IB Math Analysis & Approaches Question bank-Topic: SL 5.12 Derivative of f(x) SL Paper 1

Question

Let y = \(\frac{Inx}{x^{4}}\) for x > 0.

(a)        Show that \(\frac{dy}{dx}= \frac{1-4Inx}{x^{5}}\)

Consider the function defined  by f (x) \(\frac{Inx}{x^{4}}\) =  for x> 0 and its graph y = f (x) .

(b)        The graph of has a horizontal tangent at point P. Find the coordinates of P.                                                      [5]

(c)        Given that f ” (x) = \(\frac{20Lnx-9}{x^{6}}\) show that P is a local maximum point.                                                [3]

(d)        Solve f (x) > 0 for x > 0.                                                                                                                                                          [2]

(e) Sketch the graph of f , showing clearly the value of the x-intercept and the approximate position of point P.    [3]

Answer/Explanation

Ans

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.
Answer/Explanation

Markscheme

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

[3 marks]

a.

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

evidence of correct working     A1

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

\(x = – 5\), \(x = 1\)     (A1)

so \(x = – 5\)     A1     N2

[4 marks]

b.

METHOD 1

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

evidence of setting second derivative = 0     (M1)

e.g. \(2x + 4 = 0\)

\(x = – 2\)     A1     N2

METHOD 2

evidence of use of symmetry     (M1)

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

correct calculation     A1

e.g. \(\frac{{ – 5 + 1}}{2}\)

\(x = – 2\)     A1     N2

[3 marks]

c.

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

\(f'(3) = 16\)     A1

valid approach to finding the equation of a line     M1

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

\(y = 16x – 36\)     A1     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 \(0 \le \theta  \le \frac{\pi }{2}\) .

Write down an expression in terms of \(\theta \) for

(i)     \(x\) ;

(ii)    \(y\) .

[2]
a.

Let the area of the rectangle be A.

Show that \(A = 18\sin 2\theta \) .

[3]
b.

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

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

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

[8]
c.
Answer/Explanation

Markscheme

(i) \(x = 3\cos \theta \)     A1     N1 

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

[2 marks]

a.

finding area     (M1)

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

substituting     A1

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

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

\(A = 18\sin 2\theta \)     AG     N0

[3 marks]

b.

(i) \(\frac{{{\rm{d}}A}}{{{\rm{d}}\theta }} = 36\cos 2\theta \)     A2     N2 

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

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

\(2\theta  = \frac{\pi }{2}\)     (A1)

\(\theta  = \frac{\pi }{4}\)     A1     N2

(iii) valid reason (seen anywhere)     R1

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

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

evidence of substituting \(\frac{\pi }{4}\)     M1

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

\(\theta  = \frac{\pi }{4}\) produces the maximum area     AG     N0

[8 marks]

c.
 

Question

Let \(f(x) = {{\rm{e}}^{ – 3x}}\) and \(g(x) = \sin \left( {x – \frac{\pi }{3}} \right)\) .

Write down

(i)     \(f'(x)\) ;

(ii)    \(g'(x)\) .

[2]
a.

Let \(h(x) = {{\rm{e}}^{ – 3x}}\sin \left( {x – \frac{\pi }{3}} \right)\) . Find the exact value of \(h’\left( {\frac{\pi }{3}} \right)\) .

[4]
b.
Answer/Explanation

Markscheme

(i) \( – 3{{\rm{e}}^{ – 3x}}\)     A1     N1

(ii) \(\cos \left( {x – \frac{\pi }{3}} \right)\)     A1     N1

[4 marks]

a.

evidence of choosing product rule     (M1)

e.g. \(uv’ + vu’\)

correct expression     A1

e.g. \( – 3{{\rm{e}}^{ – 3x}}\sin \left( {x – \frac{\pi }{3}} \right) + {{\rm{e}}^{ – 3x}}\cos \left( {x – \frac{\pi }{3}} \right)\)

complete correct substitution of \(x = \frac{\pi }{3}\)     (A1)

e.g. \( – 3{{\rm{e}}^{ – 3\frac{\pi }{3}}}\sin \left( {\frac{\pi }{3} – \frac{\pi }{3}} \right) + {{\rm{e}}^{ – 3\frac{\pi }{3}}}\cos \left( {\frac{\pi }{3} – \frac{\pi }{3}} \right)\)        

\(h’\left( {\frac{\pi }{3}} \right) = {{\rm{e}}^{ – \pi }}\)     A1     N3

[4 marks]

b.

Question

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

Find \(f'(x)\) .

[2]
a.

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

[4]
b.
Answer/Explanation

Markscheme

\(f'(x) = 2x – \frac{p}{{{x^2}}}\)     A1A1     N2

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

[2 marks]

a.

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

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

correct equation     A1

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

\(p = – 16\)     A1     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)     Find the coordinates of A.

(ii)    Show that \(f'(x) = 0\) at A.

[7]
a.

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

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

(ii)    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.
Answer/Explanation

Markscheme

(i) coordinates of A are \((0{\text{, }} – 2)\)     A1A1     N2

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

evidence of correct approach     (M1)

e.g. quotient rule, chain rule

finding \(f'(x)\)     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 \(f'(x) = 0\) )     M1

at A \(f'(x) = 0\)     AG     N0

[7 marks]

a.

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

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

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

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

then the graph must have a local maximum     AG

(ii) reference to \(f”(x) = 0\) at point of inflexion     (R1)

recognizing that the second derivative is never 0     A1     N2

e.g. \(40(3{x^2} + 4) \ne 0\) , \(3{x^2} + 4 \ne 0\) , \({x^2} \ne  – \frac{4}{3}\) , the numerator 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 \(y = 3\)     A1     N1

e.g. getting closer to the line \(y = 3\) , horizontal asymptote at \(y = 3\)

[1 mark]

c.

correct inequalities, \(y \le – 2\) , \(y > 3\) , FT from (a)(i) and (c)     A1A1     N2

[2 marks]

d.

Question

Let \(f(x) = \frac{1}{2}{x^3} – {x^2} – 3x\) . Part of the graph of f is shown below.


There is a maximum point at A and a minimum point at B(3, − 9) .

Find the coordinates of A.

[8]
a.

Write down the coordinates of

(i)     the image of B after reflection in the y-axis;

(ii)    the image of B after translation by the vector \(\left( {\begin{array}{*{20}{c}}
{ – 2}\\
5
\end{array}} \right)\) ;

(iii)   the image of B after reflection in the x-axis followed by a horizontal stretch with scale factor \(\frac{1}{2}\) .

[6]
b(i), (ii) and (iii).
Answer/Explanation

Markscheme

\(f(x) = {x^2} – 2x – 3\)     A1A1A1

evidence of solving \(f'(x) = 0\)     (M1)

e.g. \({x^2} – 2x – 3 = 0\)

evidence of correct working     A1

e.g. \((x + 1)(x – 3)\) ,  \(\frac{{2 \pm \sqrt {16} }}{2}\)

\(x =  – 1\) (ignore \(x = 3\) )     (A1)

evidence of substituting their negative x-value into \(f(x)\)     (M1)

e.g. \(\frac{1}{3}{( – 1)^3} – {( – 1)^2} – 3( – 1)\) , \( – \frac{1}{3} – 1 + 3\)

\(y = \frac{5}{3}\)     A1

coordinates are \(\left( { – 1,\frac{5}{3}} \right)\)     N3

[8 marks]

a.

(i) \(( – 3{\text{, }} – 9)\)     A1     N1

(ii) \((1{\text{, }} – 4)\)     A1A1    N2

(iii) reflection gives \((3{\text{, }}9)\)     (A1)

stretch gives \(\left( {\frac{3}{2}{\text{, }}9} \right)\)     A1A1     N3

[6 marks]

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

Question

Let \(f(x) = k{x^4}\) . The point \({\text{P}}(1{\text{, }}k)\) lies on the curve of f . At P, the normal to the curve is parallel to \(y = – \frac{1}{8}x\) . Find the value of k.

Answer/Explanation

Markscheme

gradient of tangent \(= 8\) (seen anywhere)     (A1)

\(f'(x) = 4k{x^3}\) (seen anywhere)     A1

recognizing the gradient of the tangent is the derivative     (M1)

setting the derivative equal to 8     (A1)

e.g. \(4k{x^3} = 8\) , \(k{x^3} = 2\)

substituting \(x = 1\) (seen anywhere)     (M1)

\(k = 2\)     A1    N4

[6 marks]

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.
Answer/Explanation

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.

Question

The following diagram shows part of the graph of a quadratic function f .


The x-intercepts are at \(( – 4{\text{, }}0)\) and \((6{\text{, }}0)\) , and the y-intercept is at \((0{\text{, }}240)\) .

Write down \(f(x)\) in the form \(f(x) = – 10(x – p)(x – q)\) .

[2]
a.

Find another expression for \(f(x)\) in the form \(f(x) = – 10{(x – h)^2} + k\) .

[4]
b.

Show that \(f(x)\) can also be written in the form \(f(x) = 240 + 20x – 10{x^2}\) .

[2]
c.

A particle moves along a straight line so that its velocity, \(v{\text{ m}}{{\text{s}}^{ – 1}}\) , at time t seconds is given by \(v = 240 + 20t – 10{t^2}\) , for \(0 \le t \le 6\) .

(i)     Find the value of t when the speed of the particle is greatest.

(ii)    Find the acceleration of the particle when its speed is zero.

[7]
d(i) and (ii).
Answer/Explanation

Markscheme

\(f(x) = – 10(x + 4)(x – 6)\)     A1A1     N2

[2 marks]

a.

METHOD 1

attempting to find the x-coordinate of maximum point     (M1)

e.g. averaging the x-intercepts, sketch, \(y’ = 0\) , axis of symmetry

attempting to find the y-coordinate of maximum point     (M1)

e.g. \(k = – 10(1 + 4)(1 – 6)\)

\(f(x) = – 10{(x – 1)^2} + 250\)     A1A1     N4

METHOD 2

attempt to expand \(f(x)\)     (M1)

e.g. \( – 10({x^2} – 2x – 24)\)

attempt to complete the square     (M1)

e.g. \( – 10({(x – 1)^2} – 1 – 24)\)

\(f(x) = – 10{(x – 1)^2} + 250\)     A1A1     N4

[4 marks]

b.

attempt to simplify     (M1)

e.g. distributive property, \( – 10(x – 1)(x – 1) + 250\)

correct simplification     A1

e.g. \( – 10({x^2} – 6x + 4x – 24)\) , \( – 10({x^2} – 2x + 1) + 250\)

\(f(x) = 240 + 20x – 10{x^2}\)     AG     N0

[2 marks]

c.

(i) valid approach     (M1)

e.g. vertex of parabola, \(v'(t) = 0\)

\(t = 1\)     A1     N2

(ii) recognizing \(a(t) = v'(t)\)     (M1)

\(a(t) = 20 – 20t\)     A1A1

speed is zero \( \Rightarrow t = 6\)     (A1)

\(a(6) = – 100\) (\({\text{m}}{{\text{s}}^{ – 2}}\))     A1     N3

[7 marks]

d(i) and (ii).

Question

The following diagram shows the graph of \(f(x) = a\sin (b(x – c)) + d\) , for \(2 \le x \le 10\) .


There is a maximum point at P(4, 12) and a minimum point at Q(8, −4) .

Use the graph to write down the value of

(i)     a ;

(ii)    c ;

(iii)   d .

[3]
a(i), (ii) and (iii).

Show that \(b = \frac{\pi }{4}\) .

[2]
b.

Find \(f'(x)\) .

[3]
c.

At a point R, the gradient is \( – 2\pi \) . Find the x-coordinate of R.

[6]
d.
Answer/Explanation

Markscheme

(i) \(a = 8\)     A1     N1

(ii) \(c = 2\)     A1     N1

(iii) \(d = 4\)     A1     N1

[3 marks]

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

METHOD 1

recognizing that period \( = 8\)     (A1)

correct working     A1

e.g. \(8 = \frac{{2\pi }}{b}\) , \(b = \frac{{2\pi }}{8}\)

\(b = \frac{\pi }{4}\)     AG     N0

METHOD 2

attempt to substitute     M1

e.g. \(12 = 8\sin (b(4 – 2)) + 4\)

correct working     A1

e.g. \(\sin 2b = 1\)

\(b = \frac{\pi }{4}\)     AG     N0

[2 marks]

b.

evidence of attempt to differentiate or choosing chain rule     (M1)

e.g. \(\cos \frac{\pi }{4}(x – 2)\) , \(\frac{\pi }{4} \times 8\)

\(f'(x) = 2\pi \cos \left( {\frac{\pi }{4}(x – 2)} \right)\) (accept \(2\pi \cos \frac{\pi }{4}(x – 2)\) )     A2     N3

[3 marks]

c.

recognizing that gradient is \(f'(x)\)     (M1)

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

correct equation     A1

e.g. \( – 2\pi  = 2\pi \cos \left( {\frac{\pi }{4}(x – 2)} \right)\) , \( – 1 = \cos \left( {\frac{\pi }{4}(x – 2)} \right)\)

correct working     (A1)

e.g. \({\cos ^{ – 1}}( – 1) = \frac{\pi }{4}(x – 2)\)

using \({\cos ^{ – 1}}( – 1) = \pi \) (seen anywhere)     (A1)

e.g. \(\pi  = \frac{\pi }{4}(x – 2)\)

simplifying     (A1)

e.g. \(4 = (x – 2)\)

\(x = 6\)     A1     N4

[6 marks]

d.

Question

Let \(f(x) = {{\rm{e}}^{6x}}\) .

Write down \(f'(x)\) .

[1]
a.

The tangent to the graph of f at the point \({\text{P}}(0{\text{, }}b)\) has gradient m .

(i)     Show that \(m = 6\) .

(ii)    Find b .

[4]
b(i) and (ii).

Hence, write down the equation of this tangent.

[1]
c.
Answer/Explanation

Markscheme

\(f'(x) = 6{{\rm{e}}^{6x}}\)     A1     N1

[1 mark]

a.

(i) evidence of valid approach     (M1)

e.g. \(f'(0)\) ,  \(6{{\rm{e}}^{6 \times 0}}\)

correct manipulation     A1

e.g. \(6{{\rm{e}}^0}\) , \(6 \times 1\)

\(m = 6\)    AG     N0

(ii) evidence of finding \(f(0)\)     (M1)

e.g. \(y = {{\rm{e}}^{6(0)}}\)

\(b = 1\)     A1     N2

[4 marks]

b(i) and (ii).

\(y = 6x + 1\)     A1     N1

[1 mark]

c.

Question

Consider \(f(x) = {x^2}\sin x\) .

Find \(f'(x)\) .

[4]
a.

Find the gradient of the curve of \(f\) at \(x = \frac{\pi }{2}\) .

[3]
b.
Answer/Explanation

Markscheme

evidence of choosing product rule     (M1)

eg   \(uv’ + vu’\)

correct derivatives (must be seen in the product rule) \(\cos x\) , \(2x\)     (A1)(A1)

\(f'(x) = {x^2}\cos x + 2x\sin x\)     A1 N4

[4 marks]

a.

substituting \(\frac{\pi }{2}\) into their \(f'(x)\)     (M1)

eg   \(f’\left( {\frac{\pi }{2}} \right)\) , \({\left( {\frac{\pi }{2}} \right)^2}\cos \left( {\frac{\pi }{2}} \right) + 2\left( {\frac{\pi }{2}} \right)\sin \left( {\frac{\pi }{2}} \right)\)

correct values for both \(\sin \frac{\pi }{2}\) and \(\cos \frac{\pi }{2}\) seen in \(f'(x)\)     (A1)

eg   \(0 + 2\left( {\frac{\pi }{2}} \right) \times 1\) 

\(f’\left( {\frac{\pi }{2}} \right) = \pi \)     A1 N2

[3 marks]

b.

Question

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

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.

Find \(g(4)\) .

[3]
b.

(i)     Write down the value of \(h\) .

(ii)     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.
Answer/Explanation

Markscheme

\(f'(x) = \cos x + x – 2\)     A1A1A1     N3

Note: Award A1 for each term.

[3 marks]

a.

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

recognize symmetry     (M1)

eg vertex, sketch

\(g(4) = 5\)     A1     N3

[3 marks]

b.

(i)     \(h = 2\)     A1 N1

(ii)     substituting into \(g(x) = a{(x – 2)^2} + 3\) (not the vertex)     (M1)

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

working towards solution     (A1)

eg   \(5 = 4a + 3\) , \(4a = 2\)

\(a = \frac{1}{2}\)     A1     N2

[4 marks]

c.

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

correct derivative of \(g\)     A1A1

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

evidence of equating both derivatives     (M1)

eg   \(f’ = g’\)

correct equation     (A1)

eg   \(\cos x + x – 2 = x – 2\)

working towards a solution     (A1)

eg   \(\cos x = 0\) , combining like terms

\(x = \frac{\pi }{2}\)    A1     N0

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

[6 marks]

d.

Question

Let \(f(x) = \frac{{{{(\ln x)}^2}}}{2}\), for \(x > 0\).

Let \(g(x) = \frac{1}{x}\). The following diagram shows parts of the graphs of \(f’\) and g.

The graph of \(f’\) has an x-intercept at \(x = p\).

Show that \(f'(x) = \frac{{\ln x}}{x}\).

[2]
a.

There is a minimum on the graph of \(f\). Find the \(x\)-coordinate of this minimum.

[3]
b.

Write down the value of \(p\).

[2]
c.

The graph of \(g\) intersects the graph of \(f’\) when \(x = q\).

Find the value of \(q\).

[3]
d.

The graph of \(g\) intersects the graph of \(f’\) when \(x = q\).

Let \(R\) be the region enclosed by the graph of \(f’\), the graph of \(g\) and the line \(x = p\).

Show that the area of \(R\) is \(\frac{1}{2}\).

[5]
e.
Answer/Explanation

Markscheme

METHOD 1

correct use of chain rule     A1A1

eg     \(\frac{{2\ln x}}{2} \times \frac{1}{x},{\text{ }}\frac{{2\ln x}}{{2x}}\)

Note: Award A1 for \(\frac{{2\ln x}}{{2x}}\), A1 for \( \times \frac{1}{x}\).

\(f'(x) = \frac{{\ln x}}{x}\)     AG     N0

[2 marks]

METHOD 2

correct substitution into quotient rule, with derivatives seen     A1

eg     \(\frac{{2 \times 2\ln x \times \frac{1}{x} – 0 \times {{(\ln x)}^2}}}{4}\)

correct working     A1

eg     \(\frac{{4\ln x \times \frac{1}{x}}}{4}\)

\(f'(x) = \frac{{\ln x}}{x}\)     AG     N0

[2 marks]

a.

setting derivative \( = 0\)     (M1)

eg     \(f'(x) = 0,{\text{ }}\frac{{\ln x}}{x} = 0\)

correct working     (A1)

eg     \(\ln x = 0,{\text{ }}x = {{\text{e}}^0}\)

\(x = 1\)     A1     N2

[3 marks] 

b.

intercept when \(f'(x) = 0\)     (M1)

\(p = 1\)     A1     N2

[2 marks]

c.

equating functions     (M1)

eg     \(f’ = g,{\text{ }}\frac{{\ln x}}{x} = \frac{1}{x}\)

correct working     (A1)

eg     \(\ln x = 1\)

\(q = {\text{e   (accept }}x = {\text{e)}}\)     A1     N2

[3 marks]

d.

evidence of integrating and subtracting functions (in any order, seen anywhere)     (M1)

eg     \(\int_q^e {\left( {\frac{1}{x} – \frac{{\ln x}}{x}} \right){\text{d}}x{\text{, }}\int {f’ – g} } \)

correct integration \(\ln x – \frac{{{{(\ln x)}^2}}}{2}\)     A2

substituting limits into their integrated function and subtracting (in any order)     (M1)

eg     \((\ln {\text{e}} – \ln 1) – \left( {\frac{{{{(\ln {\text{e}})}^2}}}{2} – \frac{{{{(\ln 1)}^2}}}{2}} \right)\)

Note: Do not award M1 if the integrated function has only one term.

correct working     A1

eg     \((1 – 0) – \left( {\frac{1}{2} – 0} \right),{\text{ }}1 – \frac{1}{2}\)

\({\text{area}} = \frac{1}{2}\)     AG     N0

Notes: Candidates may work with two separate integrals, and only combine them at the end. Award marks in line with the markscheme.

[5 marks]

e.

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.
Answer/Explanation

Markscheme

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

 

Note:     Award A1 if only 1 error.

 

[2 marks]

a.

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

eg     \({b^2} – 4ac\)

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

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

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

recognizing discriminant less or equal than zero     R1

eg     \(\Delta  \leqslant 0,{\text{ }}4{p^2} – 12pq \leqslant 0\)

correct working that clearly leads to the required answer     A1

eg     \({p^2} – 3pq \leqslant 0,{\text{ }}4{p^2} \leqslant 12pq\)

\({p^2} \leqslant 3pq\)     AG     N0

[5 marks]

b.

Question

A function \(f\) has its derivative given by \(f'(x) = 3{x^2} – 2kx – 9\), where \(k\) is a constant.

Find \(f”(x)\).

[2]
a.

The graph of \(f\) has a point of inflexion when \(x = 1\).

Show that \(k = 3\).

[3]
b.

Find \(f'( – 2)\).

[2]
c.

Find the equation of the tangent to the curve of \(f\) at \(( – 2,{\text{ }}1)\), giving your answer in the form \(y = ax + b\).

[4]
d.

Given that \(f'( – 1) = 0\), explain why the graph of \(f\) has a local maximum when \(x =  – 1\).

[3]
e.
Answer/Explanation

Markscheme

\(f”(x) = 6x – 2k\)     A1A1     N2

[2 marks]

a.

substituting \(x = 1\) into \(f”\)     (M1)

eg\(\;\;\;f”(1),{\text{ }}6(1) – 2k\)

recognizing \(f”(x) = 0\;\;\;\)(seen anywhere)     M1

correct equation     A1

eg\(\;\;\;6 – 2k = 0\)

\(k = 3\)     AG     N0

[3 marks]

b.

correct substitution into \(f'(x)\)     (A1)

eg\(\;\;\;3{( – 2)^2} – 6( – 2) – 9\)

\(f'( – 2) = 15\)     A1     N2

[2 marks]

c.

recognizing gradient value (may be seen in equation)     M1

eg\(\;\;\;a = 15,{\text{ }}y = 15x + b\)

attempt to substitute \(( – 2,{\text{ }}1)\) into equation of a straight line     M1

eg\(\;\;\;1 = 15( – 2) + b,{\text{ }}(y – 1) = m(x + 2),{\text{ }}(y + 2) = 15(x – 1)\)

correct working     (A1)

eg\(\;\;\;31 = b,{\text{ }}y = 15x + 30 + 1\)

\(y = 15x + 31\)     A1     N2

[4 marks]

d.

METHOD 1 (\({{\text{2}}^{{\text{nd}}}}\) derivative)

recognizing \(f” < 0\;\;\;\)(seen anywhere)     R1

substituting \(x =  – 1\) into \(f”\)     (M1)

eg\(\;\;\;f”( – 1),{\text{ }}6( – 1) – 6\)

\(f”( – 1) =  – 12\)     A1

therefore the graph of \(f\) has a local maximum when \(x =  – 1\)     AG     N0

METHOD 2 (\({{\text{1}}^{{\text{st}}}}\) derivative)

recognizing change of sign of \(f'(x)\;\;\;\)(seen anywhere)     R1

eg\(\;\;\;\)sign chart\(\;\;\;\)

correct value of \(f’\) for \( – 1 < x < 3\)     A1

eg\(\;\;\;f'(0) =  – 9\)

correct value of \(f’\) for \(x\) value to the left of \( – 1\)     A1

eg\(\;\;\;f'( – 2) = 15\)

therefore the graph of \(f\) has a local maximum when \(x =  – 1\)     AG     N0

[3 marks]

Total [14 marks]

e.

Question

Let \(f'(x) = \frac{{6 – 2x}}{{6x – {x^2}}}\), for \(0 < x < 6\).

The graph of \(f\) has a maximum point at P.

The \(y\)-coordinate of P is \(\ln 27\).

Find the \(x\)-coordinate of P.

[3]
a.

Find \(f(x)\), expressing your answer as a single logarithm.

[8]
b.

The graph of \(f\) is transformed by a vertical stretch with scale factor \(\frac{1}{{\ln 3}}\). The image of P under this transformation has coordinates \((a,{\text{ }}b)\).

Find the value of \(a\) and of \(b\), where \(a,{\text{ }}b \in \mathbb{N}\).

[[N/A]]
c.
Answer/Explanation

Markscheme

recognizing \(f'(x) = 0\)     (M1)

correct working     (A1)

eg\(\,\,\,\,\,\)\(6 – 2x = 0\)

\(x = 3\)    A1     N2

[3 marks]

a.

evidence of integration     (M1)

eg\(\,\,\,\,\,\)\(\int {f’,{\text{ }}\int {\frac{{6 – 2x}}{{6x – {x^2}}}{\text{d}}x} } \)

using substitution     (A1)

eg\(\,\,\,\,\,\)\(\int {\frac{1}{u}{\text{d}}u} \) where \(u = 6x – {x^2}\)

correct integral     A1

eg\(\,\,\,\,\,\)\(\ln (u) + c,{\text{ }}\ln (6x – {x^2})\)

substituting \((3,{\text{ }}\ln 27)\) into their integrated expression (must have \(c\))     (M1)

eg\(\,\,\,\,\,\)\(\ln (6 \times 3 – {3^2}) + c = \ln 27,{\text{ }}\ln (18 – 9) + \ln k = \ln 27\)

correct working     (A1)

eg\(\,\,\,\,\,\)\(c = \ln 27 – \ln 9\)

EITHER

\(c = \ln 3\)    (A1)

attempt to substitute their value of \(c\) into \(f(x)\)     (M1)

eg\(\,\,\,\,\,\)\(f(x) = \ln (6x – {x^2}) + \ln 3\)     A1     N4

OR

attempt to substitute their value of \(c\) into \(f(x)\)     (M1)

eg\(\,\,\,\,\,\)\(f(x) = \ln (6x – {x^2}) + \ln 27 – \ln 9\)

correct use of a log law     (A1)

eg\(\,\,\,\,\,\)\(f(x) = \ln (6x – {x^2}) + \ln \left( {\frac{{27}}{9}} \right),{\text{ }}f(x) = \ln \left( {27(6x – {x^2})} \right) – \ln 9\)

\(f(x) = \ln \left( {3(6x – {x^2})} \right)\)    A1     N4

[8 marks]

b.

\(a = 3\)    A1     N1

correct working     A1

eg\(\,\,\,\,\,\)\(\frac{{\ln 27}}{{\ln 3}}\)

correct use of log law     (A1)

eg\(\,\,\,\,\,\)\(\frac{{3\ln 3}}{{\ln 3}},{\text{ }}{\log _3}27\)

\(b = 3\)    A1     N2

[4 marks]

c.

Question

Fred makes an open metal container in the shape of a cuboid, as shown in the following diagram.

M16/5/MATME/SP1/ENG/TZ2/09

The container has height \(x{\text{ m}}\), width \(x{\text{ m}}\) and length \(y{\text{ m}}\). The volume is \(36{\text{ }}{{\text{m}}^3}\).

Let \(A(x)\) be the outside surface area of the container.

Show that \(A(x) = \frac{{108}}{x} + 2{x^2}\).

[4]
a.

Find \(A'(x)\).

[2]
b.

Given that the outside surface area is a minimum, find the height of the container.

[5]
c.

Fred paints the outside of the container. A tin of paint covers a surface area of \({\text{10 }}{{\text{m}}^{\text{2}}}\) and costs $20. Find the total cost of the tins needed to paint the container.

[5]
d.
Answer/Explanation

Markscheme

correct substitution into the formula for volume     A1

eg\(\,\,\,\,\,\)\(36 = y \times x \times x\)

valid approach to eliminate \(y\) (may be seen in formula/substitution)     M1

eg\(\,\,\,\,\,\)\(y = \frac{{36}}{{{x^2}}},{\text{ }}xy = \frac{{36}}{x}\)

correct expression for surface area     A1

eg\(\,\,\,\,\,\)\(xy + xy + xy + {x^2} + {x^2},{\text{ area}} = 3xy + 2{x^2}\)

correct expression in terms of \(x\) only     A1

eg\(\,\,\,\,\,\)\(3x\left( {\frac{{36}}{{{x^2}}}} \right) + 2{x^2},{\text{ }}{x^2} + {x^2} + \frac{{36}}{x} + \frac{{36}}{x} + \frac{{36}}{x},{\text{ }}2{x^2} + 3\left( {\frac{{36}}{x}} \right)\)

\(A(x) = \frac{{108}}{x} + 2{x^2}\)    AG     N0

[4 marks]

a.

\(A'(x) =  – \frac{{108}}{{{x^2}}} + 4x,{\text{ }}4x – 108{x^{ – 2}}\)    A1A1     N2

Note:     Award A1 for each term.

[2 marks]

b.

recognizing that minimum is when \(A'(x) = 0\)     (M1)

correct equation     (A1)

eg\(\,\,\,\,\,\)\( – \frac{{108}}{{{x^2}}} + 4x = 0,{\text{ }}4x = \frac{{108}}{{{x^2}}}\)

correct simplification     (A1)

eg\(\,\,\,\,\,\)\( – 108 + 4{x^3} = 0,{\text{ }}4{x^3} = 108\)

correct working     (A1)

eg\(\,\,\,\,\,\)\({x^3} = 27\)

\({\text{height}} = 3{\text{ (m) }}({\text{accept }}x = 3)\)    A1     N2

[5 marks]

c.

attempt to find area using their height     (M1)

eg\(\,\,\,\,\,\)\(\frac{{108}}{3} + 2{(3)^2},{\text{ }}9 + 9 + 12 + 12 + 12\)

minimum surface area \( = 54{\text{ }}{{\text{m}}^{\text{2}}}\) (may be seen in part (c))     A1

attempt to find the number of tins     (M1)

eg\(\,\,\,\,\,\)\(\frac{{54}}{{10}},{\text{ }}5.4\)

6 (tins)     (A1)

$120     A1     N3

[5 marks]

d.

Question

Let \(f(x) = \cos x\).

Let \(g(x) = {x^k}\), where \(k \in {\mathbb{Z}^ + }\).

Let \(k = 21\) and \(h(x) = \left( {{f^{(19)}}(x) \times {g^{(19)}}(x)} \right)\).

(i)     Find the first four derivatives of \(f(x)\).

(ii)     Find \({f^{(19)}}(x)\).

[4]
a.

(i)     Find the first three derivatives of \(g(x)\).

(ii)     Given that \({g^{(19)}}(x) = \frac{{k!}}{{(k – p)!}}({x^{k – 19}})\), find \(p\).

[5]
b.

(i)     Find \(h'(x)\).

(ii)     Hence, show that \(h'(\pi ) = \frac{{ – 21!}}{2}{\pi ^2}\).

[7]
c.
Answer/Explanation

Markscheme

(i)     \(f'(x) =  – \sin x,{\text{ }}f”(x) =  – \cos x,{\text{ }}{f^{(3)}}(x) = \sin x,{\text{ }}{f^{(4)}}(x) = \cos x\)     A2     N2

(ii)     valid approach     (M1)

eg\(\,\,\,\,\,\)recognizing that 19 is one less than a multiple of 4, \({f^{(19)}}(x) = {f^{(3)}}(x)\)

\({f^{(19)}}(x) = \sin x\)     A1     N2

[4 marks]

a.

(i)     \(g'(x) = k{x^{k – 1}}\)

\(g”(x) = k(k – 1){x^{k – 2}},{\text{ }}{g^{(3)}}(x) = k(k – 1)(k – 2){x^{k – 3}}\)     A1A1     N2

(ii)     METHOD 1

correct working that leads to the correct answer, involving the correct expression for the 19th derivative     A2

eg\(\,\,\,\,\,\)\(k(k – 1)(k – 2) \ldots (k – 18) \times \frac{{(k – 19)!}}{{(k – 19)!}},{{\text{ }}_k}{P_{19}}\)

\(p = 19\) (accept \(\frac{{k!}}{{(k – 19)!}}{x^{k – 19}}\))     A1     N1

METHOD 2

correct working involving recognizing patterns in coefficients of first three derivatives (may be seen in part (b)(i)) leading to a general rule for 19th coefficient     A2

eg\(\,\,\,\,\,\)\(g” = 2!\left( {\begin{array}{*{20}{c}} k \\ 2 \end{array}} \right),{\text{ }}k(k – 1)(k – 2) = \frac{{k!}}{{(k – 3)!}},{\text{ }}{g^{(3)}}(x){ = _k}{P_3}({x^{k – 3}})\)

\({g^{(19)}}(x) = 19!\left( {\begin{array}{*{20}{c}} k \\ {19} \end{array}} \right),{\text{ }}19! \times \frac{{k!}}{{(k – 19)! \times 19!}},{{\text{ }}_k}{P_{19}}\)

\(p = 19\) (accept \(\frac{{k!}}{{(k – 19)!}}{x^{k – 19}}\))     A1     N1

[5 marks]

b.

(i)     valid approach using product rule     (M1)

eg\(\,\,\,\,\,\)\(uv’ + vu’,{\text{ }}{f^{(19)}}{g^{(20)}} + {f^{(20)}}{g^{(19)}}\)

correct 20th derivatives (must be seen in product rule)     (A1)(A1)

eg\(\,\,\,\,\,\)\({g^{(20)}}(x) = \frac{{21!}}{{(21 – 20)!}}x,{\text{ }}{f^{(20)}}(x) = \cos x\)

\(h'(x) = \sin x(21!x) + \cos x\left( {\frac{{21!}}{2}{x^2}} \right){\text{ }}\left( {{\text{accept }}\sin x\left( {\frac{{21!}}{{1!}}x} \right) + \cos x\left( {\frac{{21!}}{{2!}}{x^2}} \right)} \right)\)    A1     N3

(ii)     substituting \(x = \pi \) (seen anywhere)     (A1)

eg\(\,\,\,\,\,\)\({f^{(19)}}(\pi ){g^{(20)}}(\pi ) + {f^{(20)}}(\pi ){g^{(19)}}(\pi ),{\text{ }}\sin \pi \frac{{21!}}{{1!}}\pi  + \cos \pi \frac{{21!}}{{2!}}{\pi ^2}\)

evidence of one correct value for \(\sin \pi \) or \(\cos \pi \) (seen anywhere)     (A1)

eg\(\,\,\,\,\,\)\(\sin \pi  = 0,{\text{ }}\cos \pi  =  – 1\)

evidence of correct values substituted into \(h'(\pi )\)     A1

eg\(\,\,\,\,\,\)\(21!(\pi )\left( {0 – \frac{\pi }{{2!}}} \right),{\text{ }}21!(\pi )\left( { – \frac{\pi }{2}} \right),{\text{ }}0 + ( – 1)\frac{{21!}}{2}{\pi ^2}\)

Note: If candidates write only the first line followed by the answer, award A1A0A0.

\(\frac{{ – 21!}}{2}{\pi ^2}\)     AG     N0

[7 marks]

c.

Question

Let \(f(x) = {x^2}\). The following diagram shows part of the graph of \(f\).

M17/5/MATME/SP1/ENG/TZ2/10

The line \(L\) is the tangent to the graph of \(f\) at the point \({\text{A}}( – k,{\text{ }}{k^2})\), and intersects the \(x\)-axis at point B. The point C is \(( – k,{\text{ }}0)\).

The region \(R\) is enclosed by \(L\), the graph of \(f\), and the \(x\)-axis. This is shown in the following diagram.

M17/5/MATME/SP1/ENG/TZ2/10.d

Write down \(f'(x)\).

[1]
a.i.

Find the gradient of \(L\).

[2]
a.ii.

Show that the \(x\)-coordinate of B is \( – \frac{k}{2}\).

[5]
b.

Find the area of triangle ABC, giving your answer in terms of \(k\).

[2]
c.

Given that the area of triangle ABC is \(p\) times the area of \(R\), find the value of \(p\).

[7]
d.
Answer/Explanation

Markscheme

\(f'(x) = 2x\)     A1     N1

[1 mark]

a.i.

attempt to substitute \(x =  – k\) into their derivative     (M1)

gradient of \(L\) is \( – 2k\)     A1     N2

[2 marks]

a.ii.

METHOD 1 

attempt to substitute coordinates of A and their gradient into equation of a line     (M1)

eg\(\,\,\,\,\,\)\({k^2} =  – 2k( – k) + b\)

correct equation of \(L\) in any form     (A1)

eg\(\,\,\,\,\,\)\(y – {k^2} =  – 2k(x + k),{\text{ }}y =  – 2kx – {k^2}\)

valid approach     (M1)

eg\(\,\,\,\,\,\)\(y = 0\)

correct substitution into \(L\) equation     A1

eg\(\,\,\,\,\,\)\( – {k^2} =  – 2kx – 2{k^2},{\text{ }}0 =  – 2kx – {k^2}\)

correct working     A1

eg\(\,\,\,\,\,\)\(2kx =  – {k^2}\)

\(x =  – \frac{k}{2}\)     AG     N0

METHOD 2

valid approach     (M1)

eg\(\,\,\,\,\,\)\({\text{gradient}} = \frac{{{y_2} – {y_1}}}{{{x_2} – {x_1}}},{\text{ }} – 2k = \frac{{{\text{rise}}}}{{{\text{run}}}}\)

recognizing \(y = 0\) at B     (A1)

attempt to substitute coordinates of A and B into slope formula     (M1)

eg\(\,\,\,\,\,\)\(\frac{{{k^2} – 0}}{{ – k – x}},{\text{ }}\frac{{ – {k^2}}}{{x + k}}\)

correct equation     A1

eg\(\,\,\,\,\,\)\(\frac{{{k^2} – 0}}{{ – k – x}} =  – 2k,{\text{ }}\frac{{ – {k^2}}}{{x + k}} =  – 2k,{\text{ }} – {k^2} =  – 2k(x + k)\)

correct working     A1

eg\(\,\,\,\,\,\)\(2kx =  – {k^2}\)

\(x =  – \frac{k}{2}\)     AG     N0

[5 marks]

b.

valid approach to find area of triangle     (M1)

eg\(\,\,\,\,\,\)\(\frac{1}{2}({k^2})\left( {\frac{k}{2}} \right)\)

area of \({\text{ABC}} = \frac{{{k^3}}}{4}\)     A1     N2

[2 marks]

c.

METHOD 1 (\(\int {f – {\text{triangle}}} \))

valid approach to find area from \( – k\) to 0     (M1)

eg\(\,\,\,\,\,\)\(\int_{ – k}^0 {{x^2}{\text{d}}x,{\text{ }}\int_0^{ – k} f } \)

correct integration (seen anywhere, even if M0 awarded)     A1

eg\(\,\,\,\,\,\)\(\frac{{{x^3}}}{3},{\text{ }}\left[ {\frac{1}{3}{x^3}} \right]_{ – k}^0\)

substituting their limits into their integrated function and subtracting     (M1)

eg\(\,\,\,\,\,\)\(0 – \frac{{{{( – k)}^3}}}{3}\), area from \( – k\) to 0 is \(\frac{{{k^3}}}{3}\)

Note:     Award M0 for substituting into original or differentiated function.

attempt to find area of \(R\)     (M1)

eg\(\,\,\,\,\,\)\(\int_{ – k}^0 {f(x){\text{d}}x – {\text{ triangle}}} \)

correct working for \(R\)     (A1)

eg\(\,\,\,\,\,\)\(\frac{{{k^3}}}{3} – \frac{{{k^3}}}{4},{\text{ }}R = \frac{{{k^3}}}{{12}}\)

correct substitution into \({\text{triangle}} = pR\)     (A1)

eg\(\,\,\,\,\,\)\(\frac{{{k^3}}}{4} = p\left( {\frac{{{k^3}}}{3} – \frac{{{k^3}}}{4}} \right),{\text{ }}\frac{{{k^3}}}{4} = p\left( {\frac{{{k^3}}}{{12}}} \right)\)

\(p = 3\)     A1     N2

METHOD 2 (\(\int {(f – L)} \))

valid approach to find area of \(R\)     (M1)

eg\(\,\,\,\,\,\)\(\int_{ – k}^{ – \frac{k}{2}} {{x^2} – ( – 2kx – {k^2}){\text{d}}x + \int_{ – \frac{k}{2}}^0 {{x^2}{\text{d}}x,{\text{ }}\int_{ – k}^{ – \frac{k}{2}} {(f – L) + \int_{ – \frac{k}{2}}^0 f } } } \)

correct integration (seen anywhere, even if M0 awarded)     A2

eg\(\,\,\,\,\,\)\(\frac{{{x^3}}}{3} + k{x^2} + {k^2}x,{\text{ }}\left[ {\frac{{{x^3}}}{3} + k{x^2} + {k^2}x} \right]_{ – k}^{ – \frac{k}{2}} + \left[ {\frac{{{x^3}}}{3}} \right]_{ – \frac{k}{2}}^0\)

substituting their limits into their integrated function and subtracting     (M1)

eg\(\,\,\,\,\,\)\(\left( {\frac{{{{\left( { – \frac{k}{2}} \right)}^3}}}{3} + k{{\left( { – \frac{k}{2}} \right)}^2} + {k^2}\left( { – \frac{k}{2}} \right)} \right) – \left( {\frac{{{{( – k)}^3}}}{3} + k{{( – k)}^2} + {k^2}( – k)} \right) + (0) – \left( {\frac{{{{\left( { – \frac{k}{2}} \right)}^3}}}{3}} \right)\)

Note:     Award M0 for substituting into original or differentiated function.

correct working for \(R\)     (A1)

eg\(\,\,\,\,\,\)\(\frac{{{k^3}}}{{24}} + \frac{{{k^3}}}{{24}},{\text{ }} – \frac{{{k^3}}}{{24}} + \frac{{{k^3}}}{4} – \frac{{{k^3}}}{2} + \frac{{{k^3}}}{3} – {k^3} + {k^3} + \frac{{{k^3}}}{{24}},{\text{ }}R = \frac{{{k^3}}}{{12}}\)

correct substitution into \({\text{triangle}} = pR\)     (A1)

eg\(\,\,\,\,\,\)\(\frac{{{k^3}}}{4} = p\left( {\frac{{{k^3}}}{{24}} + \frac{{{k^3}}}{{24}}} \right),{\text{ }}\frac{{{k^3}}}{4} = p\left( {\frac{{{k^3}}}{{12}}} \right)\)

\(p = 3\)     A1     N2

[7 marks]

d.

Question

Let \(f(x) = {x^2} – x\), for \(x \in \mathbb{R}\). The following diagram shows part of the graph of \(f\).

N17/5/MATME/SP1/ENG/TZ0/08

The graph of \(f\) crosses the \(x\)-axis at the origin and at the point \({\text{P}}(1,{\text{ }}0)\).

The line L is the normal to the graph of f at P.

The line \(L\) intersects the graph of \(f\) at another point Q, as shown in the following diagram.

N17/5/MATME/SP1/ENG/TZ0/08.c.d

Show that \(f’(1) = 1\).

[3]
a.

Find the equation of \(L\) in the form \(y = ax + b\).

[3]
b.

Find the \(x\)-coordinate of Q.

[4]
c.

Find the area of the region enclosed by the graph of \(f\) and the line \(L\).

[6]
d.
Answer/Explanation

Markscheme

\(f’(x) = 2x – 1\)     A1A1

correct substitution     A1

eg\(\,\,\,\,\,\)\(2(1) – 1,{\text{ }}2 – 1\)

\(f’(1) = 1\)     AG     N0

[3 marks]

a.

correct approach to find the gradient of the normal     (A1)

eg\(\,\,\,\,\,\)\(\frac{{ – 1}}{{f'(1)}},{\text{ }}{m_1}{m_2} =  – 1,{\text{ slope}} =  – 1\)

attempt to substitute correct normal gradient and coordinates into equation of a line     (M1)

eg\(\,\,\,\,\,\)\(y – 0 =  – 1(x – 1),{\text{ }}0 =  – 1 + b,{\text{ }}b = 1,{\text{ }}L =  – x + 1\)

\(y =  – x + 1\)     A1     N2

[3 marks]

b.

equating expressions     (M1)

eg\(\,\,\,\,\,\)\(f(x) = L,{\text{ }} – x + 1 = {x^2} – x\)

correct working (must involve combining terms)     (A1)

eg\(\,\,\,\,\,\)\({x^2} – 1 = 0,{\text{ }}{x^2} = 1,{\text{ }}x = 1\)

\(x =  – 1\,\,\,\,\,\left( {{\text{accept }}Q( – 1,{\text{ }}2)} \right)\)     A2     N3

[4 marks]

c.

valid approach     (M1)

eg\(\,\,\,\,\,\)\(\int {L – f,{\text{ }}\int_{ – 1}^1 {(1 – {x^2}){\text{d}}x} } \), splitting area into triangles and integrals

correct integration     (A1)(A1)

eg\(\,\,\,\,\,\)\(\left[ {x – \frac{{{x^3}}}{3}} \right]_{ – 1}^1,{\text{ }} – \frac{{{x^3}}}{3} – \frac{{{x^2}}}{2} + \frac{{{x^2}}}{2} + x\)

substituting their limits into their integrated function and subtracting (in any order)     (M1)

eg\(\,\,\,\,\,\)\(1 – \frac{1}{3} – \left( { – 1 – \frac{{ – 1}}{3}} \right)\)

Note:     Award M0 for substituting into original or differentiated function.

area \( = \frac{4}{3}\)     A2     N3

[6 marks]

d.

Question

A closed cylindrical can with radius r centimetres and height h centimetres has a volume of 20\(\pi \) cm3.

The material for the base and top of the can costs 10 cents per cm2 and the material for the curved side costs 8 cents per cm2. The total cost of the material, in cents, is C.

Express h in terms of r.

[2]
a.

Show that \(C = 20\pi {r^2} + \frac{{320\pi }}{r}\).

[4]
b.

Given that there is a minimum value for C, find this minimum value in terms of \(\pi \).

[9]
c.
Answer/Explanation

Markscheme

correct equation for volume      (A1)
eg  \(\pi {r^2}h = 20\pi \)

\(h = \frac{{20}}{{{r^2}}}\)     A1 N2

[2 marks]

a.

attempt to find formula for cost of parts      (M1)
eg  10 × two circles, 8 × curved side

correct expression for cost of two circles in terms of r (seen anywhere)      A1
eg  \(2\pi {r^2} \times 10\)

correct expression for cost of curved side (seen anywhere)      (A1)
eg  \(2\pi r \times h \times 8\)

correct expression for cost of curved side in terms of     A1
eg  \(8 \times 2\pi r \times \frac{{20}}{{{r^2}}},\,\,\frac{{320\pi }}{{{r^2}}}\)

\(C = 20\pi {r^2} + \frac{{320\pi }}{r}\)      AG N0

[4 marks]

b.

recognize \(C’ = 0\) at minimum       (R1)
eg  \(C’ = 0,\,\,\frac{{{\text{d}}C}}{{{\text{d}}r}} = 0\)

correct differentiation (may be seen in equation)

\(C’ = 40\pi r – \frac{{320\pi }}{{{r^2}}}\)        A1A1

correct equation      A1
eg  \(40\pi r – \frac{{320\pi }}{{{r^2}}} = 0,\,\,40\pi r\frac{{320\pi }}{{{r^2}}}\)

correct working     (A1)
eg  \(40{r^3} = 320,\,\,{r^3} = 8\)

r = 2 (m)     A1

attempt to substitute their value of r into C
eg  \(20\pi  \times 4 + 320 \times \frac{\pi }{2}\)     (M1)

correct working
eg  \(80\pi  + 160\pi \)        (A1)

\(240\pi \) (cents)      A1 N3

Note: Do not accept 753.6, 753.98 or 754, even if 240\(\pi \) is seen.

[9 marks]

c.

Question

Given that \(f(x) = \frac{1}{x}\) , answer the following.

Find the first four derivatives of \(f(x)\) .

[4]
a.

Write an expression for \({f^{(n)}}(x)\) in terms of x and n .

[3]
b.
Answer/Explanation

Markscheme

\(f'(x) = – {x^{ – 2}}\) (or \( – \frac{1}{{{x^2}}}\) )     A1     N1

\(f”(x) = 2{x^{ – 3}}\) (or \(\frac{2}{{{x^3}}}\) )     A1     N1

\(f”'(x) = – 6{x^{ – 4}}\) (or \( – \frac{6}{{{x^4}}}\) )     A1     N1

\({f^{(4)}}(x) = 24{x^{ – 5}}\) (or \(\frac{{24}}{{{x^5}}}\) )     A1     N1

[4 marks]

a.

\({f^{(n)}}(x) = \frac{{{{( – 1)}^n}n!}}{{{x^{n + 1}}}}\) or \({( – 1)^n}n!({x^{ – (n + 1)}})\)     A1A1A1     N3

[3 marks]

b.

Question

Let \(f(x) = \cos x + \sqrt 3 \sin x\) , \(0 \le x \le 2\pi \) . The following diagram shows the graph of \(f\) .


The \(y\)-intercept is at (\(0\), \(1\)) , there is a minimum point at A (\(p\), \(q\)) and a maximum point at B.

Find \(f'(x)\) .

[2]
a.

Hence

(i)     show that \(q = – 2\) ;

(ii)    verify that A is a minimum point.

[10]
b(i) and (ii).

Find the maximum value of \(f(x)\) .

[3]
c.

The function \(f(x)\) can be written in the form \(r\cos (x – a)\) .

Write down the value of r and of a .

[2]
d.
Answer/Explanation

Markscheme

\(f'(x) = – \sin x + \sqrt 3 \cos x\)     A1A1     N2

[2 marks]

a.

(i) at A, \(f'(x) = 0\)     R1

correct working     A1

e.g. \(\sin x = \sqrt 3 \cos x\)

\(\tan x = \sqrt 3 \)     A1

\(x = \frac{\pi }{3}\) , \(\frac{{4\pi }}{3}\)     A1

attempt to substitute their x into \(f(x)\)     M1

e.g. \(\cos \left( {\frac{{4\pi }}{3}} \right) + \sqrt 3 \sin \left( {\frac{{4\pi }}{3}} \right)\)

correct substitution     A1

e.g. \( – \frac{1}{2} + \sqrt 3 \left( { – \frac{{\sqrt 3 }}{2}} \right)\)

correct working that clearly leads to \( – 2\)     A1

e.g. \( – \frac{1}{2} – \frac{3}{2}\)

 \(q = – 2\)     AG     N0

(ii) correct calculations to find \(f'(x)\) either side of \(x = \frac{{4\pi }}{3}\)     A1A1

e.g. \(f'(\pi ) = 0 – \sqrt 3 \) ,  \(f'(2\pi ) = 0 + \sqrt 3 \)   

\(f'(x)\) changes sign from negative to positive     R1

so A is a minimum     AG     N0

[10 marks]

b(i) and (ii).

max when \(x = \frac{\pi }{3}\)     R1

correctly substituting \(x = \frac{\pi }{3}\) into \(f(x)\)     A1

e.g. \(\frac{1}{2} + \sqrt 3 \left( {\frac{{\sqrt 3 }}{2}} \right)\)

max value is 2     A1     N1

[3 marks]

c.

\(r = 2\) , \(a = \frac{\pi }{3}\)     A1A1     N2

[2 marks]

d.

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