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.
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

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

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

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.
Answer/Explanation

Markscheme

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

Note: Award A1 for 1, A2 for \(- 2\cos 2t\) .

[3 marks]

a.

evidence of valid approach     (M1)

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

correct working     A1

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

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

\(t = \frac{{5\pi }}{6}\)     A1     N3 

[4 marks]

b.

evidence of valid approach     (M1)

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

correct substitution     A1

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

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

\(s'(t) > 0\)     AG     N0

[3 marks]

c.

evidence of approach using s or integral of \(s’\)     (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 subtracting     (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 substitution     A1

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

distance is \(\frac{{2\pi }}{3} + \sqrt 3 \)     A1A1     N3

Note: Award A1 for \(\frac{{2\pi }}{3}\) , A1 for \(\sqrt 3 \) .

[5 marks]

d.

Question

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

Find \(f'(x)\) .

[5]
a.

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

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

[4]
b.

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

[7]
c.
Answer/Explanation

Markscheme

METHOD 1

evidence of choosing quotient rule     (M1)

e.g. \(\frac{{u’v – uv’}}{{{v^2}}}\)

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

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

\(f'(x) = \frac{6}{{{{(x + 1)}^2}}}\)    A1     N4

[5 marks]

METHOD 2

evidence of choosing product rule     (M1)

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

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

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

\(f'(x) = \frac{6}{{{{(x + 1)}^2}}}\)   A1     N4

[5 marks]

a.

METHOD 1

evidence of choosing chain rule     (M1)

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

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

correct substitution into chain rule     A1

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

working that clearly leads to the answer     A1

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

\(g'(x) = \frac{1}{{x(x + 1)}}\)     AG     N0

[4 marks]

METHOD 2

attempt to subtract logs     (M1)

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

correct derivatives (must be seen in correct expression)     A1A1

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

working that clearly leads to the answer     A1

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

\(g'(x) = \frac{1}{{x(x + 1)}}\)     AG     N0

[4 marks]

b.

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

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

recognizing that integral of derivative will give original function     (R1)

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

correct substitution and subtraction     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 \(\ln 4\)     (M1) 

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 logs     A1

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

correct working     (A1)

e.g. \(6k = 4k + 4\) , \(2k = 4\)

\(k = 2\)    A1     N4

[7 marks]

c.

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

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

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.

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