IB DP Maths Topic 6.2 The product and quotient rules SL Paper 2

 

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Question

Let \(f(x) = {{\rm{e}}^x}(1 – {x^2})\) .

Part of the graph of \(y = f(x)\), for \( – 6 \le x \le 2\) , is shown below. The x-coordinates of the local minimum and maximum points are r and s respectively.


Show that \(f'(x) = {{\rm{e}}^x}(1 – 2x – {x^2})\) . 

[3]
a.

Write down the equation of the horizontal asymptote.

[1]
b.

Write down the value of r and of s.

[4]
c.

Let L be the normal to the curve of f at \({\text{P}}(0{\text{, }}1)\) . Show that L has equation \(x + y = 1\) .

[4]
d.

Let R be the region enclosed by the curve \(y = f(x)\) and the line L.

(i)     Find an expression for the area of R.

(ii)    Calculate the area of R.

[5]
e(i) and (ii).
Answer/Explanation

Markscheme

evidence of using the product rule     M1

\(f'(x) = {{\rm{e}}^x}(1 – {x^2}) + {{\rm{e}}^x}( – 2x)\)     A1A1

Note: Award A1 for \({{\rm{e}}^x}(1 – {x^2})\) , A1 for \({{\rm{e}}^x}( – 2x)\) .

\(f'(x) = {{\rm{e}}^x}(1 – 2x – {x^2})\)     AG     N0

[3 marks]

a.

\(y = 0\)     A1     N1

[1 mark]

b.

at the local maximum or minimum point

\(f'(x) = 0\) \(({{\rm{e}}^x}(1 – 2x – {x^2}) = 0)\)     (M1)

\( \Rightarrow 1 – 2x – {x^2} = 0\)     (M1)

\(r = – 2.41\) \(s = 0.414\)     A1A1     N2N2

[4 marks]

c.

\(f'(0) = 1\)     A1

gradient of the normal \(= – 1\)     A1

evidence of substituting into an equation for a straight line     (M1)

correct substitution     A1

e.g. \(y – 1 = – 1(x – 0)\) , \(y – 1 = – x\) , \(y = – x + 1\)

\(x + y = 1\)     AG     N0

[4 marks]

d.

(i) intersection points at \(x = 0\) and \(x = 1\) (may be seen as the limits)     (A1)

approach involving subtraction and integrals     (M1)

fully correct expression     A2     N4

e.g. \(\int_0^1 {\left( {{{\rm{e}}^x}(1 – {x^2}) – (1 – x)} \right)} {\rm{d}}x\) , \(\int_0^1 {f(x){\rm{d}}x – \int_0^1 {(1 – x){\rm{d}}x} } \)

(ii) area \(R = 0.5\)     A1     N1

[5 marks]

e(i) and (ii).

Question

Let \(f(x) = {{\rm{e}}^{2x}}\cos x\) , \( – 1 \le x \le 2\) .

Show that \(f'(x) = {{\rm{e}}^{2x}}(2\cos x – \sin x)\) .

[3]
a.

Let the line L be the normal to the curve of f at \(x = 0\) .

Find the equation of L .

[5]
b.

The graph of f and the line L intersect at the point (0, 1) and at a second point P.

(i)     Find the x-coordinate of P.

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

[6]
c(i) and (ii).
Answer/Explanation

Markscheme

correctly finding the derivative of  \({{\rm{e}}^{2x}}\) , i.e. \(2{{\rm{e}}^{2x}}\)     A1

correctly finding the derivative of  \(\cos x\) , i.e. \( – \sin x\)     A1

evidence of using the product rule, seen anywhere     M1

e.g. \(f'(x) = 2{{\rm{e}}^{2x}}\cos x – {{\rm{e}}^{2x}}\sin x\)

\(f'(x) = 2{{\rm{e}}^{2x}}(2\cos x – \sin x)\)     AG     N0

[3 marks]

a.

evidence of finding \(f(0) = 1\) , seen anywhere     A1

attempt to find the gradient of f     (M1)

e.g. substituting \(x = 0\) into \(f'(x)\)

value of the gradient of f     A1

e.g. \(f'(0) = 2\) , equation of tangent is \(y = 2x + 1\)

gradient of normal \( = – \frac{1}{2}\)     (A1)

\(y – 1 = – \frac{1}{2}x\left( {y = – \frac{1}{2}x + 1} \right)\)     A1     N3

[5 marks]

b.

(i) evidence of equating correct functions     M1

e.g. \({{\rm{e}}^{2x}}\cos x = – \frac{1}{2}x + 1\) , sketch showing intersection of graphs    

\(x = 1.56\)     A1     N1

(ii) evidence of approach involving subtraction of integrals/areas     (M1)

e.g. \(\int {\left[ {f(x) – g(x)} \right]} {\rm{d}}x\) , \(\int {f(x)} {\rm{d}}x – {\text{area under trapezium}}\)

fully correct integral expression     A2

e.g. \(\int_0^{1.56} {\left[ {{{\rm{e}}^{2x}}\cos x – \left( { – \frac{1}{2}x + 1} \right)} \right]} {\rm{d}}x\) , \(\int_0^{1.56} {{{\rm{e}}^{2x}}\cos x} {\rm{d}}x – 0.951 \ldots \)

\({\rm{area}} = 3.28\)     A1     N2

[6 marks]

c(i) and (ii).

Question

Let \(f(x) = \cos 2x\) and \(g(x) = \ln (3x – 5)\) .

Find \(f'(x)\) .

[2]
a.

Find \(g'(x)\) .

[2]
b.

Let \(h(x) = f(x) \times g(x)\) . Find \(h'(x)\) .

[2]
c.
Answer/Explanation

Markscheme

(a) \(f'(x) = – \sin 2x \times 2( = – 2\sin 2x)\)     A1A1     N2

Note: Award A1 for 2, A1 for \( – \sin 2x\) .

[2 marks]

a.

\(g'(x) = 3 \times \frac{1}{{3x – 5}}\) \(\left( { = \frac{3}{{3x – 5}}} \right)\)     A1A1     N2

Note: Award A1 for 3, A1 for \(\frac{1}{{3x – 5}}\) .

[2 marks]

b.

evidence of using product rule     (M1)

\(h'(x) = (\cos 2x)\left( {\frac{3}{{3x – 5}}} \right) + \ln (3x – 5)( – 2\sin 2x)\)     A1     N2 

[2 marks]

c.

Question

Let \(f(x) = x\cos x\) , for \(0 \le x \le 6\) .

Find \(f'(x)\) .

[3]
a.

On the grid below, sketch the graph of \(y = f'(x)\) .


[4]
b.
Answer/Explanation

Markscheme

evidence of choosing the product rule     (M1)

e.g. \(x \times ( – \sin x) + 1 \times \cos x\)

\(f'(x) = \cos x – x\sin x\)     A1A1     N3

[3 marks]

a.


     A1A1A1A1     N4

Note: Award A1 for correct domain, \(0 \le x \le 6\) with endpoints in circles, A1 for approximately correct shape, A1 for local minimum in circle, A1 for local maximum in circle.

[4 marks]

b.

Question

Consider \(f(x) = x\ln (4 – {x^2})\) , for \( – 2 < x < 2\) . The graph of f is given below.


Let P and Q be points on the curve of f where the tangent to the graph of f is parallel to the x-axis.

(i)     Find the x-coordinate of P and of Q.

(ii)    Consider \(f(x) = k\) . Write down all values of k for which there are exactly two solutions.

[5]
a(i) and (ii).

Let \(g(x) = {x^3}\ln (4 – {x^2})\) , for \( – 2 < x < 2\) .

Show that \(g'(x) = \frac{{ – 2{x^4}}}{{4 – {x^2}}} + 3{x^2}\ln (4 – {x^2})\) .

[4]
b.

Let \(g(x) = {x^3}\ln (4 – {x^2})\) , for \( – 2 < x < 2\) .

Sketch the graph of \(g’\) .

[2]
c.

Let \(g(x) = {x^3}\ln (4 – {x^2})\) , for \( – 2 < x < 2\) .

Consider \(g'(x) = w\) . Write down all values of w for which there are exactly two solutions.

[3]
d.
Answer/Explanation

Markscheme

(i) \( – 1.15{\text{, }}1.15\)     A1A1     N2

(ii) recognizing that it occurs at P and Q     (M1)

e.g. \(x = – 1.15\) , \(x = 1.15\)

\(k = – 1.13\) , \(k = 1.13\)     A1A1     N3

[5 marks]

a(i) and (ii).

evidence of choosing the product rule     (M1)

e.g. \(uv’ + vu’\)

derivative of \({x^3}\) is \(3{x^2}\)     (A1)

derivative of \(\ln (4 – {x^2})\) is \(\frac{{ – 2x}}{{4 – {x^2}}}\)     (A1)

correct substitution     A1

e.g. \({x^3} \times \frac{{ – 2x}}{{4 – {x^2}}} + \ln (4 – {x^2}) \times 3{x^2}\)

\(g'(x) = \frac{{ – 2{x^4}}}{{4 – {x^2}}} + 3{x^2}\ln (4 – {x^2})\)     AG     N0

[4 marks]

b.


     A1A1     N2

[2 marks]

c.

\(w = 2.69\) , \(w < 0\)     A1A2     N2

[3 marks]

d.

Question

Let \(f(x) = \frac{{20x}}{{{{\rm{e}}^{0.3x}}}}\) , for \(0 \le x \le 20\) .

Sketch the graph of f .

[3]
a.

(i)     Write down the x-coordinate of the maximum point on the graph of f .

(ii)    Write down the interval where f is increasing.

[3]
b(i) and (ii).

Show that \(f'(x) = \frac{{20 – 6x}}{{{{\rm{e}}^{0.3x}}}}\) .

[5]
c.

Find the interval where the rate of change of f is increasing.

[4]
d.
Answer/Explanation

Markscheme


     A1A1A1     N3

Note: Award A1 for approximately correct shape with inflexion/change of curvature, A1 for maximum skewed to the left, A1 for asymptotic behaviour to the right.

[3 marks]

a.

(i) \(x = 3.33\)     A1     N1

(ii) correct interval, with right end point \(3\frac{1}{3}\)     A1A1     N2

e.g. \(0 < x \le 3.33\) , \(0 \le x < 3\frac{1}{3}\)

Note: Accept any inequalities in the right direction.

[3 marks]

b(i) and (ii).

valid approach     (M1)

e.g. quotient rule, product rule

2 correct derivatives (must be seen in product or quotient rule)     (A1)(A1)

e.g. \(20\) , \(0.3{{\rm{e}}^{0.3x}}\) or \( – 0.3{{\rm{e}}^{ – 0.3x}}\)

correct substitution into product or quotient rule     A1

e.g. \(\frac{{20{{\rm{e}}^{0.3x}} – 20x(0.3){{\rm{e}}^{0.3x}}}}{{{{({{\rm{e}}^{0.3x}})}^2}}}\) , \(20{{\rm{e}}^{ – 0.3x}} + 20x( – 0.3){{\rm{e}}^{ – 0.3x}}\)

correct working     A1

e.g. \(\frac{{20{{\rm{e}}^{0.3x}} – 6x{{\rm{e}}^{0.3x}}}}{{{{\rm{e}}^{0.6x}}}}\) , \(\frac{{{{\rm{e}}^{0.3x}}(20 – 20x(0.3))}}{{{{{\rm{(}}{{\rm{e}}^{0.3x}})}^2}}}\) , \({{\rm{e}}^{ – 0.3x}}(20 + 20x( – 0.3))\)

\(f'(x) = \frac{{20 – 6x}}{{{{\rm{e}}^{0.3x}}}}\)     AG     N0

[5 marks]

c.

consideration of \(f’\) or \(f”\)     (M1)

valid reasoning     R1

e.g. sketch of \(f’\) , \(f”\) is positive, \(f” = 0\) , reference to minimum of \(f’\)

correct value \(6.6666666 \ldots \) \(\left( {6\frac{2}{3}} \right)\)     (A1)

correct interval, with both endpoints     A1     N3

e.g. \(6.67 < x \le 20\) , \(6\frac{2}{3} \le x < 20\)

[4 marks]

d.

Question

Let \(f(x) = \frac{{g(x)}}{{h(x)}}\), where \(g(2) = 18,{\text{ }}h(2) = 6,{\text{ }}g'(2) = 5\), and \(h'(2) = 2\). Find the equation of the normal to the graph of \(f\) at \(x = 2\).

Answer/Explanation

Markscheme

recognizing need to find \(f(2)\) or \(f'(2)\)     (R1)

\(f(2) = \frac{{18}}{6}\)   (seen anywhere)     (A1)

correct substitution into the quotient rule     (A1)

eg     \(\frac{{6(5) – 18(2)}}{{{6^2}}}\)

\(f'(2) =  – \frac{6}{{36}}\)     A1

gradient of normal is 6     (A1)

attempt to use the point and gradient to find equation of straight line     (M1)

eg     \(y – f(2) =  – \frac{1}{{f'(2)}}(x – 2)\)

correct equation in any form     A1     N4

eg     \(y – 3 = 6(x – 2),{\text{ }}y = 6x – 9\)

[7 marks]

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