Question: [Maximum mark: 20]
Consider the function f (x) = \(\sqrt{x^{2}-1}\), where 1 ≤ x ≤ 2.
(a) Sketch the curve y = f (x), clearly indicating the coordinates of the endpoints.
(b) (i) Show that the inverse function of f is given by f-1(x) = \(\sqrt{x^{2}+1}\).
(ii) State the domain and range of f -1.
The curve y = f (x) is rotated 2π about the y-axis to form a solid of revolution that is used to model a water container.
(c) (i) Show that the volume, Vm3, of water in the container when it is filled to a height of h metres is given by V = \(\pi \left ( \frac{1}{3}h^{3}+h \right )\).
(ii) Hence, determine the maximum volume of the container.
At t = 0, the container is empty. Water is then added to the container at a constant rate of 0.4m3 s-1.
(d) Find the time it takes to fill the container to its maximum volume.
(e) Find the rate of change of the height of the water when the container is filled to half its maximum volume.
▶️Answer/Explanation
Ans:
correct shape (concave down) within the given domain 1≤ x ≤ 2
(1, 0) and \(\left ( 2, \sqrt{3} \right )\) (=(2, 1.73))
Note: The coordinates of endpoints may be seen on the graph or marked on the axes.
Question
Consider the curve defined by the equation \({x^2} + \sin y – xy = 0\) .
a.Find the gradient of the tangent to the curve at the point \((\pi ,{\text{ }}\pi )\) .[6]
b.Hence, show that \(\tan \theta = \frac{1}{{1 + 2\pi }}\), where \(\theta \) is the acute angle between the tangent to the curve at \((\pi ,{\text{ }}\pi )\) and the line y = x .[3]
▶️Answer/Explanation
Markscheme
attempt to differentiate implicitly M1
\(2x + \cos y\frac{{{\text{d}}y}}{{{\text{d}}x}} – y – x\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0\) A1A1
Note: A1 for differentiating \({x^2}\) and sin y ; A1 for differentiating xy.
substitute x and y by \(\pi \) M1
\(2\pi – \frac{{{\text{d}}y}}{{{\text{d}}x}} – \pi – \pi \frac{{{\text{d}}y}}{{{\text{d}}x}} = 0 \Rightarrow \frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{\pi }{{1 + \pi }}\) M1A1
Note: M1 for attempt to make dy/dx the subject. This could be seen earlier.
[6 marks]
\(\theta = \frac{\pi }{4} – \arctan \frac{\pi }{{1 + \pi }}\) (or seen the other way) M1
\(\tan \theta = \tan \left( {\frac{\pi }{4} – \arctan \frac{\pi }{{1 + \pi }}} \right) = \frac{{1 – \frac{\pi }{{1 + \pi }}}}{{1 + \frac{\pi }{{1 + \pi }}}}\) M1A1
\(\tan \theta = \frac{1}{{1 + 2\pi }}\) AG
[3 marks]
Examiners report
Part a) proved an easy 6 marks for most candidates, while the majority failed to make any headway with part b), with some attempting to find the equation of their line in the form y = mx + c . Only the best candidates were able to see their way through to the given answer.
Part a) proved an easy 6 marks for most candidates, while the majority failed to make any headway with part b), with some attempting to find the equation of their line in the form y = mx + c . Only the best candidates were able to see their way through to the given answer.
Question
Paint is poured into a tray where it forms a circular pool with a uniform thickness of 0.5 cm. If the paint is poured at a constant rate of \(4{\text{ c}}{{\text{m}}^3}{{\text{s}}^{ – 1}}\), find the rate of increase of the radius of the circle when the radius is 20 cm.
▶️Answer/Explanation
Markscheme
\(V = 0.5\pi {r^2}\) (A1)
EITHER
\(\frac{{dV}}{{dr}} = \pi r\) A1
\(\frac{{dV}}{{dt}} = 4\) (A1)
applying chain rule M1
for example \(\frac{{dr}}{{dt}} = \frac{{dV}}{{dt}} \times \frac{{dr}}{{dV}}\)
OR
\(\frac{{dV}}{{dt}} = \pi r\frac{{dr}}{{dt}}\) M1A1
\(\frac{{dV}}{{dt}} = 4\) (A1)
THEN
\(\frac{{dr}}{{dt}} = 4 \times \frac{1}{{\pi r}}\) A1
when \(r = 20,{\text{ }}\frac{{dr}}{{dt}} = \frac{4}{{20\pi }}{\text{ or }}\frac{1}{{5\pi }}{\text{ (cm}}\,{{\text{s}}^{ – 1}})\) A1
Note: Allow h instead of 0.5 up until the final A1.
[6 marks]
Examiners report
There was a large variety of methods used in this question, with most candidates choosing to implicitly differentiate the expression for volume in terms of r.
Question
A curve has equation \(\arctan {x^2} + \arctan {y^2} = \frac{\pi }{4}\).
(a) Find \(\frac{{{\text{d}}y}}{{{\text{d}}x}}\) in terms of x and y.
(b) Find the gradient of the curve at the point where \(x = \frac{1}{{\sqrt 2 }}\) and \(y < 0\).
▶️Answer/Explanation
Markscheme
(a) METHOD 1
\(\frac{{2x}}{{1 + {x^4}}} + \frac{{2y}}{{1 + {y^4}}}\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0\) M1A1A1
Note: Award M1 for implicit differentiation, A1 for LHS and A1 for RHS.
\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = – \frac{{x\left( {1 + {y^4}} \right)}}{{y\left( {1 + {x^4}} \right)}}\) A1
METHOD 2
\({y^2} = \tan \left( {\frac{\pi }{4} – \arctan {x^2}} \right)\)
\( = \frac{{\tan \frac{\pi }{4} – \tan \left( {\arctan {x^2}} \right)}}{{1 + \left( {\tan \frac{\pi }{4}} \right)\left( {\tan \left( {\arctan {x^2}} \right)} \right)}}\) (M1)
\( = \frac{{1 – {x^2}}}{{1 + {x^2}}}\) A1
\(2y\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{ – 2x\left( {1 + {x^2}} \right) – 2x\left( {1 – {x^2}} \right)}}{{{{\left( {1 + {x^2}} \right)}^2}}}\) M1
\(2y\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{ – 4x}}{{{{\left( {1 + {x^2}} \right)}^2}}}\)
\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = – \frac{{2x}}{{y{{\left( {1 + {x^2}} \right)}^2}}}\) A1
\(\left( { = \frac{{2x\sqrt {1 + {x^2}} }}{{\sqrt {1 – {x^2}} {{\left( {1 + {x^2}} \right)}^2}}}} \right)\)
[4 marks]
(b) \({y^2} = \tan \left( {\frac{\pi }{4} – \arctan \frac{1}{2}} \right)\) (M1)
\( = \frac{{\tan \frac{\pi }{4} – \tan \left( {\arctan \frac{1}{2}} \right)}}{{1 + \left( {\tan \frac{\pi }{4}} \right)\left( {\tan \left( {\arctan \frac{1}{2}} \right)} \right)}}\) (M1)
Note: The two M1s may be awarded for working in part (a).
\( = \frac{{1 – \frac{1}{2}}}{{1 + \frac{1}{2}}} = \frac{1}{3}\) A1
\(y = – \frac{1}{{\sqrt 3 }}\) A1
substitution into \(\frac{{{\text{d}}y}}{{{\text{d}}x}}\)
\( = \frac{{4\sqrt 6 }}{9}\) A1
Note: Accept \(\frac{{8\sqrt 3 }}{{9\sqrt 2 }}\) etc.
[5 marks]
Total [9 marks]