Show thatÂ $$A = 18\sin 2\theta$$ .

[3]
b.

(i)Â Â Â Â  Find $$\frac{{{\rm{d}}A}}{{{\rm{d}}\theta }}$$ .

(ii)Â Â Â  Hence, find the exact value of $$\theta$$ which maximizes the area of the rectangle.

(iii)Â Â  Use the second derivative to justify that this value of $$\theta$$ does give a maximum.

[8]
c.

Markscheme

(i) $$x = 3\cos \theta$$Â Â Â  Â A1Â Â Â Â  N1Â

(ii) $$y = 3\sin \theta$$Â Â Â Â Â A1 Â  Â  N1

[2 marks]

a.

finding areaÂ Â Â Â  (M1)

e.g.Â $$A = 2x \times 2y$$ , $$A = 8 \times \frac{1}{2}bh$$Â

substitutingÂ Â Â Â  A1

e.g. $$A = 4 \times 3\sin \thetaÂ \times 3\cos \theta$$ , $$8 \times \frac{1}{2} \times 3\cos \thetaÂ \times 3\sin \theta$$

$$A = 18(2\sin \theta \cos \theta )$$Â Â Â  A1

$$A = 18\sin 2\theta$$Â Â Â Â  AGÂ Â Â Â  N0

[3 marks]

b.

(i) $$\frac{{{\rm{d}}A}}{{{\rm{d}}\theta }} = 36\cos 2\theta$$Â Â Â Â Â A2Â Â Â Â  N2Â

(ii) for setting derivative equal to 0Â Â Â  Â (M1)

e.g. $$36\cos 2\thetaÂ = 0$$ , $$\frac{{{\rm{d}}A}}{{{\rm{d}}\theta }} = 0$$

$$2\thetaÂ = \frac{\pi }{2}$$Â Â Â Â  (A1)

$$\thetaÂ = \frac{\pi }{4}$$Â Â Â Â  A1Â Â Â Â  N2

(iii) valid reason (seen anywhere)Â Â Â Â  R1

e.g. at $$\frac{\pi }{4}$$, $$\frac{{{{\rm{d}}^2}A}}{{{\rm{d}}{\theta ^2}}} < 0$$Â ; maximum when $$f”(x) < 0$$

finding second derivative $$\frac{{{{\rm{d}}^2}A}}{{{\rm{d}}{\theta ^2}}} =Â – 72\sin 2\theta$$Â Â Â  Â A1

evidence of substituting $$\frac{\pi }{4}$$Â Â Â  Â M1

e.g. $$– 72\sin \left( {2 \times \frac{\pi }{4}} \right)$$ , $$– 72\sin \left( {\frac{\pi }{2}} \right)$$ , $$– 72$$

$$\thetaÂ = \frac{\pi }{4}$$ produces the maximum areaÂ Â Â Â  AGÂ Â Â Â  N0

[8 marks]

c.

Question

Consider $$f(x) = {x^2} + \frac{p}{x}$$ , $$x \ne 0$$ , where p is a constant.

FindÂ $$f'(x)$$ .

[2]
a.

There is a minimum value of $$f(x)$$ when $$x = – 2$$ . Find the value of $$p$$ .

[4]
b.

Markscheme

$$f'(x) = 2x – \frac{p}{{{x^2}}}$$Â Â Â Â  A1A1Â Â Â Â  N2

Note: Award A1 for $$2x$$Â , A1 for $$– \frac{p}{{{x^2}}}$$Â .

[2 marks]

a.

evidence of equating derivative to 0 (seen anywhere)Â Â Â Â  (M1)

evidence of finding $$f'( – 2)$$Â (seen anywhere)Â Â Â Â  (M1)

correct equationÂ Â Â Â  A1

e.g. $$– 4 – \frac{p}{4} = 0$$ , $$– 16 – p = 0$$

$$p = – 16$$Â Â Â Â  A1 Â  Â  N3

[4 marks]

b.

Question

Let $$f(x) = 3 + \frac{{20}}{{{x^2} – 4}}$$ , for $$x \ne \pm 2$$ . The graph of f is given below.

The y-intercept is at the point A.

(i) Â  Â  Find the coordinates of A.

(ii)Â Â Â  Show that $$f'(x) = 0$$ at A.

[7]
a.

The second derivative $$f”(x) = \frac{{40(3{x^2} + 4)}}{{{{({x^2} – 4)}^3}}}$$Â . Use this to

(i)Â Â Â Â  justify that the graph of f has a local maximum at A;

(ii)Â Â Â  explain why the graph of f does not have a point of inflexion.

[6]
b.

Describe the behaviour of the graph of $$f$$ for large $$|x|$$ .

[1]
c.

Write down the range of $$f$$ .

[2]
d.

Markscheme

(i) coordinates of A are $$(0{\text{, }} – 2)$$ Â Â Â  A1A1Â Â Â Â  N2

(ii) derivative of $${x^2} – 4 = 2x$$ (seen anywhere)Â Â Â Â  (A1)

evidence of correct approachÂ Â Â Â  (M1)

e.g. quotient rule, chain rule

finding $$f'(x)$$Â Â Â Â Â A2

e.g. $$f'(x) = 20 \times ( – 1) \times {({x^2} – 4)^{ – 2}} \times (2x)$$ , $$\frac{{({x^2} – 4)(0) – (20)(2x)}}{{{{({x^2} – 4)}^2}}}$$

substituting $$x = 0$$ into $$f'(x)$$ (do not accept solving $$f'(x) = 0$$ ) Â Â Â  M1

at A $$f'(x) = 0$$Â Â Â Â Â AGÂ Â Â Â  N0

[7 marks]

a.

(i) reference to $$f'(x) = 0$$Â (seen anywhere)Â Â Â Â  (R1)

reference to $$f”(0)$$Â is negative (seen anywhere)Â Â Â Â  R1

evidence of substituting $$x = 0$$Â into $$f”(x)$$Â Â Â Â Â M1

finding $$f”(0) = \frac{{40 \times 4}}{{{{( – 4)}^3}}}$$ $$\left( { = – \frac{5}{2}} \right)$$Â Â Â Â Â A1

then the graph must have a local maximumÂ Â Â Â  AG

(ii) reference to $$f”(x) = 0$$Â at point of inflexionÂ Â Â Â  (R1)

recognizing that the second derivative is never 0Â Â Â Â  A1Â Â Â Â  N2

e.g. $$40(3{x^2} + 4) \ne 0$$Â , $$3{x^2} + 4 \ne 0$$Â , $${x^2} \neÂ – \frac{4}{3}$$Â , the numeratorÂ is always positive

Note: Do not accept the use of the first derivative in part (b).

[6 marks]

b.

correct (informal) statement, including reference to approaching $$y = 3$$Â Â Â Â Â A1Â Â Â Â  N1

e.g. getting closer to the line $$y = 3$$ , horizontal asymptote at $$y = 3$$

[1 mark]

c.

correct inequalities, $$y \le – 2$$Â , $$y > 3$$Â , FT from (a)(i) and (c)Â Â Â Â  A1A1Â Â Â Â  N2

[2 marks]

d.

Question

In this question, you are given that $$\cos \frac{\pi }{3} = \frac{1}{2}$$Â , and $$\sin \frac{\pi }{3} = \frac{{\sqrt 3 }}{2}$$Â .

The displacement of an object from a fixed point, O is given by $$s(t) = t – \sin 2t$$ for $$0 \le t \le \pi$$ .

Find $$s'(t)$$ .

[3]
a.

In this interval, there are only two values of t for which the object is not moving. One value is $$t = \frac{\pi }{6}$$ .

Find the other value.

[4]
b.

Show that $$s'(t) > 0$$ between these two values of t .

[3]
c.

Find the distance travelled between these two values of t .

[5]
d.

Markscheme

$$s'(t) = 1 – 2\cos 2t$$Â Â Â  A1A2 Â  Â  N3

Note: Award A1 for 1, A2 for $$- 2\cos 2t$$Â .

[3 marks]

a.

evidence of valid approachÂ Â Â Â  (M1)

e.g. setting $$s'(t) = 0$$

correct workingÂ Â Â Â  A1

e.g. $$2\cos 2t = 1$$ , $$\cos 2t = \frac{1}{2}$$

$$2t = \frac{\pi }{3}$$ , $$\frac{{5\pi }}{3}$$ , $$\ldots$$Â Â Â Â  (A1)

$$t = \frac{{5\pi }}{6}$$Â Â Â Â  A1 Â  Â  N3Â

[4 marks]

b.

evidence of valid approachÂ Â Â Â  (M1)

e.g. choosing a value in the interval $$\frac{\pi }{6} < t < \frac{{5\pi }}{6}$$

correct substitutionÂ Â Â Â  A1

e.g. $$s’\left( {\frac{\pi }{2}} \right) = 1 – 2\cos \pi$$

$$s’\left( {\frac{\pi }{2}} \right) = 3$$Â Â Â Â  A1

$$s'(t) > 0$$Â Â Â Â  AG Â  Â  N0

[3 marks]

c.

evidence of approach using sÂ or integral of $$s’$$Â Â Â  Â (M1)

e.g. $$\int {s'(t){\rm{d}}t}$$ ; $$s\left( {\frac{{5\pi }}{6}} \right)$$ , $$s\left( {\frac{\pi }{6}} \right)$$ ; $$\left[ {t – \sin 2t} \right]_{\frac{\pi }{6}}^{\frac{{5\pi }}{6}}$$

substituting values and subtractingÂ Â Â Â  (M1)

e.g. $$s\left( {\frac{{5\pi }}{6}} \right) – s\left( {\frac{\pi }{6}} \right)$$ , $$\left( {\frac{\pi }{6} – \frac{{\sqrt 3 }}{2}} \right) – \left( {\frac{{5\pi }}{6} – \left( { – \frac{{\sqrt 3 }}{2}} \right)} \right)$$

correct substitutionÂ Â Â Â  A1

e.g.Â $$\frac{{5\pi }}{6} – \sin \frac{{5\pi }}{3} – \left[ {\frac{\pi }{6} – \sin \frac{\pi }{3}} \right]$$ , $$\left( {\frac{{5\pi }}{6} – \left( { – \frac{{\sqrt 3 }}{2}} \right)} \right) – \left( {\frac{\pi }{6} – \frac{{\sqrt 3 }}{2}} \right)$$

distance is $$\frac{{2\pi }}{3} + \sqrt 3$$Â Â Â Â Â A1A1Â Â Â Â  N3

Note: Award A1 for $$\frac{{2\pi }}{3}$$Â , A1 forÂ $$\sqrt 3$$ .

[5 marks]

d.

Question

Let $$f(x) = \frac{{6x}}{{x + 1}}$$ , for $$x > 0$$ .

Find $$f'(x)$$ .

[5]
a.

LetÂ $$g(x) = \ln \left( {\frac{{6x}}{{x + 1}}} \right)$$ , for $$x > 0$$Â .

Show that $$g'(x) = \frac{1}{{x(x + 1)}}$$ .

[4]
b.

LetÂ $$h(x) = \frac{1}{{x(x + 1)}}$$ . The area enclosed by the graph of h , the x-axis and the lines $$x = \frac{1}{5}$$Â  and $$x = k$$ is $$\ln 4$$ . Given that $$k > \frac{1}{5}$$Â , find the value of k .

[7]
c.

Markscheme

METHOD 1

evidence of choosing quotient ruleÂ Â Â Â  (M1)

e.g. $$\frac{{u’v – uv’}}{{{v^2}}}$$

evidence of correct differentiation (must be seen in quotient rule)Â Â Â  Â (A1)(A1)

e.g. $$\frac{{\rm{d}}}{{{\rm{d}}x}}(6x) = 6$$ , $$\frac{{\rm{d}}}{{{\rm{d}}x}}(x + 1) = 1$$

correct substitution into quotient ruleÂ Â Â Â  A1

e.g. $$\frac{{(x + 1)6 – 6x}}{{{{(x + 1)}^2}}}$$ , $$\frac{{6x + 6 – 6x}}{{{{(x + 1)}^2}}}$$

$$f'(x) = \frac{6}{{{{(x + 1)}^2}}}$$Â Â Â  A1Â Â Â Â  N4

[5 marks]

METHOD 2

evidence of choosing product ruleÂ Â Â Â  (M1)

e.g. $$6x{(x + 1)^{ – 1}}$$ , $$uv’ + vu’$$

evidence of correct differentiation (must be seen in product rule)Â Â Â  Â (A1)(A1)

e.g. $$\frac{{\rm{d}}}{{{\rm{d}}x}}(6x) = 6$$ , $$\frac{{\rm{d}}}{{{\rm{d}}x}}{(x + 1)^{ – 1}} = – 1{(x + 1)^{ – 2}} \times 1$$

correct workingÂ Â Â Â  A1

e.g. $$6x \times – {(x + 1)^{ – 2}} + {(x + 1)^{ – 1}} \times 6$$ , $$\frac{{ – 6x + 6(x + 1)}}{{{{(x + 1)}^2}}}$$

$$f'(x) = \frac{6}{{{{(x + 1)}^2}}}$$Â Â  A1Â Â Â Â  N4

[5 marks]

a.

METHOD 1

evidence of choosing chain ruleÂ Â Â Â  (M1)

e.g. formula, $$\frac{1}{{\left( {\frac{{6x}}{{x + 1}}} \right)}} \times \left( {\frac{{6x}}{{x + 1}}} \right)$$

correct reciprocal of $$\frac{1}{{\left( {\frac{{6x}}{{x + 1}}} \right)}}$$Â is $$\frac{{x + 1}}{{6x}}$$Â (seen anywhere)Â Â Â Â  A1

correct substitution into chain ruleÂ Â Â Â  A1

e.g. $$\frac{1}{{\left( {\frac{{6x}}{{x + 1}}} \right)}} \times \frac{6}{{{{(x + 1)}^2}}}$$ , $$\left( {\frac{6}{{{{(x + 1)}^2}}}} \right)\left( {\frac{{x + 1}}{{6x}}} \right)$$

e.g. $$\left( {\frac{6}{{(x + 1)}}} \right)\left( {\frac{1}{{6x}}} \right)$$ , $$\left( {\frac{1}{{{{(x + 1)}^2}}}} \right)\left( {\frac{{x + 1}}{x}} \right)$$ , $$\frac{{6(x + 1)}}{{6x{{(x + 1)}^2}}}$$

$$g'(x) = \frac{1}{{x(x + 1)}}$$Â Â Â  Â AGÂ Â Â Â  N0

[4 marks]

METHOD 2

attempt to subtract logsÂ Â Â Â  (M1)

e.g. $$\ln a – \ln b$$ , $$\ln 6x – \ln (x + 1)$$

correct derivatives (must be seen in correct expression)Â Â Â Â  A1A1

e.g. $$\frac{6}{{6x}} – \frac{1}{{x + 1}}$$ , $$\frac{1}{x} – \frac{1}{{x + 1}}$$

e.g. $$\frac{{x + 1 – x}}{{x(x + 1)}}$$ , $$\frac{{6x + 6 – 6x}}{{6x(x + 1)}}$$ , $$\frac{{6(x + 1 – x)}}{{6x(x + 1)}}$$

$$g'(x) = \frac{1}{{x(x + 1)}}$$Â Â Â  Â AGÂ Â Â Â  N0

[4 marks]

b.

valid method using integral ofÂ  h(x) (accept missing/incorrect limits or missing $${\text{d}}x$$ )Â Â Â Â  (M1)

e.g. $${\rm{area}} = \int_{\frac{1}{5}}^k {h(x){\rm{d}}x}$$ , $$\int{\left( {\frac{1}{{x(x + 1)}}} \right)}$$Â

recognizing that integral of derivative will give original functionÂ Â Â Â  (R1)

e.g. $$\int{\left( {\frac{1}{{x(x + 1)}}} \right)} {\rm{d}}x = \ln \left( {\frac{{6x}}{{x + 1}}} \right)$$

correct substitution and subtractionÂ Â Â Â  A1

e.g. $$\ln \left( {\frac{{6k}}{{k + 1}}} \right) – \ln \left( {\frac{{6 \times \frac{1}{5}}}{{\frac{1}{5} + 1}}} \right)$$ , $$\ln \left( {\frac{{6k}}{{k + 1}}} \right) – \ln (1)$$

setting their expression equal to $$\ln 4$$ Â Â Â  (M1)Â

e.g. $$\ln \left( {\frac{{6k}}{{k + 1}}} \right) – \ln (1) = \ln 4$$ , $$\ln \left( {\frac{{6k}}{{k + 1}}} \right) = \ln 4$$ , $$\int_{\frac{1}{5}}^k {h(x){\rm{d}}x = \ln 4}$$

correct equation without logsÂ Â Â Â  A1

e.g.$$\frac{{6k}}{{k + 1}} = 4$$ , $$6k = 4(k + 1)$$Â

correct workingÂ Â Â Â  (A1)

e.g. $$6k = 4k + 4$$ , $$2k = 4$$

$$k = 2$$Â Â Â  A1 Â  Â  N4

[7 marks]

c.

Question

LetÂ $$f(x) = \sin x + \frac{1}{2}{x^2} – 2x$$ , forÂ $$0 \le x \le \pi$$ .

Let $$g$$ be a quadratic function such that $$g(0) = 5$$ . The line $$x = 2$$ is the axis of symmetry of the graph of $$g$$ .

The function $$g$$ can be expressed in the form $$g(x) = a{(x – h)^2} + 3$$ .

Find $$f'(x)$$ .

[3]
a.

FindÂ $$g(4)$$ .

[3]
b.

(i)Â Â Â Â  Write down the value of $$h$$ .

(ii)Â Â Â Â  Find the value of $$a$$ .

[4]
c.

Find the value of $$x$$ for which the tangent to the graph of $$f$$ is parallel to the tangent to the graph of $$g$$ .

[6]
d.

Markscheme

$$f'(x) = \cos x + x – 2$$Â Â Â Â  A1A1A1Â Â Â Â  N3

Note: Award A1 for each term.

[3 marks]

a.

recognizing $$g(0) = 5$$Â gives the point ($$0$$, $$5$$)Â Â Â  Â (R1)

recognize symmetryÂ Â Â Â  (M1)

eg vertex, sketch

$$g(4) = 5$$Â Â Â Â  A1Â Â Â Â  N3

[3 marks]

b.

(i)Â Â Â Â  $$h = 2$$Â Â Â Â  A1 N1

(ii)Â Â Â Â  substituting intoÂ $$g(x) = a{(x – 2)^2} + 3$$Â (not the vertex)Â Â Â Â  (M1)

eg Â  $$5 = a{(0 – 2)^2} + 3$$ , $$5 = a{(4 – 2)^2} + 3$$

working towards solutionÂ Â Â Â  (A1)

eg Â  $$5 = 4a + 3$$ , $$4a = 2$$

$$a = \frac{1}{2}$$Â Â Â Â  A1Â Â Â Â  N2

[4 marks]

c.

$$g(x) = \frac{1}{2}{(x – 2)^2} + 3 = \frac{1}{2}{x^2} – 2x + 5$$

correct derivative of $$g$$Â Â Â Â  A1A1

eg Â  $$2 \times \frac{1}{2}(x – 2)$$ , $$x – 2$$

evidence of equating both derivativesÂ Â Â Â  (M1)

eg Â  $$f’ = g’$$

correct equationÂ Â Â Â  (A1)

eg Â  $$\cos x + x – 2 = x – 2$$

working towards a solutionÂ Â Â Â  (A1)

egÂ Â  $$\cos x = 0$$Â , combining like terms

$$x = \frac{\pi }{2}$$Â Â Â  A1Â Â Â Â  N0

Note: Do not award final A1 if additional values are given.

[6 marks]

d.

Question

Let $$f(x) = p{x^3} + p{x^2} + qx$$.

Find $$f'(x)$$.

[2]
a.

Given that $$f'(x) \geqslant 0$$, show that $${p^2} \leqslant 3pq$$.

[5]
b.

Markscheme

$$f'(x) = 3p{x^2} + 2px + q$$ Â  Â Â A2 Â  Â  N2

Â

Note: Â  Â  Award A1 if only 1 error.

Â

[2 marks]

a.

evidence of discriminant (must be seen explicitly, not in quadratic formula) Â  Â Â (M1)

egÂ  Â  Â $${b^2} – 4ac$$

correct substitution into discriminant (may be seen in inequality) Â  Â Â A1

egÂ  Â  Â $${(2p)^2} – 4 \times 3p \times q,{\text{ }}4{p^2} – 12pq$$

$$f'(x) \geqslant 0$$ then $$f’$$Â has two equal roots or no roots Â  Â Â (R1)

recognizing discriminant less or equal than zero Â  Â Â R1

egÂ  Â  Â $$\Delta Â \leqslant 0,{\text{ }}4{p^2} – 12pq \leqslant 0$$

correct working that clearly leads to the required answer Â  Â Â A1

egÂ Â  Â  $${p^2} – 3pq \leqslant 0,{\text{ }}4{p^2} \leqslant 12pq$$

$${p^2} \leqslant 3pq$$ Â  Â Â AG Â  Â  N0

[5 marks]

b.

Question

Let $$f(x) = 3 + \frac{{20}}{{{x^2} – 4}}$$ , for $$x \ne \pm 2$$ . The graph of f is given below.

The y-intercept is at the point A.

(i) Â  Â  Find the coordinates of A.

(ii)Â Â Â  Show that $$f'(x) = 0$$ at A.

[7]
a.

The second derivative $$f”(x) = \frac{{40(3{x^2} + 4)}}{{{{({x^2} – 4)}^3}}}$$Â . Use this to

(i)Â Â Â Â  justify that the graph of f has a local maximum at A;

(ii)Â Â Â  explain why the graph of f does not have a point of inflexion.

[6]
b.

Describe the behaviour of the graph of $$f$$ for large $$|x|$$ .

[1]
c.

Write down the range of $$f$$ .

[2]
d.

Markscheme

(i) coordinates of A are $$(0{\text{, }} – 2)$$ Â Â Â  A1A1Â Â Â Â  N2

(ii) derivative of $${x^2} – 4 = 2x$$ (seen anywhere)Â Â Â Â  (A1)

evidence of correct approachÂ Â Â Â  (M1)

e.g. quotient rule, chain rule

finding $$f'(x)$$Â Â Â Â Â A2

e.g. $$f'(x) = 20 \times ( – 1) \times {({x^2} – 4)^{ – 2}} \times (2x)$$ , $$\frac{{({x^2} – 4)(0) – (20)(2x)}}{{{{({x^2} – 4)}^2}}}$$

substituting $$x = 0$$ into $$f'(x)$$ (do not accept solving $$f'(x) = 0$$ ) Â Â Â  M1

at A $$f'(x) = 0$$Â Â Â Â Â AGÂ Â Â Â  N0

[7 marks]

a.

(i) reference to $$f'(x) = 0$$Â (seen anywhere)Â Â Â Â  (R1)

reference to $$f”(0)$$Â is negative (seen anywhere)Â Â Â Â  R1

evidence of substituting $$x = 0$$Â into $$f”(x)$$Â Â Â Â Â M1

finding $$f”(0) = \frac{{40 \times 4}}{{{{( – 4)}^3}}}$$ $$\left( { = – \frac{5}{2}} \right)$$Â Â Â Â Â A1

then the graph must have a local maximumÂ Â Â Â  AG

(ii) reference to $$f”(x) = 0$$Â at point of inflexionÂ Â Â Â  (R1)

recognizing that the second derivative is never 0Â Â Â Â  A1Â Â Â Â  N2

e.g. $$40(3{x^2} + 4) \ne 0$$Â , $$3{x^2} + 4 \ne 0$$Â , $${x^2} \neÂ – \frac{4}{3}$$Â , the numeratorÂ is always positive

Note: Do not accept the use of the first derivative in part (b).

[6 marks]

b.

correct (informal) statement, including reference to approaching $$y = 3$$Â Â Â Â Â A1Â Â Â Â  N1

e.g. getting closer to the line $$y = 3$$ , horizontal asymptote at $$y = 3$$

[1 mark]

c.

correct inequalities, $$y \le – 2$$Â , $$y > 3$$Â , FT from (a)(i) and (c)Â Â Â Â  A1A1Â Â Â Â  N2

[2 marks]

d.