Question
Find the area between the curves \(y = 2 + x – {x^2}{\text{ and }}y = 2 – 3x + {x^2}\) .
▶️Answer/Explanation
Markscheme
\(2 + x – {x^2} = 2 – 3x + {x^2}\) M1
\( \Rightarrow 2{x^2} – 4x = 0\)
\( \Rightarrow 2x(x – 2) = 0\)
\( \Rightarrow x = 0,{\text{ }}x = 2\) A1A1
Note: Accept graphical solution.
Award M1 for correct graph and A1A1 for correctly labelled roots.
\(\therefore {\text{A}} = \int_0^2 {\left( {(2 + x – {x^2}) – (2 – 3x + {x^2})} \right){\text{d}}x} \) (M1)
\( = \int_0^2 {(4x – 2{x^2}){\text{d}}x\,\,\,\,\,{\text{or equivalent}}} \) A1
\( = \left[ {2{x^2} – \frac{{2{x^3}}}{3}} \right]_0^2\) A1
\( = \frac{8}{3}\left( { = 2\frac{2}{3}} \right)\) A1
[7 marks]
Examiners report
This was the question that gained the most correct responses. A few candidates struggled to find the limits of the integration or found a negative area.
Question
A function is defined as \(f(x) = k\sqrt x \), with \(k > 0\) and \(x \geqslant 0\) .
(a) Sketch the graph of \(y = f(x)\) .
(b) Show that f is a one-to-one function.
(c) Find the inverse function, \({f^{ – 1}}(x)\) and state its domain.
(d) If the graphs of \(y = f(x)\) and \(y = {f^{ – 1}}(x)\) intersect at the point (4, 4) find the value of k .
(e) Consider the graphs of \(y = f(x)\) and \(y = {f^{ – 1}}(x)\) using the value of k found in part (d).
(i) Find the area enclosed by the two graphs.
(ii) The line x = c cuts the graphs of \(y = f(x)\) and \(y = {f^{ – 1}}(x)\) at the points P and Q respectively. Given that the tangent to \(y = f(x)\) at point P is parallel to the tangent to \(y = {f^{ – 1}}(x)\) at point Q find the value of c .
▶️Answer/Explanation
Markscheme
(a)
A1
Note: Award A1 for correct concavity, passing through (0, 0) and increasing.
Scales need not be there.
[1 mark]
(b) a statement involving the application of the Horizontal Line Test or equivalent A1
[1 mark]
(c) \(y = k\sqrt x \)
for either \(x = k\sqrt y \) or \(x = \frac{{{y^2}}}{{{k^2}}}\) A1
\({f^{ – 1}}(x) = \frac{{{x^2}}}{{{k^2}}}\) A1
\({\text{dom}}\left( {{f^{ – 1}}(x)} \right) = \left[ {0,\infty } \right[\) A1
[3 marks]
(d) \(\frac{{{x^2}}}{{{k^2}}} = k\sqrt x \,\,\,\,\,\)or equivalent method M1
\(k = \sqrt x \)
\(k = 2\) A1
[2 marks]
(e) (i) \(A = \int_a^b {({y_1} – {y_2}){\text{d}}x} \) (M1)
\(A = \int_0^4 {\left( {2{x^{\frac{1}{2}}} – \frac{1}{4}{x^2}} \right){\text{d}}x} \) A1
\( = \left[ {\frac{4}{3}{x^{\frac{3}{2}}} – \frac{1}{{12}}{x^3}} \right]_0^4\) A1
\( = \frac{{16}}{3}\) A1
(ii) attempt to find either \(f'(x)\) or \(({f^{ – 1}})'(x)\) M1
\(f'(x) = \frac{1}{{\sqrt x }},{\text{ }}\left( {({f^{ – 1}})'(x) = \frac{x}{2}} \right)\) A1A1
\(\frac{1}{{\sqrt c }} = \frac{c}{2}\) M1
\(c = {2^{\frac{2}{3}}}\) A1
[9 marks]
Total [16 marks]
Examiners report
Many students could not sketch the function. There was confusion between the vertical and horizontal line test for one-to-one functions. A significant number of students gave long and inaccurate explanations for a one-to-one function. Finding the inverse was done very well by most students although the notation used was generally poor. The domain of the inverse was ignored by many or done incorrectly even if the sketch was correct. Many did not make the connections between the parts of the question. An example of this was the number of students who spent time finding the point of intersection in part e) even though it was given in d).
Question
▶️Answer/Explanation
Markscheme
METHOD 1
\({\text{area}} = \mathop \smallint \limits_0^{\sqrt 3 } \arctan x{\text{d}}x\) A1
attempting to integrate by parts M1
\( = \left[ {x\arctan x} \right]_0^{\sqrt 3 } – \mathop \smallint \limits_0^{\sqrt 3 } x\frac{1}{{1 + {x^2}}}{\text{d}}x\) A1A1
\( = \left[ {x\arctan x} \right]_0^{\sqrt 3 } – \left[ {\frac{1}{2}\ln \left( {1 + {x^2}} \right)} \right]_0^{\sqrt 3 }\) A1
Note: Award A1 even if limits are absent.
\( = \frac{\pi }{{\sqrt 3 }} – \frac{1}{2}\ln 4\) A1
\(\left( { = \frac{{\pi \sqrt 3 }}{3} – \ln 2} \right)\)
METHOD 2
\({\text{area}} = \frac{{\pi \sqrt 3 }}{3} – \mathop \smallint \limits_0^{\frac{\pi }{3}} \tan y{\text{d}}y\) M1A1A1
\(\ { = \frac{{\pi \sqrt 3 }}{3} + \left[ {\ln \left| {\cos y} \right|} \right]_0^{\frac{\pi }{3}}} \) M1A1
\( = \frac{{\pi \sqrt 3 }}{3} + \ln \frac{1}{2}\) \(\left( { = \frac{{\pi \sqrt 3 }}{3} – \ln 2} \right)\) A1
[6 marks]
Examiners report
Many candidates were able to write down the correct expression for the required area, although in some cases with incorrect integration limits. However, very few managed to achieve any further marks due to a number of misconceptions, in particular \(\arctan x = \cot x = \frac{{\cos x}}{{\sin x}}\). Candidates who realised they should use integration by parts were in general very successful in answering this question. It was pleasing to see a few alternative correct approaches to this question.