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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.
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.
Describe the behaviour of the graph of \(f\) for large \(|x|\) .
Write down the range of \(f\) .
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]
(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]
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]
correct inequalities, \(y \le – 2\) , \(y > 3\) , FT from (a)(i) and (c) A1A1 N2
[2 marks]
Question
Let \(f(x) = \sqrt x \) . Line L is the normal to the graph of f at the point (4, 2) .
In the diagram below, the shaded region R is bounded by the x-axis, the graph of f and the line L .
Show that the equation of L is \(y = – 4x + 18\) .
Point A is the x-intercept of L . Find the x-coordinate of A.
Find an expression for the area of R .
The region R is rotated \(360^\circ \) about the x-axis. Find the volume of the solid formed, giving your answer in terms of \(\pi \) .
Answer/Explanation
Markscheme
finding derivative (A1)
e.g. \(f'(x) = \frac{1}{2}{x^{\frac{1}{2}}},\frac{{1}}{{2\sqrt x }}\)
correct value of derivative or its negative reciprocal (seen anywhere) A1
e.g. \(\frac{1}{{2\sqrt 4 }}\) , \(\frac{1}{4}\)
gradient of normal = \(\frac{1}{{{\text{gradient of tangent}}}}\) (seen anywhere) A1
e.g. \( – \frac{1}{{f'(4)}} = – 4\) , \( – 2\sqrt x \)
substituting into equation of line (for normal) M1
e.g. \(y – 2 = – 4(x – 4)\)
\(y = – 4x + 18\) AG N0
[4 marks]
recognition that \(y = 0\) at A (M1)
e.g. \( – 4x + 18 = 0\)
\(x = \frac{{18}}{4}\) \(\left( { = \frac{9}{2}} \right)\) A1 N2
[2 marks]
splitting into two appropriate parts (areas and/or integrals) (M1)
correct expression for area of R A2 N3
e.g. area of R = \(\int_0^4 {\sqrt x } {\rm{d}}x + \int_4^{4.5} {( – 4x + 18){\rm{d}}x} \) , \(\int_0^4 {\sqrt x } {\rm{d}}x + \frac{1}{2} \times 0.5 \times 2\) (triangle)
Note: Award A1 if dx is missing.
[3 marks]
correct expression for the volume from \(x = 0\) to \(x = 4\) (A1)
e.g. \(V = \int_0^4 {\pi \left[ {f{{(x)}^2}} \right]} {\rm{d}}x\) , \({\int_0^4 {\pi \sqrt x } ^2}{\rm{d}}x\) , \(\int_0^4 {\pi x{\rm{d}}x} \)
\(V = \left[ {\frac{1}{2}\pi {x^2}} \right]_0^4\) A1
\(V = \pi \left( {\frac{1}{2} \times 16 – \frac{1}{2} \times 0} \right)\) (A1)
\(V = 8\pi \) A1
finding the volume from \(x = 4\) to \(x = 4.5\)
EITHER
recognizing a cone (M1)
e.g. \(V = \frac{1}{3}\pi {r^2}h\)
\(V = \frac{1}{3}\pi {(2)^2} \times \frac{1}{2}\) (A1)
\( = \frac{{2\pi }}{3}\) A1
total volume is \(8\pi + \frac{2}{3}\pi \) \(\left( { = \frac{{26}}{3}\pi } \right)\) A1 N4
OR
\(V = \pi \int_4^{4.5} {{{( – 4x + 18)}^2}{\rm{d}}x} \) (M1)
\( = \int_4^{4.5} {\pi (16{x^2} – 144x + 324){\rm{d}}x} \)
\( = \pi \left[ {\frac{{16}}{3}{x^3} – 72{x^2} + 324x} \right]_4^{4.5}\) A1
\( = \frac{{2\pi }}{3}\) A1
total volume is \(8\pi + \frac{2}{3}\pi \) \(\left( { = \frac{{26}}{3}\pi } \right)\) A1 N4
[8 marks]
Question
Let \(f(x) = 6 + 6\sin x\) . Part of the graph of f is shown below.
The shaded region is enclosed by the curve of f , the x-axis, and the y-axis.
Solve for \(0 \le x < 2\pi \)
(i) \(6 + 6\sin x = 6\) ;
(ii) \(6 + 6\sin x = 0\) .
Write down the exact value of the x-intercept of f , for \(0 \le x < 2\pi \) .
The area of the shaded region is k . Find the value of k , giving your answer in terms of \(\pi \) .
Let \(g(x) = 6 + 6\sin \left( {x – \frac{\pi }{2}} \right)\) . The graph of f is transformed to the graph of g.
Give a full geometric description of this transformation.
Let \(g(x) = 6 + 6\sin \left( {x – \frac{\pi }{2}} \right)\) . The graph of f is transformed to the graph of g.
Given that \(\int_p^{p + \frac{{3\pi }}{2}} {g(x){\rm{d}}x} = k\) and \(0 \le p < 2\pi \) , write down the two values of p.
Answer/Explanation
Markscheme
(i) \(\sin x = 0\) A1
\(x = 0\) , \(x = \pi \) A1A1 N2
(ii) \(\sin x = – 1\) A1
\(x = \frac{{3\pi }}{2}\) A1 N1
[5 marks]
\(\frac{{3\pi }}{2}\) A1 N1
[1 mark]
evidence of using anti-differentiation (M1)
e.g. \(\int_0^{\frac{{3\pi }}{2}} {(6 + 6\sin x){\rm{d}}x} \)
correct integral \(6x – 6\cos x\) (seen anywhere) A1A1
correct substitution (A1)
e.g. \(6\left( {\frac{{3\pi }}{2}} \right) – 6\cos \left( {\frac{{3\pi }}{2}} \right) – ( – 6\cos 0)\) , \(9\pi – 0 + 6\)
\(k = 9\pi + 6\) A1A1 N3
[6 marks]
translation of \(\left( {\begin{array}{*{20}{c}}
{\frac{\pi }{2}}\\
0
\end{array}} \right)\) A1A1 N2
[2 marks]
recognizing that the area under g is the same as the shaded region in f (M1)
\(p = \frac{\pi }{2}\) , \(p = 0\) A1A1 N3
[3 marks]
Question
Consider \(f(x) = 2k{x^2} – 4kx + 1\) , for \(k \ne 0\) . The equation \(f(x) = 0\) has two equal roots.
Find the value of k .
The line \(y = p\) intersects the graph of f . Find all possible values of p .
Answer/Explanation
Markscheme
valid approach (M1)
e.g. \({b^2} – 4ac\) , \(\Delta = 0\) , \({( – 4k)^2} – 4(2k)(1)\)
correct equation A1
e.g. \({( – 4k)^2} – 4(2k)(1) = 0\) , \(16{k^2} = 8k\) , \(2{k^2} – k = 0\)
correct manipulation A1
e.g. \(8k(2k – 1)\) , \(\frac{{8 \pm \sqrt {64} }}{{32}}\)
\(k = \frac{1}{2}\) A2 N3
[5 marks]
recognizing vertex is on the x-axis M1
e.g. (1, 0) , sketch of parabola opening upward from the x-axis
\(p \ge 0\) A1 N1
[2 marks]
Question
Let \(f(x) = \frac{x}{{ – 2{x^2} + 5x – 2}}\) for \( – 2 \le x \le 4\) , \(x \ne \frac{1}{2}\) , \(x \ne 2\) . The graph of \(f\) is given below.
The graph of \(f\) has a local minimum at A(\(1\), \(1\)) and a local maximum at B.
Use the quotient rule to show that \(f'(x) = \frac{{2{x^2} – 2}}{{{{( – 2{x^2} + 5x – 2)}^2}}}\) .
Hence find the coordinates of B.
Given that the line \(y = k\) does not meet the graph of f , find the possible values of k .
Answer/Explanation
Markscheme
correct derivatives applied in quotient rule (A1)A1A1
\(1\), \( – 4x + 5\)
Note: Award (A1) for 1, A1 for \( – 4x\) and A1 for \(5\), only if it is clear candidates are using the quotient rule.
correct substitution into quotient rule A1
e.g. \(\frac{{1 \times ( – 2{x^2} + 5x – 2) – x( – 4x + 5)}}{{{{( – 2{x^2} + 5x – 2)}^2}}}\) , \(\frac{{ – 2{x^2} + 5x – 2 – x( – 4x + 5)}}{{{{( – 2{x^2} + 5x – 2)}^2}}}\)
correct working (A1)
e.g. \(\frac{{ – 2{x^2} + 5x – 2 – ( – 4{x^2} + 5x)}}{{{{( – 2{x^2} + 5x – 2)}^2}}}\)
expression clearly leading to the answer A1
e.g. \(\frac{{ – 2{x^2} + 5x – 2 + 4{x^2} – 5x}}{{{{( – 2{x^2} + 5x – 2)}^2}}}\)
\(f'(x) = \frac{{2{x^2} – 2}}{{{{( – 2{x^2} + 5x – 2)}^2}}}\) AG N0
[6 marks]
evidence of attempting to solve \(f'(x) = 0\) (M1)
e.g. \(2{x^2} – 2 = 0\)
evidence of correct working A1
e.g. \({x^2} = 1,\frac{{ \pm \sqrt {16} }}{4}{\text{, }}2(x – 1)(x + 1)\)
correct solution to quadratic (A1)
e.g. \(x = \pm 1\)
correct x-coordinate \(x = – 1\) (may be seen in coordinate form \(\left( { – 1,\frac{1}{9}} \right)\) ) A1 N2
attempt to substitute \( – 1\) into f (do not accept any other value) (M1)
e.g. \(f( – 1) = \frac{{ – 1}}{{ – 2 \times {{( – 1)}^2} + 5 \times ( – 1) – 2}}\)
correct working
e.g. \(\frac{{ – 1}}{{ – 2 – 5 – 2}}\) A1
correct y-coordinate \(y = \frac{1}{9}\) (may be seen in coordinate form \(\left( { – 1,\frac{1}{9}} \right)\) ) A1 N2
[7 marks]
recognizing values between max and min (R1)
\(\frac{1}{9} < k < 1\) A2 N3
[3 marks]
Question
Let \(f(x) = 3x – 2\) and \(g(x) = \frac{5}{{3x}}\), for \(x \ne 0\).
Let \(h(x) = \frac{5}{{x + 2}}\), for \(x \geqslant 0\). The graph of h has a horizontal asymptote at \(y = 0\).
Find \({f^{ – 1}}(x)\).
Show that \(\left( {g \circ {f^{ – 1}}} \right)(x) = \frac{5}{{x + 2}}\).
Find the \(y\)-intercept of the graph of \(h\).
Hence, sketch the graph of \(h\).
For the graph of \({h^{ – 1}}\), write down the \(x\)-intercept;
For the graph of \({h^{ – 1}}\), write down the equation of the vertical asymptote.
Given that \({h^{ – 1}}(a) = 3\), find the value of \(a\).
Answer/Explanation
Markscheme
interchanging \(x\) and \(y\) (M1)
eg \(x = 3y – 2\)
\({f^{ – 1}}(x) = \frac{{x + 2}}{3}{\text{ }}\left( {{\text{accept }}y = \frac{{x + 2}}{3},{\text{ }}\frac{{x + 2}}{3}} \right)\) A1 N2
[2 marks]
attempt to form composite (in any order) (M1)
eg \(g\left( {\frac{{x + 2}}{3}} \right),{\text{ }}\frac{{\frac{5}{{3x}} + 2}}{3}\)
correct substitution A1
eg \(\frac{5}{{3\left( {\frac{{x + 2}}{3}} \right)}}\)
\(\left( {g \circ {f^{ – 1}}} \right)(x) = \frac{5}{{x + 2}}\) AG N0
[2 marks]
valid approach (M1)
eg \(h(0),{\text{ }}\frac{5}{{0 + 2}}\)
\(y = \frac{5}{2}{\text{ }}\left( {{\text{accept (0, 2.5)}}} \right)\) A1 N2
[2 marks]
A1A2 N3
Notes: Award A1 for approximately correct shape (reciprocal, decreasing, concave up).
Only if this A1 is awarded, award A2 for all the following approximately correct features: y-intercept at \((0, 2.5)\), asymptotic to x-axis, correct domain \(x \geqslant 0\).
If only two of these features are correct, award A1.
[3 marks]
\(x = \frac{5}{2}{\text{ }}\left( {{\text{accept (2.5, 0)}}} \right)\) A1 N1
[1 mark]
\(x = 0\) (must be an equation) A1 N1
[1 mark]
METHOD 1
attempt to substitute \(3\) into \(h\) (seen anywhere) (M1)
eg \(h(3),{\text{ }}\frac{5}{{3 + 2}}\)
correct equation (A1)
eg \(a = \frac{5}{{3 + 2}},{\text{ }}h(3) = a\)
\(a = 1\) A1 N2
[3 marks]
METHOD 2
attempt to find inverse (may be seen in (d)) (M1)
eg \(x = \frac{5}{{y + 2}},{\text{ }}{h^{ – 1}} = \frac{5}{x} – 2,{\text{ }}\frac{5}{x} + 2\)
correct equation, \(\frac{5}{x} – 2 = 3\) (A1)
\(a = 1\) A1 N2
[3 marks]
Question
Consider \(f(x) = {x^2} + qx + r\). The graph of \(f\) has a minimum value when \(x = – 1.5\).
The distance between the two zeros of \(f\) is 9.
Show that the two zeros are 3 and \( – 6\).
Find the value of \(q\) and of \(r\).
Answer/Explanation
Markscheme
recognition that the \(x\)-coordinate of the vertex is \( – 1.5\) (seen anywhere) (M1)
eg\(\,\,\,\,\,\)axis of symmetry is \( – 1.5\), sketch, \(f'( – 1.5) = 0\)
correct working to find the zeroes A1
eg\(\,\,\,\,\,\)\( – 1.5 \pm 4.5\)
\(x = – 6\) and \(x = 3\) AG N0
[2 marks]
METHOD 1 (using factors)
attempt to write factors (M1)
eg\(\,\,\,\,\,\)\((x – 6)(x + 3)\)
correct factors A1
eg\(\,\,\,\,\,\)\((x – 3)(x + 6)\)
\(q = 3,{\text{ }}r = – 18\) A1A1 N3
METHOD 2 (using derivative or vertex)
valid approach to find \(q\) (M1)
eg\(\,\,\,\,\,\)\(f'( – 1.5) = 0,{\text{ }} – \frac{q}{{2a}} = – 1.5\)
\(q = 3\) A1
correct substitution A1
eg\(\,\,\,\,\,\)\({3^2} + 3(3) + r = 0,{\text{ }}{( – 6)^2} + 3( – 6) + r = 0\)
\(r = – 18\) A1
\(q = 3,{\text{ }}r = – 18\) N3
METHOD 3 (solving simultaneously)
valid approach setting up system of two equations (M1)
eg\(\,\,\,\,\,\)\(9 + 3q + r = 0,{\text{ }}36 – 6q + r = 0\)
one correct value
eg\(\,\,\,\,\,\)\(q = 3,{\text{ }}r = – 18\) A1
correct substitution A1
eg\(\,\,\,\,\,\)\({3^2} + 3(3) + r = 0,{\text{ }}{( – 6)^2} + 3( – 6) + r = 0,{\text{ }}{3^2} + 3q – 18 = 0,{\text{ }}36 – 6q – 18 = 0\)
second correct value A1
eg\(\,\,\,\,\,\)\(q = 3,{\text{ }}r = – 18\)
\(q = 3,{\text{ }}r = – 18\) N3
[4 marks]
Question
The following diagram shows part of the graph of a quadratic function \(f\).
The vertex is at \((3,{\text{ }} – 1)\) and the \(x\)-intercepts at 2 and 4.
The function \(f\) can be written in the form \(f(x) = {(x – h)^2} + k\).
The function can also be written in the form \(f(x) = (x – a)(x – b)\).
Write down the value of \(h\) and of \(k\).
Write down the value of \(a\) and of \(b\).
Find the \(y\)-intercept.
Answer/Explanation
Markscheme
\(h = 3,{\text{ }}k = – 1\) A1A1 N2
[2 marks]
\(a = 2,{\text{ }}b = 4{\text{ }}({\text{or }}a = 4,{\text{ }}b = 2)\) A1A1 N2
[2 marks]
attempt to substitute \(x = 0\) into their \(f\) (M1)
eg\(\,\,\,\,\,\)\({(0 – 3)^2} – 1,{\text{ }}(0 – 2)(0 – 4)\)
\(y = 8\) A1 N2
[2 marks]
Question
A quadratic function \(f\) can be written in the form \(f(x) = a(x – p)(x – 3)\). The graph of \(f\) has axis of symmetry \(x = 2.5\) and \(y\)-intercept at \((0,{\text{ }} – 6)\)
Find the value of \(p\).
Find the value of \(a\).
The line \(y = kx – 5\) is a tangent to the curve of \(f\). Find the values of \(k\).
Answer/Explanation
Markscheme
METHOD 1 (using x-intercept)
determining that 3 is an \(x\)-intercept (M1)
eg\(\,\,\,\,\,\)\(x – 3 = 0\), 
valid approach (M1)
eg\(\,\,\,\,\,\)\(3 – 2.5,{\text{ }}\frac{{p + 3}}{2} = 2.5\)
\(p = 2\) A1 N2
METHOD 2 (expanding f (x))
correct expansion (accept absence of \(a\)) (A1)
eg\(\,\,\,\,\,\)\(a{x^2} – a(3 + p)x + 3ap,{\text{ }}{x^2} – (3 + p)x + 3p\)
valid approach involving equation of axis of symmetry (M1)
eg\(\,\,\,\,\,\)\(\frac{{ – b}}{{2a}} = 2.5,{\text{ }}\frac{{a(3 + p)}}{{2a}} = \frac{5}{2},{\text{ }}\frac{{3 + p}}{2} = \frac{5}{2}\)
\(p = 2\) A1 N2
METHOD 3 (using derivative)
correct derivative (accept absence of \(a\)) (A1)
eg\(\,\,\,\,\,\)\(a(2x – 3 – p),{\text{ }}2x – 3 – p\)
valid approach (M1)
eg\(\,\,\,\,\,\)\(f’(2.5) = 0\)
\(p = 2\) A1 N2
[3 marks]
attempt to substitute \((0,{\text{ }} – 6)\) (M1)
eg\(\,\,\,\,\,\)\( – 6 = a(0 – 2)(0 – 3),{\text{ }}0 = a( – 8)( – 9),{\text{ }}a{(0)^2} – 5a(0) + 6a = – 6\)
correct working (A1)
eg\(\,\,\,\,\,\)\( – 6 = 6a\)
\(a = – 1\) A1 N2
[3 marks]
METHOD 1 (using discriminant)
recognizing tangent intersects curve once (M1)
recognizing one solution when discriminant = 0 M1
attempt to set up equation (M1)
eg\(\,\,\,\,\,\)\(g = f,{\text{ }}kx – 5 = – {x^2} + 5x – 6\)
rearranging their equation to equal zero (M1)
eg\(\,\,\,\,\,\)\({x^2} – 5x + kx + 1 = 0\)
correct discriminant (if seen explicitly, not just in quadratic formula) A1
eg\(\,\,\,\,\,\)\({(k – 5)^2} – 4,{\text{ }}25 – 10k + {k^2} – 4\)
correct working (A1)
eg\(\,\,\,\,\,\)\(k – 5 = \pm 2,{\text{ }}(k – 3)(k – 7) = 0,{\text{ }}\frac{{10 \pm \sqrt {100 – 4 \times 21} }}{2}\)
\(k = 3,{\text{ }}7\) A1A1 N0
METHOD 2 (using derivatives)
attempt to set up equation (M1)
eg\(\,\,\,\,\,\)\(g = f,{\text{ }}kx – 5 = – {x^2} + 5x – 6\)
recognizing derivative/slope are equal (M1)
eg\(\,\,\,\,\,\)\(f’ = {m_T},{\text{ }}f’ = k\)
correct derivative of \(f\) (A1)
eg\(\,\,\,\,\,\)\( – 2x + 5\)
attempt to set up equation in terms of either \(x\) or \(k\) M1
eg\(\,\,\,\,\,\)\(( – 2x + 5)x – 5 = – {x^2} + 5x – 6,{\text{ }}k\left( {\frac{{5 – k}}{2}} \right) – 5 = – {\left( {\frac{{5 – k}}{2}} \right)^2} + 5\left( {\frac{{5 – k}}{2}} \right) – 6\)
rearranging their equation to equal zero (M1)
eg\(\,\,\,\,\,\)\({x^2} – 1 = 0,{\text{ }}{k^2} – 10k + 21 = 0\)
correct working (A1)
eg\(\,\,\,\,\,\)\(x = \pm 1,{\text{ }}(k – 3)(k – 7) = 0,{\text{ }}\frac{{10 \pm \sqrt {100 – 4 \times 21} }}{2}\)
\(k = 3,{\text{ }}7\) A1A1 N0
[8 marks]
Question
Let \(f(x) = {x^2}\). The following diagram shows part of the graph of \(f\).
The line \(L\) is the tangent to the graph of \(f\) at the point \({\text{A}}( – k,{\text{ }}{k^2})\), and intersects the \(x\)-axis at point B. The point C is \(( – k,{\text{ }}0)\).
The region \(R\) is enclosed by \(L\), the graph of \(f\), and the \(x\)-axis. This is shown in the following diagram.
Write down \(f'(x)\).
Find the gradient of \(L\).
Show that the \(x\)-coordinate of B is \( – \frac{k}{2}\).
Find the area of triangle ABC, giving your answer in terms of \(k\).
Given that the area of triangle ABC is \(p\) times the area of \(R\), find the value of \(p\).
Answer/Explanation
Markscheme
\(f'(x) = 2x\) A1 N1
[1 mark]
attempt to substitute \(x = – k\) into their derivative (M1)
gradient of \(L\) is \( – 2k\) A1 N2
[2 marks]
METHOD 1
attempt to substitute coordinates of A and their gradient into equation of a line (M1)
eg\(\,\,\,\,\,\)\({k^2} = – 2k( – k) + b\)
correct equation of \(L\) in any form (A1)
eg\(\,\,\,\,\,\)\(y – {k^2} = – 2k(x + k),{\text{ }}y = – 2kx – {k^2}\)
valid approach (M1)
eg\(\,\,\,\,\,\)\(y = 0\)
correct substitution into \(L\) equation A1
eg\(\,\,\,\,\,\)\( – {k^2} = – 2kx – 2{k^2},{\text{ }}0 = – 2kx – {k^2}\)
correct working A1
eg\(\,\,\,\,\,\)\(2kx = – {k^2}\)
\(x = – \frac{k}{2}\) AG N0
METHOD 2
valid approach (M1)
eg\(\,\,\,\,\,\)\({\text{gradient}} = \frac{{{y_2} – {y_1}}}{{{x_2} – {x_1}}},{\text{ }} – 2k = \frac{{{\text{rise}}}}{{{\text{run}}}}\)
recognizing \(y = 0\) at B (A1)
attempt to substitute coordinates of A and B into slope formula (M1)
eg\(\,\,\,\,\,\)\(\frac{{{k^2} – 0}}{{ – k – x}},{\text{ }}\frac{{ – {k^2}}}{{x + k}}\)
correct equation A1
eg\(\,\,\,\,\,\)\(\frac{{{k^2} – 0}}{{ – k – x}} = – 2k,{\text{ }}\frac{{ – {k^2}}}{{x + k}} = – 2k,{\text{ }} – {k^2} = – 2k(x + k)\)
correct working A1
eg\(\,\,\,\,\,\)\(2kx = – {k^2}\)
\(x = – \frac{k}{2}\) AG N0
[5 marks]
valid approach to find area of triangle (M1)
eg\(\,\,\,\,\,\)\(\frac{1}{2}({k^2})\left( {\frac{k}{2}} \right)\)
area of \({\text{ABC}} = \frac{{{k^3}}}{4}\) A1 N2
[2 marks]
METHOD 1 (\(\int {f – {\text{triangle}}} \))
valid approach to find area from \( – k\) to 0 (M1)
eg\(\,\,\,\,\,\)\(\int_{ – k}^0 {{x^2}{\text{d}}x,{\text{ }}\int_0^{ – k} f } \)
correct integration (seen anywhere, even if M0 awarded) A1
eg\(\,\,\,\,\,\)\(\frac{{{x^3}}}{3},{\text{ }}\left[ {\frac{1}{3}{x^3}} \right]_{ – k}^0\)
substituting their limits into their integrated function and subtracting (M1)
eg\(\,\,\,\,\,\)\(0 – \frac{{{{( – k)}^3}}}{3}\), area from \( – k\) to 0 is \(\frac{{{k^3}}}{3}\)
Note: Award M0 for substituting into original or differentiated function.
attempt to find area of \(R\) (M1)
eg\(\,\,\,\,\,\)\(\int_{ – k}^0 {f(x){\text{d}}x – {\text{ triangle}}} \)
correct working for \(R\) (A1)
eg\(\,\,\,\,\,\)\(\frac{{{k^3}}}{3} – \frac{{{k^3}}}{4},{\text{ }}R = \frac{{{k^3}}}{{12}}\)
correct substitution into \({\text{triangle}} = pR\) (A1)
eg\(\,\,\,\,\,\)\(\frac{{{k^3}}}{4} = p\left( {\frac{{{k^3}}}{3} – \frac{{{k^3}}}{4}} \right),{\text{ }}\frac{{{k^3}}}{4} = p\left( {\frac{{{k^3}}}{{12}}} \right)\)
\(p = 3\) A1 N2
METHOD 2 (\(\int {(f – L)} \))
valid approach to find area of \(R\) (M1)
eg\(\,\,\,\,\,\)\(\int_{ – k}^{ – \frac{k}{2}} {{x^2} – ( – 2kx – {k^2}){\text{d}}x + \int_{ – \frac{k}{2}}^0 {{x^2}{\text{d}}x,{\text{ }}\int_{ – k}^{ – \frac{k}{2}} {(f – L) + \int_{ – \frac{k}{2}}^0 f } } } \)
correct integration (seen anywhere, even if M0 awarded) A2
eg\(\,\,\,\,\,\)\(\frac{{{x^3}}}{3} + k{x^2} + {k^2}x,{\text{ }}\left[ {\frac{{{x^3}}}{3} + k{x^2} + {k^2}x} \right]_{ – k}^{ – \frac{k}{2}} + \left[ {\frac{{{x^3}}}{3}} \right]_{ – \frac{k}{2}}^0\)
substituting their limits into their integrated function and subtracting (M1)
eg\(\,\,\,\,\,\)\(\left( {\frac{{{{\left( { – \frac{k}{2}} \right)}^3}}}{3} + k{{\left( { – \frac{k}{2}} \right)}^2} + {k^2}\left( { – \frac{k}{2}} \right)} \right) – \left( {\frac{{{{( – k)}^3}}}{3} + k{{( – k)}^2} + {k^2}( – k)} \right) + (0) – \left( {\frac{{{{\left( { – \frac{k}{2}} \right)}^3}}}{3}} \right)\)
Note: Award M0 for substituting into original or differentiated function.
correct working for \(R\) (A1)
eg\(\,\,\,\,\,\)\(\frac{{{k^3}}}{{24}} + \frac{{{k^3}}}{{24}},{\text{ }} – \frac{{{k^3}}}{{24}} + \frac{{{k^3}}}{4} – \frac{{{k^3}}}{2} + \frac{{{k^3}}}{3} – {k^3} + {k^3} + \frac{{{k^3}}}{{24}},{\text{ }}R = \frac{{{k^3}}}{{12}}\)
correct substitution into \({\text{triangle}} = pR\) (A1)
eg\(\,\,\,\,\,\)\(\frac{{{k^3}}}{4} = p\left( {\frac{{{k^3}}}{{24}} + \frac{{{k^3}}}{{24}}} \right),{\text{ }}\frac{{{k^3}}}{4} = p\left( {\frac{{{k^3}}}{{12}}} \right)\)
\(p = 3\) A1 N2
[7 marks]