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Question
The following diagram shows the graphs of \(f(x) = \ln (3x – 2) + 1\) and \(g(x) = – 4\cos (0.5x) + 2\) , for \(1 \le x \le 10\) .
Let A be the area of the region enclosed by the curves of f and g.
(i) Find an expression for A.
(ii) Calculate the value of A.
(i) Find \(f'(x)\) .
(ii) Find \(g'(x)\) .
There are two values of x for which the gradient of f is equal to the gradient of g. Find both these values of x.
Answer/Explanation
Markscheme
(i) intersection points \(x = 3.77\) , \(x = 8.30\) (may be seen as the limits) (A1)(A1)
approach involving subtraction and integrals (M1)
fully correct expression A2
e.g. \(\int_{3.77}^{8.30} {(( – 4\cos (0.5x) + 2) – (\ln (3x – 2) + 1)){\rm{d}}x} \) , \(\int_{3.77}^{8.30} {g(x){\rm{d}}x – } \int_{3.77}^{8.30} {f(x){\rm{d}}x} \)
(ii) \(A = 6.46\) A1 N1
[6 marks]
(i) \(f'(x) = \frac{3}{{3x – 2}}\) A1A1 N2
Note: Award A1 for numerator (3), A1 for denominator (\({3x – 2}\)) , but penalize 1 mark for additional terms.
(ii) \(g'(x) = 2\sin (0.5x)\) A1A1 N2
Note: Award A1 for 2, A1 for \(\sin (0.5x)\) , but penalize 1 mark for additional terms.
[4 marks]
evidence of using derivatives for gradients (M1)
correct approach (A1)
e.g. \(f'(x) = g'(x)\) , points of intersection
\(x = 1.43\) , \(x = 6.10\) A1A1 N2N2
[4 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
Consider the curve \(y = \ln (3x – 1)\) . Let P be the point on the curve where \(x = 2\) .
Write down the gradient of the curve at P.
The normal to the curve at P cuts the x-axis at R. Find the coordinates of R.
Answer/Explanation
Markscheme
gradient is \(0.6\) A2 N2
[2 marks]
at R, \(y = 0\) (seen anywhere) A1
at \(x = 2\) , \(y = \ln 5\) \(( = 1.609 \ldots )\) (A1)
gradient of normal \( = – 1.6666 \ldots \) (A1)
evidence of finding correct equation of normal A1
e.g. \(y = \ln 5 = – \frac{5}{3}(x – 2)\) , \(y = – 1.67x + c\)
\(x = 2.97\) (accept 2.96) A1
coordinates of R are (2.97,0) N3
[5 marks]
Question
Let \(f(x) = 3\sin x + 4\cos x\) , for \( – 2\pi \le x \le 2\pi \) .
Sketch the graph of f .
Write down
(i) the amplitude;
(ii) the period;
(iii) the x-intercept that lies between \( – \frac{\pi }{2}\) and 0.
Hence write \(f(x)\) in the form \(p\sin (qx + r)\) .
Write down one value of x such that \(f'(x) = 0\) .
Write down the two values of k for which the equation \(f(x) = k\) has exactly two solutions.
Let \(g(x) = \ln (x + 1)\) , for \(0 \le x \le \pi \) . There is a value of x, between \(0\) and \(1\), for which the gradient of f is equal to the gradient of g. Find this value of x.
Answer/Explanation
Markscheme
A1A1A1 N3
Note: Award A1 for approximately sinusoidal shape, A1 for end points approximately correct \(( – 2\pi {\text{, }}4)\) \((2\pi {\text{, }}4)\), A1 for approximately correct position of graph, (y-intercept \((0{\text{, }}4)\), maximum to right of y-axis).
[3 marks]
(i) 5 A1 N1
(ii) \(2\pi \) (6.28) A1 N1
(iii) \( – 0.927\) A1 N1
[3 marks]
\(f(x) = 5\sin (x + 0.927)\) (accept \(p = 5\) , \(q = 1\) , \(r = 0.927\) ) A1A1A1 N3
[3 marks]
evidence of correct approach (M1)
e.g. max/min, sketch of \(f'(x)\) indicating roots
one 3 s.f. value which rounds to one of \( – 5.6\), \( – 2.5\), \(0.64\), \(3.8\) A1 N2
[2 marks]
\(k = – 5\) , \(k = 5\) A1A1 N2
[2 marks]
METHOD 1
graphical approach (but must involve derivative functions) M1
e.g.
each curve A1A1
\(x = 0.511\) A2 N2
METHOD 2
\(g'(x) = \frac{1}{{x + 1}}\) A1
\(f'(x) = 3\cos x – 4\sin x\) \((5\cos (x + 0.927))\) A1
evidence of attempt to solve \(g'(x) = f'(x)\) M1
\(x = 0.511\) A2 N2
[5 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 \(f(x) = A{{\rm{e}}^{kx}} + 3\) . Part of the graph of f is shown below.
The y-intercept is at (0, 13) .
Show that \(A = 10\) .
Given that \(f(15) = 3.49\) (correct to 3 significant figures), find the value of k.
(i) Using your value of k , find \(f'(x)\) .
(ii) Hence, explain why f is a decreasing function.
(iii) Write down the equation of the horizontal asymptote of the graph f .
Let \(g(x) = – {x^2} + 12x – 24\) .
Find the area enclosed by the graphs of f and g .
Answer/Explanation
Markscheme
substituting (0, 13) into function M1
e.g. \(13 = A{{\rm{e}}^0} + 3\)
\(13 = A + 3\) A1
\(A = 10\) AG N0
[2 marks]
substituting into \(f(15) = 3.49\) A1
e.g. \(3.49 = 10{{\rm{e}}^{15k}} + 3\) , \(0.049 = {{\rm{e}}^{15k}}\)
evidence of solving equation (M1)
e.g. sketch, using \(\ln \)
\(k = – 0.201\) (accept \(\frac{{\ln 0.049}}{{15}}\) ) A1 N2
[3 marks]
(i) \(f(x) = 10{{\rm{e}}^{ – 0.201x}} + 3\)
\(f(x) = 10{{\rm{e}}^{ – 0.201x}} \times – 0.201\) \(( = – 2.01{{\rm{e}}^{ – 0.201x}})\) A1A1A1 N3
Note: Award A1 for \(10{{\rm{e}}^{ – 0.201x}}\) , A1 for \( \times – 0.201\) , A1 for the derivative of 3 is zero.
(ii) valid reason with reference to derivative R1 N1
e.g. \(f'(x) < 0\) , derivative always negative
(iii) \(y = 3\) A1 N1
[5 marks]
finding limits \(3.8953 \ldots \), \(8.6940 \ldots \) (seen anywhere) A1A1
evidence of integrating and subtracting functions (M1)
correct expression A1
e.g. \(\int_{3.90}^{8.69} {g(x) – f(x){\rm{d}}x} \) , \(\int_{3.90}^{8.69} {\left[ {\left( { – {x^2} + 12x – 24} \right) – \left( {10{{\rm{e}}^{ – 0.201x}} + 3} \right)} \right]} {\rm{d}}x\)
area \(= 19.5\) A2 N4
[6 marks]
Question
The following diagram shows the graph of \(f(x) = {{\rm{e}}^{ – {x^2}}}\) .
The points A, B, C, D and E lie on the graph of f . Two of these are points of inflexion.
Identify the two points of inflexion.
(i) Find \(f'(x)\) .
(ii) Show that \(f”(x) = (4{x^2} – 2){{\rm{e}}^{ – {x^2}}}\) .
Find the x-coordinate of each point of inflexion.
Use the second derivative to show that one of these points is a point of inflexion.
Answer/Explanation
Markscheme
B, D A1A1 N2
[2 marks]
(i) \(f'(x) = – 2x{{\rm{e}}^{ – {x^2}}}\) A1A1 N2
Note: Award A1 for \({{\rm{e}}^{ – {x^2}}}\) and A1 for \( – 2x\) .
(ii) finding the derivative of \( – 2x\) , i.e. \( – 2\) (A1)
evidence of choosing the product rule (M1)
e.g. \( – 2{{\rm{e}}^{ – {x^2}}}\) \( – 2x \times – 2x{{\rm{e}}^{ – {x^2}}}\)
\( – 2{{\rm{e}}^{ – {x^2}}} + 4{x^2}{{\rm{e}}^{ – {x^2}}}\) A1
\(f”(x) = (4{x^2} – 2){{\rm{e}}^{ – {x^2}}}\) AG N0
[5 marks]
valid reasoning R1
e.g. \(f”(x) = 0\)
attempting to solve the equation (M1)
e.g. \((4{x^2} – 2) = 0\) , sketch of \(f”(x)\)
\(p = 0.707\) \(\left( { = \frac{1}{{\sqrt 2 }}} \right)\) , \(q = – 0.707\) \(\left( { = – \frac{1}{{\sqrt 2 }}} \right)\) A1A1 N3
[4 marks]
evidence of using second derivative to test values on either side of POI M1
e.g. finding values, reference to graph of \(f”\) , sign table
correct working A1A1
e.g. finding any two correct values either side of POI,
checking sign of \(f”\) on either side of POI
reference to sign change of \(f”(x)\) R1 N0
[4 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 \(f(x) = \cos ({{\rm{e}}^x})\) , for \( – 2 \le x \le 2\) .
Find \(f'(x)\) .
On the grid below, sketch the graph of \(f'(x)\) .
Answer/Explanation
Markscheme
\(f'(x) = – {{\rm{e}}^x}\sin ({{\rm{e}}^x})\) A1A1 N2
[2 marks]
A1A1A1A1 N4
Note: Award A1 for shape that must have the correct domain (from \( – 2\) to \( + 2\) ) and correct range (from \( – 6\) to \(4\) ), A1 for minimum in circle, A1 for maximum in circle and A1 for intercepts in circles.
[4 marks]
Question
Let \(f(x) = \frac{{100}}{{(1 + 50{{\rm{e}}^{ – 0.2x}})}}\) . Part of the graph of \(f\) is shown below.
Write down \(f(0)\) .
Solve \(f(x) = 95\) .
Find the range of \(f\) .
Show that \(f'(x) = \frac{{1000{{\rm{e}}^{ – 0.2x}}}}{{{{(1 + 50{{\rm{e}}^{ – 0.2x}})}^2}}}\) .
Find the maximum rate of change of \(f\) .
Answer/Explanation
Markscheme
\(f(0) = \frac{{100}}{{51}}\) (exact), \(1.96\) A1 N1
[1 mark]
setting up equation (M1)
eg \(95 = \frac{{100}}{{1 + 50{{\rm{e}}^{ – 0.2x}}}}\) , sketch of graph with horizontal line at \(y = 95\)
\(x = 34.3\) A1 N2
[2 marks]
upper bound of \(y\) is \(100\) (A1)
lower bound of \(y\) is \(0\) (A1)
range is \(0 < y < 100\) A1 N3
[3 marks]
METHOD 1
setting function ready to apply the chain rule (M1)
eg \(100{(1 + 50{{\rm{e}}^{ – 0.2x}})^{ – 1}}\)
evidence of correct differentiation (must be substituted into chain rule) (A1)(A1)
eg \(u’ = – 100{(1 + 50{{\rm{e}}^{ – 0.2x}})^{ – 2}}\) , \(v’ = (50{{\rm{e}}^{ – 0.2x}})( – 0.2)\)
correct chain rule derivative A1
eg \(f'(x) = – 100{(1 + 50{{\rm{e}}^{ – 0.2x}})^{ – 2}}(50{{\rm{e}}^{ – 0.2x}})( – 0.2)\)
correct working clearly leading to the required answer A1
eg \(f'(x) = 1000{{\rm{e}}^{ – 0.2x}}{(1 + 50{{\rm{e}}^{ – 0.2x}})^{ – 2}}\)
\(f'(x) = \frac{{1000{{\rm{e}}^{ – 0.2x}}}}{{{{(1 + 50{{\rm{e}}^{ – 0.2x}})}^2}}}\) AG N0
METHOD 2
attempt to apply the quotient rule (accept reversed numerator terms) (M1)
eg \(\frac{{vu’ – uv’}}{{{v^2}}}\) , \(\frac{{uv’ – vu’}}{{{v^2}}}\)
evidence of correct differentiation inside the quotient rule (A1)(A1)
eg \(f'(x) = \frac{{(1 + 50{{\rm{e}}^{ – 0.2x}})(0) – 100(50{{\rm{e}}^{ – 0.2x}} \times – 0.2)}}{{{{(1 + 50{{\rm{e}}^{ – 0.2x}})}^2}}}\) , \(\frac{{100( – 10){{\rm{e}}^{ – 0.2x}} – 0}}{{{{(1 + 50{{\rm{e}}^{ – 0.2x}})}^2}}}\)
any correct expression for derivative (\(0\) may not be explicitly seen) (A1)
eg \(\frac{{ – 100(50{{\rm{e}}^{ – 0.2x}} \times – 0.2)}}{{{{(1 + 50{{\rm{e}}^{ – 0.2x}})}^2}}}\)
correct working clearly leading to the required answer A1
eg \(f'(x) = \frac{{0 – 100( – 10){{\rm{e}}^{ – 0.2x}}}}{{{{(1 + 50{{\rm{e}}^{ – 0.2x}})}^2}}}\) , \(\frac{{ – 100( – 10){{\rm{e}}^{ – 0.2x}}}}{{{{(1 + 50{{\rm{e}}^{ – 0.2x}})}^2}}}\)
\(f'(x) = \frac{{{\rm{1000}}{{\rm{e}}^{ – 0.2x}}}}{{{{(1 + 50{{\rm{e}}^{ – 0.2x}})}^2}}}\) AG N0
[5 marks]
METHOD 1
sketch of \(f'(x)\) (A1)
eg
recognizing maximum on \(f'(x)\) (M1)
eg dot on max of sketch
finding maximum on graph of \(f'(x)\) A1
eg (\(19.6\), \(5\)) , \(x = 19.560 \ldots \)
maximum rate of increase is \(5\) A1 N2
METHOD 2
recognizing \(f”(x) = 0\) (M1)
finding any correct expression for \(f”(x) = 0\) (A1)
eg \(\frac{{{{(1 + 50{{\rm{e}}^{ – 0.2x}})}^2}( – 200{{\rm{e}}^{ – 0.2x}}) – (1000{{\rm{e}}^{ – 0.2x}})(2(1 + 50{{\rm{e}}^{ – 0.2x}})( – 10{{\rm{e}}^{ – 0.2x}}))}}{{{{(1 + 50{{\rm{e}}^{ – 0.2x}})}^4}}}\)
finding \(x = 19.560 \ldots \) A1
maximum rate of increase is \(5\) A1 N2
[4 marks]
Question
Let \(f(x) = {{\rm{e}}^{\frac{x}{4}}}\) and \(g(x) = mx\) , where \(m \ge 0\) , and \( – 5 \le x \le 5\) . Let \(R\) be the region enclosed by the \(y\)-axis, the graph of \(f\) , and the graph of \(g\) .
Let \(m = 1\).
(i) Sketch the graphs of \(f\) and \(g\) on the same axes.
(ii) Find the area of \(R\) .
Find the area of \(R\) .
Consider all values of \(m\) such that the graphs of \(f\) and \(g\) intersect. Find the value of \(m\) that gives the greatest value for the area of \(R\) .
Answer/Explanation
Markscheme
(i)
A1A1 N2
Notes: Award A1 for the graph of \(f\) positive, increasing and concave up.
Award A1 for graph of \(g\) increasing and linear with \(y\)-intercept of \(0\).
Penalize one mark if domain is not [\( – 5\), \(5\)] and/or if \(f\) and \(g\) do not intersect in the first quadrant.
[2 marks]
(ii)
attempt to find intersection of the graphs of \(f\) and \(g\) (M1)
eg \({{\rm{e}}^{\frac{x}{4}}} = x\)
\(x = 1.42961 \ldots \) A1
valid attempt to find area of \(R\) (M1)
eg \(\int {(x – {{\rm{e}}^{\frac{x}{4}}}} ){\rm{d}}x\) , \(\int_0^1 {(g – f)} \) , \(\int {(f – g)} \)
area \( = 0.697\) A2 N3
[5 marks]
attempt to find intersection of the graphs of \(f\) and \(g\) (M1)
eg \({{\rm{e}}^{\frac{x}{4}}} = x\)
\(x = 1.42961 \ldots \) A1
valid attempt to find area of \(R\) (M1)
eg \(\int {(x – {{\rm{e}}^{\frac{x}{4}}}} ){\rm{d}}x\) , \(\int_0^1 {(g – f)} \) , \(\int {(f – g)} \)
area \( = 0.697\) A2 N3
[5 marks]
recognize that area of \(R\) is a maximum at point of tangency (R1)
eg \(m = f'(x)\)
equating functions (M1)
eg \(f(x) = g(x)\) , \({{\rm{e}}^{\frac{x}{4}}} = mx\)
\(f'(x) = \frac{1}{4}{{\rm{e}}^{\frac{x}{4}}}\) (A1)
equating gradients (A1)
eg \(f'(x) = g'(x)\) , \(\frac{1}{4}{{\rm{e}}^{\frac{x}{4}}} = m\)
attempt to solve system of two equations for \(x\) (M1)
eg \(\frac{1}{4}{{\rm{e}}^{\frac{x}{4}}} \times x = {{\rm{e}}^{\frac{x}{4}}}\)
\(x = 4\) (A1)
attempt to find \(m\) (M1)
eg \(f'(4)\) , \(\frac{1}{4}{{\rm{e}}^{\frac{4}{4}}}\)
\(m = \frac{1}{4}e\) (exact), \(0.680\) A1 N3
[8 marks]
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 \(f\).
Find \(f'(x)\).
Write down the value of \(b\).
Given that \(g'(1) = – e\), find the value of \(a\).
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.
Answer/Explanation
Markscheme
\(y = 2\) (correct equation only) A2 N2
[2 marks]
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]
correct equation for the asymptote of \(g\)
eg\(\,\,\,\,\,\)\(y = b\) (A1)
\(b = 2\) A1 N2
[2 marks]
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]
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,
correct solution (may be seen in sketch) (A1)
eg\(\,\,\,\,\,\)\(x = 2,{\text{ }}(2,{\text{ }} – 1)\)
gradient is \( – 1\) A1 N3
[4 marks]
Question
Let \(f(x) = {({x^2} + 3)^7}\). Find the term in \({x^5}\) in the expansion of the derivative, \(f’(x)\).
Answer/Explanation
Markscheme
METHOD 1
derivative of \(f(x)\) A2
\(7{({x^2} + 3)^6}(x2)\)
recognizing need to find \({x^4}\) term in \({({x^2} + 3)^6}\) (seen anywhere) R1
eg\(\,\,\,\,\,\)\(14x{\text{ (term in }}{x^4})\)
valid approach to find the terms in \({({x^2} + 3)^6}\) (M1)
eg\(\,\,\,\,\,\)\(\left( {\begin{array}{*{20}{c}} 6 \\ r \end{array}} \right){({x^2})^{6 – r}}{(3)^r},{\text{ }}{({x^2})^6}{(3)^0} + {({x^2})^5}{(3)^1} + \ldots \), Pascal’s triangle to 6th row
identifying correct term (may be indicated in expansion) (A1)
eg\(\,\,\,\,\,\)\({\text{5th term, }}r = 2,{\text{ }}\left( {\begin{array}{*{20}{c}} 6 \\ 4 \end{array}} \right),{\text{ }}{({x^2})^2}{(3)^4}\)
correct working (may be seen in expansion) (A1)
eg\(\,\,\,\,\,\)\(\left( {\begin{array}{*{20}{c}} 6 \\ 4 \end{array}} \right){({x^2})^2}{(3)^4},{\text{ }}15 \times {3^4},{\text{ }}14x \times 15 \times 81{({x^2})^2}\)
\(17010{x^5}\) A1 N3
METHOD 2
recognition of need to find \({x^6}\) in \({({x^2} + 3)^7}\) (seen anywhere) R1
valid approach to find the terms in \({({x^2} + 3)^7}\) (M1)
eg\(\,\,\,\,\,\)\(\left( {\begin{array}{*{20}{c}} 7 \\ r \end{array}} \right){({x^2})^{7 – r}}{(3)^r},{\text{ }}{({x^2})^7}{(3)^0} + {({x^2})^6}{(3)^1} + \ldots \), Pascal’s triangle to 7th row
identifying correct term (may be indicated in expansion) (A1)
eg\(\,\,\,\,\,\)6th term, \(r = 3,{\text{ }}\left( {\begin{array}{*{20}{c}} 7 \\ 3 \end{array}} \right),{\text{ (}}{{\text{x}}^2}{)^3}{(3)^4}\)
correct working (may be seen in expansion) (A1)
eg\(\,\,\,\,\,\)\(\left( {\begin{array}{*{20}{c}} 7 \\ 4 \end{array}} \right){{\text{(}}{{\text{x}}^2})^3}{(3)^4},{\text{ }}35 \times {3^4}\)
correct term (A1)
\(2835{x^6}\)
differentiating their term in \({x^6}\) (M1)
eg\(\,\,\,\,\,\)\((2835{x^6})’,{\text{ (6)(2835}}{{\text{x}}^5})\)
\(17010{x^5}\) A1 N3
[7 marks]