Question
A family of cubic functions is defined as \({f_k}(x) = {k^2}{x^3} – k{x^2} + x,{\text{ }}k \in {\mathbb{Z}^ + }\) .
(a) Express in terms of k
(i) \({{f’}_k}(x){\text{ and }}{{f”}_k}(x)\) ;
(ii) the coordinates of the points of inflexion \({P_k}\) on the graphs of \({f_k}\) .
(b) Show that all \({P_k}\) lie on a straight line and state its equation.
(c) Show that for all values of k, the tangents to the graphs of \({f_k}\) at \({P_k}\) are parallel, and find the equation of the tangent lines.
Answer/Explanation
Markscheme
(a) (i) \({{f’}_k}(x) = 3{k^2}{x^2} – 2kx + 1\) A1
\({{f”}_k}(x) = 6{k^2}x – 2k\) A1
(ii) Setting \(f”(x) = 0\) M1
\( \Rightarrow 6{k^2}x – 2k = 0 \Rightarrow x = \frac{1}{{3k}}\) A1
\(f\left( {\frac{1}{{3k}}} \right) = {k^2}{\left( {\frac{1}{{3k}}} \right)^3} – k{\left( {\frac{1}{{3k}}} \right)^2} + \left( {\frac{1}{{3k}}} \right)\) M1
\( = \frac{7}{{27k}}\) A1
Hence, \({P_k}{\text{ is }}\left( {\frac{1}{{3k}},\frac{7}{{27k}}} \right)\)
[6 marks]
(b) Equation of the straight line is \(y = \frac{7}{9}x\) A1
As this equation is independent of k, all \({P_k}\) lie on this straight line R1
[2 marks]
(c) Gradient of tangent at \({P_k}\) :
\(f'({P_k}) = f’\left( {\frac{1}{{3k}}} \right) = 3{k^2}{\left( {\frac{1}{{3k}}} \right)^2} – 2k\left( {\frac{1}{{3k}}} \right) + 1 = \frac{2}{3}\) M1A1
As the gradient is independent of k, the tangents are parallel. R1
\(\frac{7}{{27k}} = \frac{2}{3} \times \frac{1}{{3k}} + c \Rightarrow c = \frac{1}{{27k}}\) (A1)
The equation is \(y = \frac{2}{3}x + \frac{1}{{27k}}\) A1
[5 marks]
Total [13 marks]
Examiners report
Many candidates scored the full 6 marks for part (a). The main mistake evidenced was to treat k as a variable, and hence use the product rule to differentiate. Of the many candidates who attempted parts (b) and (c), few scored the R1 marks in either part, but did manage to get the equations of the straight lines.
Question
Consider the curve with equation \(f(x) = {{\text{e}}^{ – 2{x^2}}}{\text{ for }}x < 0\) .
Find the coordinates of the point of inflexion and justify that it is a point of inflexion.
Answer/Explanation
Markscheme
METHOD 1
EITHER
Using the graph of \(y = f'(x)\) (M1)
A1
The maximum of \(f'(x)\) occurs at x = −0.5 . A1
OR
Using the graph of \(y = f”(x)\). (M1)
A1
The zero of \(f”(x)\) occurs at x = −0.5 . A1
THEN
Note: Do not award this A1 for stating x = ±0.5 as the final answer for x .
\(f( – 0.5) = 0.607( = {{\text{e}}^{ – 0.5}})\) A2
Note: Do not award this A1 for also stating (0.5, 0.607) as a coordinate.
EITHER
Correctly labelled graph of \(f'(x)\) for \(x < 0\) denoting the maximum \(f'(x)\) R1
(e.g. \(f'( – 0.6) = 1.17\) and \(f'( – 0.4) = 1.16\) stated) A1 N2
OR
Correctly labelled graph of \(f”(x)\) for \(x < 0\) denoting the maximum \(f'(x)\) R1
(e.g. \(f”( – 0.6) = 0.857\) and \(f”( – 0.4) = – 1.05\) stated) A1 N2
OR
\(f'(0.5) \approx 1.21\). \(f'(x) < 1.21\) just to the left of \(x = – \frac{1}{2}\)
and \(f'(x) < 1.21\) just to the right of \(x = – \frac{1}{2}\) R1
(e.g. \(f'( – 0.6) = 1.17\) and \(f'( – 0.4) = 1.16\) stated) A1 N2
OR
\(f”(x) > 0\) just to the left of \(x = – \frac{1}{2}\) and \(f”(x) < 0\) just to the right of \(x = – \frac{1}{2}\) R1
(e.g. \(f”( – 0.6) = 0.857\) and \(f”( – 0.4) = – 1.05\) stated) A1 N2
[7 marks]
METHOD 2
\(f'(x) = – 4x{{\text{e}}^{ – 2{x^2}}}\) A1
\(f”(x) = – 4{{\text{e}}^{ – 2{x^2}}} + 16{x^2}{{\text{e}}^{ – 2{x^2}}}\,\,\,\,\,\left( { = (16{x^2} – 4){{\text{e}}^{ – 2{x^2}}}} \right)\) A1
Attempting to solve \(f”(x) = 0\) (M1)
\(x = – \frac{1}{2}\) A1
Note: Do not award this A1 for stating \(x = \pm \frac{1}{2}\) as the final answer for x .
\(f\left( { – \frac{1}{2}} \right) = \frac{1}{{\sqrt {\text{e}} }}{\text{ }}( = 0.607)\) A1
Note: Do not award this A1 for also stating \(\left( {\frac{1}{2},\frac{1}{{\sqrt {\text{e}} }}} \right)\) as a coordinate.
EITHER
Correctly labelled graph of \(f'(x)\) for \(x < 0\) denoting the maximum \(f'(x)\) R1
(e.g. \(f'( – 0.6) = 1.17\) and \(f'( – 0.4) = 1.16\) stated) A1 N2
OR
Correctly labelled graph of \(f”(x)\) for \(x < 0\) denoting the maximum \(f'(x)\) R1
(e.g. \(f”( – 0.6) = 0.857\) and \(f”( – 0.4) = – 1.05\) stated) A1 N2
OR
\(f'(0.5) \approx 1.21\). \(f'(x) < 1.21\) just to the left of \(x = – \frac{1}{2}\)
and \(f'(x) < 1.21\) just to the right of \(x = – \frac{1}{2}\) R1
(e.g. \(f'( – 0.6) = 1.17\) and \(f'( – 0.4) = 1.16\) stated) A1 N2
OR
\(f”(x) > 0\) just to the left of \(x = – \frac{1}{2}\) and \(f”(x) < 0\) just to the right of \(x = – \frac{1}{2}\) R1
(e.g. \(f”( – 0.6) = 0.857\) and \(f”( – 0.4) = – 1.05\) stated) A1 N2
[7 marks]
Examiners report
Most candidates adopted an algebraic approach rather than a graphical approach. Most candidates found \(f'(x)\) correctly, however when attempting to find \(f”(x)\), a surprisingly large number either made algebraic errors using the product rule or seemingly used an incorrect form of the product rule. A large number ignored the domain restriction and either expressed \(x = \pm \frac{1}{2}\) as the x-coordinates of the point of inflection or identified \(x = \frac{1}{2}\) rather than \(x = – \frac{1}{2}\). Most candidates were unsuccessful in their attempts to justify the existence of the point of inflection.
Question
The function f is defined by \(f(x) = x\sqrt {9 – {x^2}} + 2\arcsin \left( {\frac{x}{3}} \right)\).
(a) Write down the largest possible domain, for each of the two terms of the function, f , and hence state the largest possible domain, D , for f .
(b) Find the volume generated when the region bounded by the curve y = f(x) , the x-axis, the y-axis and the line x = 2.8 is rotated through \(2\pi \) radians about the x-axis.
(c) Find \(f'(x)\) in simplified form.
(d) Hence show that \(\int_{ – p}^p {\frac{{11 – 2{x^2}}}{{\sqrt {9 – {x^2}} }}} {\text{d}}x = 2p\sqrt {9 – {p^2}} + 4\arcsin \left( {\frac{p}{3}} \right)\), where \(p \in D\) .
(e) Find the value of p which maximises the value of the integral in (d).
(f) (i) Show that \(f”(x) = \frac{{x(2{x^2} – 25)}}{{{{(9 – {x^2})}^{\frac{3}{2}}}}}\).
(ii) Hence justify that f(x) has a point of inflexion at x = 0 , but not at \(x = \pm \sqrt {\frac{{25}}{2}} \) .
Answer/Explanation
Markscheme
(a) For \(x\sqrt {9 – {x^2}} \), \( – 3 \leqslant x \leqslant 3\) and for \(2\arcsin \left( {\frac{x}{3}} \right)\), \( – 3 \leqslant x \leqslant 3\) A1
\( \Rightarrow D{\text{ is }} – 3 \leqslant x \leqslant 3\) A1
[2 marks]
(b) \(V = \pi \int_0^{2.8} {{{\left( {x\sqrt {9 – {x^2}} = 2\arcsin \frac{x}{3}} \right)}^2}{\text{d}}x} \) M1A1
= 181 A1
[3 marks]
(c) \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = {(9 – {x^2})^{\frac{1}{2}}} – \frac{{{x^2}}}{{{{(9 – {x^2})}^{\frac{1}{2}}}}} + \frac{{\frac{2}{3}}}{{\sqrt {1 – \frac{{{x^2}}}{9}} }}\) M1A1
\( = {(9 – {x^2})^{\frac{1}{2}}} – \frac{{{x^2}}}{{{{(9 – {x^2})}^{\frac{1}{2}}}}} + \frac{2}{{{{(9 – {x^2})}^{\frac{1}{2}}}}}\) A1
\( = \frac{{9 – {x^2} – {x^2} + 2}}{{{{(9 – {x^2})}^{\frac{1}{2}}}}}\) A1
\( = \frac{{11 – 2{x^2}}}{{\sqrt {9 – {x^2}} }}\) A1
[5 marks]
(d) \(\int_{ – p}^p {\frac{{11 – 2{x^2}}}{{\sqrt {9 – {x^2}} }}{\text{d}}x = \left[ {x\sqrt {9 – {x^2}} + 2\arcsin \frac{x}{3}} \right]_{ – p}^p} \) M1
\( = p\sqrt {9 – {p^2}} + 2\arcsin \frac{p}{3} + p\sqrt {9 – {p^2}} + 2\arcsin \frac{p}{3}\) A1
\( = 2p\sqrt {9 – {p^2}} + 4\arcsin \left( {\frac{p}{3}} \right)\) AG
[2 marks]
(e) \(11 – 2{p^2} = 0\) M1
\(p = 2.35\,\,\,\,\,\left( {\sqrt {\frac{{11}}{2}} } \right)\) A1
Note: Award A0 for \(p = \pm 2.35\) .
[2 marks]
(f) (i) \(f”(x) = \frac{{{{(9 – {x^2})}^{\frac{1}{2}}}( – 4x) + x(11 – 2{x^2}){{(9 – {x^2})}^{ – \frac{1}{2}}}}}{{9 – {x^2}}}\) M1A1
\( = \frac{{ – 4x(9 – {x^2}) + x(11 – 2{x^2})}}{{{{(9 – {x^2})}^{\frac{3}{2}}}}}\) A1
\( = \frac{{ – 36x + 4{x^3} + 11x – 2{x^3}}}{{{{(9 – {x^2})}^{\frac{3}{2}}}}}\) A1
\( = \frac{{x(2{x^2} – 25)}}{{{{(9 – {x^2})}^{\frac{3}{2}}}}}\) AG
(ii) EITHER
When \(0 < x < 3\), \(f”(x) < 0\). When \( – 3 < x < 0\), \(f”(x) > 0\). A1
OR
\(f”(0) = 0\) A1
THEN
Hence \(f”(x)\) changes sign through x = 0 , giving a point of inflexion. R1
EITHER
\(x = \pm \sqrt {\frac{{25}}{2}} \) is outside the domain of f. R1
OR
\(x = \pm \sqrt {\frac{{25}}{2}} \) is not a root of \(f”(x) = 0\) . R1
[7 marks]
Total [21 marks]
Examiners report
It was disappointing to note that some candidates did not know the domain for arcsin. Most candidates knew what to do in (b) but sometimes the wrong answer was obtained due to the calculator being in the wrong mode. In (c), the differentiation was often disappointing with \(\arcsin \left( {\frac{x}{3}} \right)\) causing problems. In (f)(i), some candidates who failed to do (c) guessed the correct form of \(f'(x)\) (presumably from (d)) and then went on to find \(f”(x)\) correctly. In (f)(ii), the justification of a point of inflexion at x = 0 was sometimes incorrect – for example, some candidates showed simply that \(f'(x)\) is positive on either side of the origin which is not a valid reason.
Question
Let \(f(x) = \frac{{a + b{{\text{e}}^x}}}{{a{{\text{e}}^x} + b}}\), where \(0 < b < a\).
(a) Show that \(f'(x) = \frac{{({b^2} – {a^2}){{\text{e}}^x}}}{{{{(a{{\text{e}}^x} + b)}^2}}}\).
(b) Hence justify that the graph of f has no local maxima or minima.
(c) Given that the graph of f has a point of inflexion, find its coordinates.
(d) Show that the graph of f has exactly two asymptotes.
(e) Let a = 4 and b =1. Consider the region R enclosed by the graph of \(y = f(x)\), the y-axis and the line with equation \(y = \frac{1}{2}\).
Find the volume V of the solid obtained when R is rotated through \(2\pi \) about the x-axis.
Answer/Explanation
Markscheme
(a) \(f'(x) = \frac{{b{{\text{e}}^x}(a{{\text{e}}^x} + b) – a{{\text{e}}^x}(a + b{{\text{e}}^x})}}{{{{(a{{\text{e}}^x} + b)}^2}}}\) M1A1
\( = \frac{{ab{{\text{e}}^{2x}} + {b^2}{{\text{e}}^x} – {a^2}{{\text{e}}^x} – ab{{\text{e}}^{2x}}}}{{{{(a{{\text{e}}^x} + b)}^2}}}\) A1
\( = \frac{{({b^2} – {a^2}){{\text{e}}^x}}}{{{{(a{{\text{e}}^x} + b)}^2}}}\) AG
[3 marks]
(b) EITHER
\(f'(x) = 0 \Rightarrow ({b^2} – {a^2}){{\text{e}}^x} = 0 \Rightarrow b = \pm a{\text{ or }}{{\text{e}}^x} = 0\) A1
which is impossible as \(0 < b < a\) and \({{\text{e}}^x} > 0\) for all \(x \in \mathbb{R}\) R1
OR
\(f'(x) < 0\) for all \(x \in \mathbb{R}\) since \(0 < b < a\) and \({{\text{e}}^x} > 0\) for all \(x \in \mathbb{R}\) A1R1
OR
\(f'(x)\) cannot be equal to zero because \({{\text{e}}^x}\) is never equal to zero A1R1
[2 marks]
(c) EITHER
\(f”(x) = \frac{{({b^2} – {a^2}){{\text{e}}^x}{{(a{{\text{e}}^x} + b)}^2} – 2a{{\text{e}}^x}(a{{\text{e}}^x} + b)({b^2} – {a^2}){{\text{e}}^x}}}{{{{(a{{\text{e}}^x} + b)}^4}}}\) M1A1A1
Note: Award A1 for each term in the numerator.
\( = \frac{{({b^2} – {a^2}){{\text{e}}^x}(a{{\text{e}}^x} + b – 2a{{\text{e}}^x})}}{{{{(a{{\text{e}}^x} + b)}^3}}}\)
\( = \frac{{({b^2} – {a^2})(b – a{{\text{e}}^x}){{\text{e}}^x}}}{{{{(a{{\text{e}}^x} + b)}^3}}}\)
OR
\(f'(x) = ({b^2} – {a^2}){{\text{e}}^x}{(a{{\text{e}}^x} + b)^{ – 2}}\)
\(f”(x) = ({b^2} – {a^2}){{\text{e}}^x}{(a{{\text{e}}^x} + b)^{ – 2}} + ({b^2} – {a^2}){{\text{e}}^x}( – 2a{{\text{e}}^x}){(a{{\text{e}}^x} + b)^{ – 3}}\) M1A1A1
Note: Award A1 for each term.
\( = ({b^2} – {a^2}){{\text{e}}^x}{(a{{\text{e}}^x} + b)^{ – 3}}\left( {(a{{\text{e}}^x} + b) – 2a{{\text{e}}^x}} \right)\)
\( = ({b^2} – {a^2}){{\text{e}}^x}{(a{{\text{e}}^x} + b)^{ – 3}}(b – a{{\text{e}}^x})\)
THEN
\(f”(x) = 0 \Rightarrow b – a{{\text{e}}^x} = 0 \Rightarrow x = \ln \frac{b}{a}\) M1A1
\(f\left( {\ln \frac{b}{a}} \right) = \frac{{{a^2} + {b^2}}}{{2ab}}\) A1
coordinates are \(\left( {\ln \frac{b}{a},\frac{{{a^2} + {b^2}}}{{2ab}}} \right)\)
[6 marks]
(d) \(\mathop {\lim }\limits_{x – \infty } f(x) = \frac{a}{b} \Rightarrow y = \frac{a}{b}\) horizontal asymptote A1
\(\mathop {\lim }\limits_{x \to + \infty } f(x) = \frac{b}{a} \Rightarrow y = \frac{b}{a}\) horizontal asymptote A1
\(0 < b < a \Rightarrow a{{\text{e}}^x} + b > 0\) for all \(x \in \mathbb{R}\) (accept \(a{{\text{e}}^x} + b \ne 0\))
so no vertical asymptotes R1
Note: Statement on vertical asymptote must be seen for R1.
[3 marks]
(e) \(y = \frac{{4 + {{\text{e}}^x}}}{{4{{\text{e}}^x} + 1}}\)
\(y = \frac{1}{2} \Leftrightarrow x = \ln \frac{7}{2}\) (or 1.25 to 3 sf) (M1)(A1)
\(V = \pi \int_0^{\ln \frac{7}{2}} {\left( {{{\left( {\frac{{4 + {{\text{e}}^x}}}{{4{{\text{e}}^x} + 1}}} \right)}^2} – \frac{1}{4}} \right){\text{d}}x} \) (M1)A1
\( = 1.09\) (3 sf) A1 N4
[5 marks]
Total [19 marks]
Examiners report
This question was well attempted by many candidates. In some cases, candidates who skipped other questions still answered, with some success, parts of this question. Part (a) was in general well done but in (b) candidates found difficulty in justifying that f’(x) was non-zero. Performance in part (c) was mixed: it was pleasing to see good levels of algebraic ability of good candidates who successfully answered this question; weaker candidates found the simplification required difficult. There were very few good answers to part (d) which showed the weaknesses of most candidates in dealing with the concept of asymptotes. In part (e) there were a large number of good attempts, with many candidates evaluating correctly the limits of the integral and a smaller number scoring full marks in this part.
Question
Consider the function \(f(x) = {x^3} – 3{x^2} – 9x + 10\) , \(x \in \mathbb{R}\).
Find the equation of the straight line passing through the maximum and minimum points of the graph \(y = f (x)\) .
Show that the point of inflexion of the graph \(y = f (x)\) lies on this straight line.
Answer/Explanation
Markscheme
\(f'(x) = 3{x^2} – 6x – 9\) (\(= 0\)) (M1)
\(\left( {x + 1} \right)\left( {x – 3} \right) = 0\)
\(x = – 1\); \(x = 3\)
(max) (−1, 15); (min) (3, −17) A1A1
Note: The coordinates need not be explicitly stated but the values need to be seen.
\(y = – 8x + 7\) A1 N2
[4 marks]
\(f”(x) = 6x – 6 = 0 \Rightarrow \) inflexion (1, −1) A1
which lies on \(y = – 8x + 7\) R1AG
[2 marks]
Examiners report
There were a significant number of completely correct answers to this question. Many candidates demonstrated a good understanding of basic differential calculus in the context of coordinate geometry whilst other used technology to find the turning points.
There were a significant number of completely correct answers to this question. Many candidates demonstrated a good understanding of basic differential calculus in the context of coordinate geometry whilst other used technology to find the turning points. There were many correct demonstrations of the “show that” in (b).
Question
Let \(f(x) = {x^4} + 0.2{x^3} – 5.8{x^2} – x + 4,{\text{ }}x \in \mathbb{R}\).
The domain of \(f\) is now restricted to \([0,{\text{ }}a]\).
Let \(g(x) = 2\sin (x – 1) – 3,{\text{ }} – \frac{\pi }{2} + 1 \leqslant x \leqslant \frac{\pi }{2} + 1\).
Find the solutions of \(f(x) > 0\).
For the curve \(y = f(x)\).
(i) Find the coordinates of both local minimum points.
(ii) Find the \(x\)-coordinates of the points of inflexion.
Write down the largest value of \(a\) for which \(f\) has an inverse. Give your answer correct to 3 significant figures.
For this value of a sketch the graphs of \(y = f(x)\) and \(y = {f^{ – 1}}(x)\) on the same set of axes, showing clearly the coordinates of the end points of each curve.
Solve \({f^{ – 1}}(x) = 1\).
Find an expression for \({g^{ – 1}}(x)\), stating the domain.
Solve \(({f^{ – 1}} \circ g)(x) < 1\).
Answer/Explanation
Markscheme
valid method eg, sketch of curve or critical values found (M1)
\(x < – 2.24,{\text{ }}x > 2.24,\) A1
\( – 1 < x < 0.8\) A1
Note: Award M1A1A0 for correct intervals but with inclusive inequalities.
[3 marks]
(i) \((1.67,{\text{ }} – 5.14),{\text{ }}( – 1.74,{\text{ }} – 3.71)\) A1A1
Note: Award A1A0 for any two correct terms.
(ii) \(f'(x) = 4{x^3} + 0.6{x^2} – 11.6x – 1\)
\(f”(x) = 12{x^2} + 1.2x – 11.6 = 0\) (M1)
\( – 1.03,{\text{ }}0.934\) A1A1
Note: M1 should be awarded if graphical method to find zeros of \(f”(x)\) or turning points of \(f'(x)\) is shown.
[5 marks]
1.67 A1
[2 marks]
M1A1A1
Note: Award M1 for reflection of their \(y = f(x)\) in the line \(y = x\) provided their \(f\) is one-one.
A1 for \((0,{\text{ }}4)\), \((4,{\text{ }}0)\) (Accept axis intercept values) A1 for the other two sets of coordinates of other end points
[2 marks]
\(x = f(1)\) M1
\( = – 1.6\) A1
[2 marks]
\(y = 2\sin (x – 1) – 3\)
\(x = 2\sin (y – 1) – 3\) (M1)
\(\left( {{g^{ – 1}}(x) = } \right){\text{ }}\arcsin \left( {\frac{{x + 3}}{2}} \right) + 1\) A1
\( – 5 \leqslant x \leqslant – 1\) A1A1
Note: Award A1 for −5 and −1, and A1 for correct inequalities if numbers are reasonable.
[8 marks]
\({f^{ – 1}}\left( {g(x)} \right) < 1\)
\(g(x) > – 1.6\) (M1)
\(x > {g^{ – 1}}( – 1.6) = 1.78\) (A1)
Note: Accept = in the above.
\(1.78 < x \leqslant \frac{\pi }{2} + 1\) A1A1
Note: A1 for \(x > 1.78\) (allow ≥) and A1 for \(x \leqslant \frac{\pi }{2} + 1\).
[4 marks]
Examiners report
Parts (a) and (b) were well answered, with considerably less success in part (c). Surprisingly few students were able to reflect the curve in \(y = x\) satisfactorily, and many were not making their sketch using the correct domain.
Parts (a) and (b) were well answered, with considerably less success in part (c). Surprisingly few students were able to reflect the curve in \(y = x\) satisfactorily, and many were not making their sketch using the correct domain.
Parts (a) and (b) were well answered, with considerably less success in part (c). Surprisingly few students were able to reflect the curve in \(y = x\) satisfactorily, and many were not making their sketch using the correct domain.
Parts (a) and (b) were well answered, with considerably less success in part (c). Surprisingly few students were able to reflect the curve in \(y = x\) satisfactorily, and many were not making their sketch using the correct domain.
Parts (a) and (b) were well answered, with considerably less success in part (c). Surprisingly few students were able to reflect the curve in \(y = x\) satisfactorily, and many were not making their sketch using the correct domain.
Part d(i) was generally well done, but there were few correct answers for d(ii).
Part d(i) was generally well done, but there were few correct answers for d(ii).
Question
The function f is defined on the domain [0, 2] by \(f(x) = \ln (x + 1)\sin (\pi x)\) .
Obtain an expression for \(f'(x)\) .
Sketch the graphs of f and \(f’\) on the same axes, showing clearly all x-intercepts.
Find the x-coordinates of the two points of inflexion on the graph of f .
Find the equation of the normal to the graph of f where x = 0.75 , giving your answer in the form y = mx + c .
Consider the points \({\text{A}}\left( {a{\text{ }},{\text{ }}f(a)} \right)\) , \({\text{B}}\left( {b{\text{ }},{\text{ }}f(b)} \right)\) and \({\text{C}}\left( {c{\text{ }},{\text{ }}f(c)} \right)\) where a , b and c \((a < b < c)\) are the solutions of the equation \(f(x) = f'(x)\) . Find the area of the triangle ABC.
Answer/Explanation
Markscheme
\(f'(x) = \frac{1}{{x + 1}}\sin (\pi x) + \pi \ln (x + 1)\cos (\pi x)\) M1A1A1
[3 marks]
A4
Note: Award A1A1 for graphs, A1A1 for intercepts.
[4 marks]
0.310, 1.12 A1A1
[2 marks]
\(f'(0.75) = – 0.839092\) A1
so equation of normal is \(y – 0.39570812 = \frac{1}{{0.839092}}(x – 0.75)\) M1
\(y = 1.19x – 0.498\) A1
[3 marks]
\({\text{A}}(0,{\text{ }}0)\)
\({\text{B(}}\overbrace {0.548 \ldots }^c,\overbrace {0.432 \ldots }^d)\) A1
\({\text{C(}}\overbrace {1.44 \ldots }^e,\overbrace { – 0.881 \ldots }^f)\) A1
Note: Accept coordinates for B and C rounded to 3 significant figures.
area \(\Delta {\text{ABC}} = \frac{1}{2}|\)(ci + dj) \( \times \) (ei + fj)\(|\) M1A1
\( = \frac{1}{2}(de – cf)\) A1
\( = 0.554\) A1
[6 marks]
Examiners report
[N/A]
[N/A]
[N/A]
[N/A]
[N/A]