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 \(f'(x) = {{\rm{e}}^x}(1 – 2x – {x^2})\) .
Write down the equation of the horizontal asymptote.
Write down the value of r and of s.
Let L be the normal to the curve of f at \({\text{P}}(0{\text{, }}1)\) . Show that L has equation \(x + y = 1\) .
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.
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]
\(y = 0\) A1 N1
[1 mark]
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]
\(f'(0) = 1\) 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]
(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]
Question
Let \(f(x) = {{\rm{e}}^{2x}}\cos x\) , \( – 1 \le x \le 2\) .
Show that \(f'(x) = {{\rm{e}}^{2x}}(2\cos x – \sin x)\) .
Let the line L be the normal to the curve of f at \(x = 0\) .
Find the equation of L .
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 .
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]
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]
(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]
Question
Let \(f(x) = \cos 2x\) and \(g(x) = \ln (3x – 5)\) .
Find \(f'(x)\) .
Find \(g'(x)\) .
Let \(h(x) = f(x) \times g(x)\) . Find \(h'(x)\) .
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]
\(g'(x) = 3 \times \frac{1}{{3x – 5}}\) \(\left( { = \frac{3}{{3x – 5}}} \right)\) A1A1 N2
Note: Award A1 for 3, A1 for \(\frac{1}{{3x – 5}}\) .
[2 marks]
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]
Question
Let \(f(x) = x\cos x\) , for \(0 \le x \le 6\) .
Find \(f'(x)\) .
On the grid below, sketch the graph of \(y = f'(x)\) .
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]
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]
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.
Let \(g(x) = {x^3}\ln (4 – {x^2})\) , for \( – 2 < x < 2\) .
Show that \(g'(x) = \frac{{ – 2{x^4}}}{{4 – {x^2}}} + 3{x^2}\ln (4 – {x^2})\) .
Let \(g(x) = {x^3}\ln (4 – {x^2})\) , for \( – 2 < x < 2\) .
Sketch the graph of \(g’\) .
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.
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]
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]
A1A1 N2
[2 marks]
\(w = 2.69\) , \(w < 0\) A1A2 N2
[3 marks]
Question
Let \(f(x) = \frac{{20x}}{{{{\rm{e}}^{0.3x}}}}\) , for \(0 \le x \le 20\) .
Sketch the graph of f .
(i) Write down the x-coordinate of the maximum point on the graph of f .
(ii) Write down the interval where f is increasing.
Show that \(f'(x) = \frac{{20 – 6x}}{{{{\rm{e}}^{0.3x}}}}\) .
Find the interval where the rate of change of f is increasing.
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]
(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]
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]
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]
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]