IB DP Maths Topic 6.1 Definition of derivative from first principles as f′(x)=limh→0(f(x+h)−f(x)h) SL Paper 2

 

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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.
Answer/Explanation

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.

Question

Let \(f(x) = \frac{1}{{x – 1}} + 2\), for \(x > 1\).

Let \(g(x) = a{e^{ – x}} + b\), for \(x \geqslant 1\). The graphs of \(f\) and \(g\) have the same horizontal asymptote.

Write down the equation of the horizontal asymptote of the graph of \(f\).

[2]
a.

Find \(f'(x)\).

[2]
b.

Write down the value of \(b\).

[2]
c.

Given that \(g'(1) =  – e\), find the value of \(a\).

[4]
d.

There is a value of \(x\), for \(1 < x < 4\), for which the graphs of \(f\) and \(g\) have the same gradient. Find this gradient.

[4]
e.
Answer/Explanation

Markscheme

\(y = 2\) (correct equation only)     A2     N2

[2 marks]

a.

valid approach     (M1)

eg\(\,\,\,\,\,\)\({(x – 1)^{ – 1}} + 2,{\text{ }}f'(x) = \frac{{0(x – 1) – 1}}{{{{(x – 1)}^2}}}\)

\( – {(x – 1)^{ – 2}},{\text{ }}f'(x) = \frac{{ – 1}}{{{{(x – 1)}^2}}}\)    A1     N2

[2 marks]

b.

correct equation for the asymptote of \(g\)

eg\(\,\,\,\,\,\)\(y = b\)     (A1)

\(b = 2\)     A1     N2

[2 marks]

c.

correct derivative of g (seen anywhere)     (A2)

eg\(\,\,\,\,\,\)\(g'(x) =  – a{{\text{e}}^{ – x}}\)

correct equation     (A1)

eg\(\,\,\,\,\,\)\( – {\text{e}} =  – a{{\text{e}}^{ – 1}}\)

7.38905

\(a = {{\text{e}}^2}{\text{ }}({\text{exact}}),{\text{ }}7.39\)     A1     N2

[4 marks]

d.

attempt to equate their derivatives     (M1)

eg\(\,\,\,\,\,\)\(f'(x) = g'(x),{\text{ }}\frac{{ – 1}}{{{{(x – 1)}^2}}} =  – a{{\text{e}}^{ – x}}\)

valid attempt to solve their equation     (M1)

eg\(\,\,\,\,\,\)correct value outside the domain of \(f\) such as 0.522 or 4.51,

M16/5/MATME/SP2/ENG/TZ2/09.e/M

correct solution (may be seen in sketch)     (A1)

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

gradient is \( – 1\)     A1     N3

[4 marks]

e.

Question

The following diagram shows the graph of \(f(x) = a\sin bx + c\), for \(0 \leqslant x \leqslant 12\).

N16/5/MATME/SP2/ENG/TZ0/10

The graph of \(f\) has a minimum point at \((3,{\text{ }}5)\) and a maximum point at \((9,{\text{ }}17)\).

The graph of \(g\) is obtained from the graph of \(f\) by a translation of \(\left( {\begin{array}{*{20}{c}} k \\ 0 \end{array}} \right)\). The maximum point on the graph of \(g\) has coordinates \((11.5,{\text{ }}17)\).

The graph of \(g\) changes from concave-up to concave-down when \(x = w\).

(i)     Find the value of \(c\).

(ii)     Show that \(b = \frac{\pi }{6}\).

(iii)     Find the value of \(a\).

[6]
a.

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

(ii)     Find \(g(x)\).

[3]
b.

(i)     Find \(w\).

(ii)     Hence or otherwise, find the maximum positive rate of change of \(g\).

[6]
c.
Answer/Explanation

Markscheme

(i)     valid approach     (M1)

eg\(\,\,\,\,\,\)\(\frac{{5 + 17}}{2}\)

\(c = 11\)    A1     N2

(ii)     valid approach     (M1)

eg\(\,\,\,\,\,\)period is 12, per \( = \frac{{2\pi }}{b},{\text{ }}9 – 3\)

\(b = \frac{{2\pi }}{{12}}\)    A1

\(b = \frac{\pi }{6}\)     AG     N0

(iii)     METHOD 1

valid approach     (M1)

eg\(\,\,\,\,\,\)\(5 = a\sin \left( {\frac{\pi }{6} \times 3} \right) + 11\), substitution of points

\(a =  – 6\)     A1     N2

METHOD 2

valid approach     (M1)

eg\(\,\,\,\,\,\)\(\frac{{17 – 5}}{2}\), amplitude is 6

\(a =  – 6\)     A1     N2

[6 marks]

a.

(i)     \(k = 2.5\)     A1     N1

(ii)     \(g(x) =  – 6\sin \left( {\frac{\pi }{6}(x – 2.5)} \right) + 11\)     A2     N2

[3 marks]

b.

(i)     METHOD 1 Using \(g\)

recognizing that a point of inflexion is required     M1

eg\(\,\,\,\,\,\)sketch, recognizing change in concavity

evidence of valid approach     (M1)

eg\(\,\,\,\,\,\)\(g”(x) = 0\), sketch, coordinates of max/min on \({g’}\)

\(w = 8.5\) (exact)     A1     N2

METHOD 2 Using \(f\)

recognizing that a point of inflexion is required     M1

eg\(\,\,\,\,\,\)sketch, recognizing change in concavity

evidence of valid approach involving translation     (M1)

eg\(\,\,\,\,\,\)\(x = w – k\), sketch, \(6 + 2.5\)

\(w = 8.5\) (exact)     A1     N2

(ii)     valid approach involving the derivative of \(g\) or \(f\) (seen anywhere)     (M1)

eg\(\,\,\,\,\,\)\(g'(w),{\text{ }} – \pi \cos \left( {\frac{\pi }{6}x} \right)\), max on derivative, sketch of derivative

attempt to find max value on derivative     M1

eg\(\,\,\,\,\,\)\( – \pi \cos \left( {\frac{\pi }{6}(8.5 – 2.5)} \right),{\text{ }}f'(6)\), dot on max of sketch

3.14159

max rate of change \( = \pi \) (exact), 3.14     A1     N2

[6 marks]

c.

Question

Let \(f(x) =  – 0.5{x^4} + 3{x^2} + 2x\). The following diagram shows part of the graph of \(f\).

M17/5/MATME/SP2/ENG/TZ2/08

There are \(x\)-intercepts at \(x = 0\) and at \(x = p\). There is a maximum at A where \(x = a\), and a point of inflexion at B where \(x = b\).

Find the value of \(p\).

[2]
a.

Write down the coordinates of A.

[2]
b.i.

Write down the rate of change of \(f\) at A.

[1]
b.ii.

Find the coordinates of B.

[4]
c.i.

Find the the rate of change of \(f\) at B.

[3]
c.ii.

Let \(R\) be the region enclosed by the graph of \(f\) , the \(x\)-axis, the line \(x = b\) and the line \(x = a\). The region \(R\) is rotated 360° about the \(x\)-axis. Find the volume of the solid formed.

[3]
d.
Answer/Explanation

Markscheme

evidence of valid approach     (M1)

eg\(\,\,\,\,\,\)\(f(x) = 0,{\text{ }}y = 0\)

2.73205

\(p = 2.73\)     A1     N2

[2 marks]

a.

1.87938, 8.11721

\((1.88,{\text{ }}8.12)\)     A2     N2

[2 marks]

b.i.

rate of change is 0 (do not accept decimals)     A1     N1

[1 marks]

b.ii.

METHOD 1 (using GDC)

valid approach     M1

eg\(\,\,\,\,\,\)\(f’’ = 0\), max/min on \(f’,{\text{ }}x =  – 1\)

sketch of either \(f’\) or \(f’’\), with max/min or root (respectively)     (A1)

\(x = 1\)     A1     N1

Substituting their \(x\) value into \(f\)     (M1)

eg\(\,\,\,\,\,\)\(f(1)\)

\(y = 4.5\)     A1     N1

METHOD 2 (analytical)

\(f’’ =  – 6{x^2} + 6\)     A1

setting \(f’’ = 0\)     (M1)

\(x = 1\)     A1     N1

substituting their \(x\) value into \(f\)     (M1)

eg\(\,\,\,\,\,\)\(f(1)\)

\(y = 4.5\)     A1     N1

[4 marks]

c.i.

recognizing rate of change is \(f’\)     (M1)

eg\(\,\,\,\,\,\)\(y’,{\text{ }}f’(1)\)

rate of change is 6     A1     N2

[3 marks]

c.ii.

attempt to substitute either limits or the function into formula     (M1)

involving \({f^2}\) (accept absence of \(\pi \) and/or \({\text{d}}x\))

eg\(\,\,\,\,\,\)\(\pi \int {{{( – 0.5{x^4} + 3{x^2} + 2x)}^2}{\text{d}}x,{\text{ }}\int_1^{1.88} {{f^2}} } \)

128.890

\({\text{volume}} = 129\)     A2     N3

[3 marks]

d.
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