Question
Let \(f(x) = {{\rm{e}}^{ – 3x}}\) and \(g(x) = \sin \left( {x – \frac{\pi }{3}} \right)\) .
Write down
(i) \(f'(x)\) ;
(ii) \(g'(x)\) .
Let \(h(x) = {{\rm{e}}^{ – 3x}}\sin \left( {x – \frac{\pi }{3}} \right)\) . Find the exact value of \(h’\left( {\frac{\pi }{3}} \right)\) .
Answer/Explanation
Markscheme
(i) \( – 3{{\rm{e}}^{ – 3x}}\) A1 N1
(ii) \(\cos \left( {x – \frac{\pi }{3}} \right)\) A1 N1
[4 marks]
evidence of choosing product rule (M1)
e.g. \(uv’ + vu’\)
correct expression A1
e.g. \( – 3{{\rm{e}}^{ – 3x}}\sin \left( {x – \frac{\pi }{3}} \right) + {{\rm{e}}^{ – 3x}}\cos \left( {x – \frac{\pi }{3}} \right)\)
complete correct substitution of \(x = \frac{\pi }{3}\) (A1)
e.g. \( – 3{{\rm{e}}^{ – 3\frac{\pi }{3}}}\sin \left( {\frac{\pi }{3} – \frac{\pi }{3}} \right) + {{\rm{e}}^{ – 3\frac{\pi }{3}}}\cos \left( {\frac{\pi }{3} – \frac{\pi }{3}} \right)\)
\(h’\left( {\frac{\pi }{3}} \right) = {{\rm{e}}^{ – \pi }}\) A1 N3
[4 marks]
Question
Let \(f(x) = \frac{{\cos x}}{{\sin x}}\) , for \(\sin x \ne 0\) .
In the following table, \(f’\left( {\frac{\pi }{2}} \right) = p\) and \(f”\left( {\frac{\pi }{2}} \right) = q\) . The table also gives approximate values of \(f'(x)\) and \(f”(x)\) near \(x = \frac{\pi }{2}\) .
Use the quotient rule to show that \(f'(x) = \frac{{ – 1}}{{{{\sin }^2}x}}\) .
Find \(f”(x)\) .
Find the value of p and of q.
Use information from the table to explain why there is a point of inflexion on the graph of f where \(x = \frac{\pi }{2}\) .
Answer/Explanation
Markscheme
\(\frac{{\rm{d}}}{{{\rm{d}}x}}\sin x = \cos x\) , \(\frac{{\rm{d}}}{{{\rm{d}}x}}\cos x = – \sin x\) (seen anywhere) (A1)(A1)
evidence of using the quotient rule M1
correct substitution A1
e.g. \(\frac{{\sin x( – \sin x) – \cos x(\cos x)}}{{{{\sin }^2}x}}\) , \(\frac{{ – {{\sin }^2}x – {{\cos }^2}x}}{{{{\sin }^2}x}}\)
\(f'(x) = \frac{{ – ({{\sin }^2}x + {{\cos }^2}x)}}{{{{\sin }^2}x}}\) A1
\(f'(x) = \frac{{ – 1}}{{{{\sin }^2}x}}\) AG N0
[5 marks]
METHOD 1
appropriate approach (M1)
e.g. \(f'(x) = – {(\sin x)^{ – 2}}\)
\(f”(x) = 2({\sin ^{ – 3}}x)(\cos x)\) \(\left( { = \frac{{2\cos x}}{{{{\sin }^3}x}}} \right)\) A1A1 N3
Note: Award A1 for \(2{\sin ^{ – 3}}x\) , A1 for \(\cos x\) .
METHOD 2
derivative of \({\sin ^2}x = 2\sin x\cos x\) (seen anywhere) A1
evidence of choosing quotient rule (M1)
e.g. \(u = – 1\) , \(v = {\sin ^2}x\) , \(f” = \frac{{{{\sin }^2}x \times 0 – ( – 1)2\sin x\cos x}}{{{{({{\sin }^2}x)}^2}}}\)
\(f”(x) = \frac{{2\sin x\cos x}}{{{{({{\sin }^2}x)}^2}}}\) \(\left( { = \frac{{2\cos x}}{{{{\sin }^3}x}}} \right)\) A1 N3
[3 marks]
evidence of substituting \(\frac{\pi }{2}\) M1
e.g. \(\frac{{ – 1}}{{{{\sin }^2}\frac{\pi }{2}}}\) , \(\frac{{2\cos \frac{\pi }{2}}}{{{{\sin }^3}\frac{\pi }{2}}}\)
\(p = – 1\) , \(q = 0\) A1A1 N1N1
[3 marks]
second derivative is zero, second derivative changes sign R1R1 N2
[2 marks]
Question
Let \(f(x) = \cos 2x\) and \(g(x) = 2{x^2} – 1\) .
Find \(f\left( {\frac{\pi }{2}} \right)\) .
Find \((g \circ f)\left( {\frac{\pi }{2}} \right)\) .
Given that \((g \circ f)(x)\) can be written as \(\cos (kx)\) , find the value of k, \(k \in \mathbb{Z}\) .
Answer/Explanation
Markscheme
\(f\left( {\frac{\pi }{2}} \right) = \cos \pi \) (A1)
\( = – 1\) A1 N2
[2 marks]
\((g \circ f)\left( {\frac{\pi }{2}} \right) = g( – 1)\) \(( = 2{( – 1)^2} – 1)\) (A1)
\(= 1\) A1 N2
[2 marks]
\((g \circ f)(x) = 2{(\cos (2x))^2} – 1\) \(( = 2{\cos ^2}(2x) – 1)\) A1
evidence of \(2{\cos ^2}\theta – 1 = \cos 2\theta \) (seen anywhere) (M1)
\((g \circ f)(x) = \cos 4x\)
\(k = 4\) A1 N2
[3 marks]
Question
Let \(h(x) = \frac{{6x}}{{\cos x}}\) . Find \(h'(0)\) .
Answer/Explanation
Markscheme
METHOD 1 (quotient)
derivative of numerator is 6 (A1)
derivative of denominator is \( – \sin x\) (A1)
attempt to substitute into quotient rule (M1)
correct substitution A1
e.g. \(\frac{{(\cos x)(6) – (6x)( – \sin x)}}{{{{(\cos x)}^2}}}\)
substituting \(x = 0\) (A1)
e.g. \(\frac{{(\cos 0)(6) – (6 \times 0)( – \sin 0)}}{{{{(\cos 0)}^2}}}\)
\(h'(0) = 6\) A1 N2
METHOD 2 (product)
\(h(x) = 6x \times {(\cos x)^{ – 1}}\)
derivative of 6x is 6 (A1)
derivative of \({(\cos x)^{ – 1}}\) is \(( – {(\cos x)^{ – 2}}( – \sin x))\) (A1)
attempt to substitute into product rule (M1)
correct substitution A1
e.g. \((6x)( – {(\cos x)^{ – 2}}( – \sin x)) + (6){(\cos x)^{ – 1}}\)
substituting \(x = 0\) (A1)
e.g. \((6 \times 0)( – {(\cos 0)^{ – 2}}( – \sin 0)) + (6){(\cos 0)^{ – 1}}\)
\(h'(0) = 6\) A1 N2
[6 marks]
Question
Let \(\int_\pi ^a {\cos 2x{\text{d}}x} = \frac{1}{2}{\text{, where }}\pi < a < 2\pi \). Find the value of \(a\).
Answer/Explanation
Markscheme
correct integration (ignore absence of limits and “\(+C\)”) (A1)
eg \(\frac{{\sin (2x)}}{2},{\text{ }}\int_\pi ^a {\cos 2x = \left[ {\frac{1}{2}\sin (2x)} \right]_\pi ^a} \)
substituting limits into their integrated function and subtracting (in any order) (M1)
eg \(\frac{1}{2}\sin (2a) – \frac{1}{2}\sin (2\pi ),{\text{ }}\sin (2\pi ) – \sin (2a)\)
\(\sin (2\pi ) = 0\) (A1)
setting their result from an integrated function equal to \(\frac{1}{2}\) M1
eg \(\frac{1}{2}\sin 2a = \frac{1}{2},{\text{ }}\sin (2a) = 1\)
recognizing \({\sin ^{ – 1}}1 = \frac{\pi }{2}\) (A1)
eg \(2a = \frac{\pi }{2},{\text{ }}a = \frac{\pi }{4}\)
correct value (A1)
eg \(\frac{\pi }{2} + 2\pi ,{\text{ }}2a = \frac{{5\pi }}{2},{\text{ }}a = \frac{\pi }{4} + \pi \)
\(a = \frac{{5\pi }}{4}\) A1 N3
[7 marks]
Question
The following diagram shows a triangle ABC and a sector BDC of a circle with centre B and radius 6 cm. The points A , B and D are on the same line.
\({\text{AB}} = 2\sqrt 3 {\text{ cm, BC}} = 6{\text{ cm, area of triangle ABC}} = 3\sqrt 3{\text{ c}}{{\text{m}}^{\text{2}}}{\rm{, A\hat BC}}\) is obtuse.
Find \({\rm{A\hat BC}}\).
Find the exact area of the sector BDC.
Answer/Explanation
Markscheme
METHOD 1
correct substitution into formula for area of triangle (A1)
eg\(\,\,\,\,\,\)\(\frac{1}{2}(6)\left( {2\sqrt 3 } \right)\sin B,{\text{ }}6\sqrt 3 \sin B,{\text{ }}\frac{1}{2}(6)\left( {2\sqrt 3 } \right)\sin B = 3\sqrt 3 \)
correct working (A1)
eg\(\,\,\,\,\,\)\(6\sqrt 3 \sin B = 3\sqrt 3 ,{\text{ }}\sin B = \frac{{3\sqrt 3 }}{{\frac{1}{2}(6)2\sqrt 3 }}\)
\(\sin B = \frac{1}{2}\) (A1)
\(\frac{\pi }{6}(30^\circ )\) (A1)
\({\rm{A\hat BC}} = \frac{{5\pi }}{6}(150^\circ )\) A1 N3
METHOD 2
(using height of triangle ABC by drawing perpendicular segment from C to AD)
correct substitution into formula for area of triangle (A1)
eg\(\,\,\,\,\,\)\(\frac{1}{2}\left( {2\sqrt 3 } \right)(h) = 3\sqrt 3 ,{\text{ }}h\sqrt 3 \)
correct working (A1)
eg\(\,\,\,\,\,\)\(h\sqrt 3 = 3\sqrt 3 \)
height of triangle is 3 A1
\({\rm{C\hat BD}} = \frac{\pi }{6}(30^\circ )\) (A1)
\({\rm{A\hat BC}} = \frac{{5\pi }}{6}(150^\circ )\) A1 N3
[5 marks]
recognizing supplementary angle (M1)
eg\(\,\,\,\,\,\)\({\rm{C\hat BD}} = \frac{\pi }{6},{\text{ sector}} = \frac{1}{2}(180 – {\rm{A\hat BC)(}}{{\text{6}}^2})\)
correct substitution into formula for area of sector (A1)
eg\(\,\,\,\,\,\)\(\frac{1}{2} \times \frac{\pi }{6} \times {6^2},{\text{ }}\pi ({6^2})\left( {\frac{{30}}{{360}}} \right)\)
\({\text{area}} = 3\pi {\text{ }}({\text{c}}{{\text{m}}^2})\) A1 N2
[3 marks]
Question
The following diagram shows triangle PQR.
Find PR.
Answer/Explanation
Markscheme
METHOD 1
evidence of choosing the sine rule (M1)
eg\(\,\,\,\,\,\)\(\frac{a}{{\sin A}} = \frac{b}{{\sin B}}\)
correct substitution A1
eg\(\,\,\,\,\,\)\(\frac{x}{{\sin 30}} = \frac{{13}}{{\sin 45}},{\text{ }}\frac{{13\sin 30}}{{\sin 45}}\)
\(\sin 30 = \frac{1}{2},{\text{ }}\sin 45 = \frac{1}{{\sqrt 2 }}\) (A1)(A1)
correct working A1
eg\(\,\,\,\,\,\)\(\frac{1}{2} \times \frac{{13}}{{\frac{1}{{\sqrt 2 }}}},{\text{ }}\frac{1}{2} \times 13 \times \frac{2}{{\sqrt 2 }},{\text{ }}13 \times \frac{1}{2} \times \sqrt 2 \)
correct answer A1 N3
eg\(\,\,\,\,\,\)\({\text{PR}} = \frac{{13\sqrt 2 }}{2},{\text{ }}\frac{{13}}{{\sqrt 2 }}{\text{ (cm)}}\)
METHOD 2 (using height of ΔPQR)
valid approach to find height of ΔPQR (M1)
eg\(\,\,\,\,\,\)\(\sin 30 = \frac{x}{{13}},{\text{ }}\cos 60 = \frac{x}{{13}}\)
\(\sin 30 = \frac{1}{2}{\text{ or }}\cos 60 = \frac{1}{2}\) (A1)
\({\text{height}} = 6.5\) A1
correct working A1
eg\(\,\,\,\,\,\)\(\sin 45 = \frac{{6.5}}{{{\text{PR}}}},{\text{ }}\sqrt {{{6.5}^2} + {{6.5}^2}} \)
correct working (A1)
eg\(\,\,\,\,\,\)\(\sin 45 = \frac{1}{{\sqrt 2 }},{\text{ }}\cos 45 = \frac{1}{{\sqrt 2 }},{\text{ }}\sqrt {\frac{{169 \times 2}}{4}} \)
correct answer A1 N3
eg\(\,\,\,\,\,\)\({\text{PR}} = \frac{{13\sqrt 2 }}{2},{\text{ }}\frac{{13}}{{\sqrt 2 }}{\text{ (cm)}}\)
[6 marks]
Question
The following table shows the probability distribution of a discrete random variable \(A\), in terms of an angle \(\theta \).
Show that \(\cos \theta = \frac{3}{4}\).
Given that \(\tan \theta > 0\), find \(\tan \theta \).
Let \(y = \frac{1}{{\cos x}}\), for \(0 < x < \frac{\pi }{2}\). The graph of \(y\)between \(x = \theta \) and \(x = \frac{\pi }{4}\) is rotated 360° about the \(x\)-axis. Find the volume of the solid formed.
Answer/Explanation
Markscheme
evidence of summing to 1 (M1)
eg\(\,\,\,\,\,\)\(\sum {p = 1} \)
correct equation A1
eg\(\,\,\,\,\,\)\(\cos \theta + 2\cos 2\theta = 1\)
correct equation in \(\cos \theta \) A1
eg\(\,\,\,\,\,\)\(\cos \theta + 2(2{\cos ^2}\theta – 1) = 1,{\text{ }}4{\cos ^2}\theta + \cos \theta – 3 = 0\)
evidence of valid approach to solve quadratic (M1)
eg\(\,\,\,\,\,\)factorizing equation set equal to \(0,{\text{ }}\frac{{ – 1 \pm \sqrt {1 – 4 \times 4 \times ( – 3)} }}{8}\)
correct working, clearly leading to required answer A1
eg\(\,\,\,\,\,\)\((4\cos \theta – 3)(\cos \theta + 1),{\text{ }}\frac{{ – 1 \pm 7}}{8}\)
correct reason for rejecting \(\cos \theta \ne – 1\) R1
eg\(\,\,\,\,\,\)\(\cos \theta \) is a probability (value must lie between 0 and 1), \(\cos \theta > 0\)
Note: Award R0 for \(\cos \theta \ne – 1\) without a reason.
\(\cos \theta = \frac{3}{4}\) AG N0
valid approach (M1)
eg\(\,\,\,\,\,\)sketch of right triangle with sides 3 and 4, \({\sin ^2}x + {\cos ^2}x = 1\)
correct working
(A1)
eg\(\,\,\,\,\,\)missing side \( = \sqrt 7 ,{\text{ }}\frac{{\frac{{\sqrt 7 }}{4}}}{{\frac{3}{4}}}\)
\(\tan \theta = \frac{{\sqrt 7 }}{3}\) A1 N2
[3 marks]
attempt to substitute either limits or the function into formula involving \({f^2}\) (M1)
eg\(\,\,\,\,\,\)\(\pi \int_\theta ^{\frac{\pi }{4}} {{f^2},{\text{ }}\int {{{\left( {\frac{1}{{\cos x}}} \right)}^2}} } \)
correct substitution of both limits and function (A1)
eg\(\,\,\,\,\,\)\(\pi \int_\theta ^{\frac{\pi }{4}} {{{\left( {\frac{1}{{\cos x}}} \right)}^2}{\text{d}}x} \)
correct integration (A1)
eg\(\,\,\,\,\,\)\(\tan x\)
substituting their limits into their integrated function and subtracting (M1)
eg\(\,\,\,\,\,\)\(\tan \frac{\pi }{4} – \tan \theta \)
Note: Award M0 if they substitute into original or differentiated function.
\(\tan \frac{\pi }{4} = 1\) (A1)
eg\(\,\,\,\,\,\)\(1 – \tan \theta \)
\(V = \pi – \frac{{\pi \sqrt 7 }}{3}\) A1 N3
[6 marks]
Question
The first two terms of an infinite geometric sequence are u1 = 18 and u2 = 12sin2 θ , where 0 < θ < 2\(\pi \) , and θ ≠ \(\pi \).
Find an expression for r in terms of θ.
Find the possible values of r.
Show that the sum of the infinite sequence is \(\frac{{54}}{{2 + {\text{cos}}\,\left( {2\theta } \right)}}\).
Find the values of θ which give the greatest value of the sum.
Answer/Explanation
Markscheme
valid approach (M1)
eg \(\frac{{{u_2}}}{{{u_1}}},\,\,\frac{{{u_1}}}{{{u_2}}}\)
\(r = \frac{{12\,{{\sin }^2}\,\theta }}{{18}}\left( { = \frac{{2\,{{\sin }^2}\,\theta }}{3}} \right)\) A1 N2
[2 marks]
recognizing that sinθ is bounded (M1)
eg 0 ≤ sin2 θ ≤ 1, −1 ≤ sinθ ≤ 1, −1 < sinθ < 1
0 < r ≤ \(\frac{2}{3}\) A2 N3
Note: If working shown, award M1A1 for correct values with incorrect inequality sign(s).
If no working shown, award N1 for correct values with incorrect inequality sign(s).
[3 marks]
correct substitution into formula for infinite sum A1
eg \(\frac{{18}}{{1 – \frac{{2\,{\text{si}}{{\text{n}}^2}\,\theta }}{3}}}\)
evidence of choosing an appropriate rule for cos 2θ (seen anywhere) (M1)
eg cos 2θ = 1 − 2 sin2 θ
correct substitution of identity/working (seen anywhere) (A1)
eg \(\frac{{18}}{{1 – \frac{2}{3}\left( {\frac{{1 – {\text{cos}}\,2\theta }}{2}} \right)}},\,\,\frac{{54}}{{3 – 2\left( {\frac{{1 – {\text{cos}}\,2\theta }}{2}} \right)}},\,\,\frac{{18}}{{\frac{{3 – 2\,{\text{si}}{{\text{n}}^2}\,\theta }}{3}}}\)
correct working that clearly leads to the given answer A1
eg \(\frac{{18 \times 3}}{{2 + \left( {1 – 2\,{\text{si}}{{\text{n}}^2}\,\theta } \right)}},\,\,\frac{{54}}{{3 – \left( {1 – {\text{cos}}\,2\theta } \right)}}\)
\(\frac{{54}}{{2 + {\text{cos}}\left( {2\theta } \right)}}\) AG N0
[4 marks]
METHOD 1 (using differentiation)
recognizing \(\frac{{{\text{d}}{S_\infty }}}{{{\text{d}}\theta }} = 0\) (seen anywhere) (M1)
finding any correct expression for \(\frac{{{\text{d}}{S_\infty }}}{{{\text{d}}\theta }}\) (A1)
eg \(\frac{{0 – 54 \times \left( { – 2\,{\text{sin}}\,2\,\theta } \right)}}{{{{\left( {2 + {\text{cos}}\,2\,\theta } \right)}^2}}},\,\, – 54{\left( {2 + {\text{cos}}\,2\,\theta } \right)^{ – 2}}\,\left( { – 2\,{\text{sin}}\,2\,\theta } \right)\)
correct working (A1)
eg sin 2θ = 0
any correct value for sin−1(0) (seen anywhere) (A1)
eg 0, \(\pi \), … , sketch of sine curve with x-intercept(s) marked both correct values for 2θ (ignore additional values) (A1)
2θ = \(\pi \), 3\(\pi \) (accept values in degrees)
both correct answers \(\theta = \frac{\pi }{2},\,\frac{{3\pi }}{2}\) A1 N4
Note: Award A0 if either or both correct answers are given in degrees.
Award A0 if additional values are given.
METHOD 2 (using denominator)
recognizing when S∞ is greatest (M1)
eg 2 + cos 2θ is a minimum, 1−r is smallest
correct working (A1)
eg minimum value of 2 + cos 2θ is 1, minimum r = \(\frac{2}{3}\)
correct working (A1)
eg \({\text{cos}}\,2\,\theta = – 1,\,\,\frac{2}{3}\,{\text{si}}{{\text{n}}^2}\,\theta = \frac{2}{3},\,\,{\text{si}}{{\text{n}}^2}\theta = 1\)
EITHER (using cos 2θ)
any correct value for cos−1(−1) (seen anywhere) (A1)
eg \(\pi \), 3\(\pi \), … (accept values in degrees), sketch of cosine curve with x-intercept(s) marked
both correct values for 2θ (ignore additional values) (A1)
2θ = \(\pi \), 3\(\pi \) (accept values in degrees)
OR (using sinθ)
sinθ = ±1 (A1)
sin−1(1) = \(\frac{\pi }{2}\) (accept values in degrees) (seen anywhere) A1
THEN
both correct answers \(\theta = \frac{\pi }{2},\,\frac{{3\pi }}{2}\) A1 N4
Note: Award A0 if either or both correct answers are given in degrees.
Award A0 if additional values are given.
[6 marks]
Question
The expression \(6\sin x\cos x\) can be expressed in the form \(a\sin bx\) .
Find the value of a and of b .
Hence or otherwise, solve the equation \(6\sin x\cos x = \frac{3}{2}\) , for \(\frac{\pi }{4} \le x \le \frac{\pi }{2}\) .
Answer/Explanation
Markscheme
recognizing double angle M1
e.g. \(3 \times 2\sin x\cos x\) , \(3\sin 2x\)
\(a = 3\) , \(b = 2\) A1A1 N3
[3 marks]
substitution \(3\sin 2x = \frac{3}{2}\) M1
\(\sin 2x = \frac{1}{2}\) A1
finding the angle A1
e.g. \(\frac{\pi }{6}\) , \(2x = \frac{{5\pi }}{6}\)
\(x = \frac{{5\pi }}{{12}}\) A1 N2
Note: Award A0 if other values are included.
[4 marks]
Question
Let \(f(x) = \cos x + \sqrt 3 \sin x\) , \(0 \le x \le 2\pi \) . The following diagram shows the graph of \(f\) .
The \(y\)-intercept is at (\(0\), \(1\)) , there is a minimum point at A (\(p\), \(q\)) and a maximum point at B.
Find \(f'(x)\) .
Hence
(i) show that \(q = – 2\) ;
(ii) verify that A is a minimum point.
Find the maximum value of \(f(x)\) .
The function \(f(x)\) can be written in the form \(r\cos (x – a)\) .
Write down the value of r and of a .
Answer/Explanation
Markscheme
\(f'(x) = – \sin x + \sqrt 3 \cos x\) A1A1 N2
[2 marks]
(i) at A, \(f'(x) = 0\) R1
correct working A1
e.g. \(\sin x = \sqrt 3 \cos x\)
\(\tan x = \sqrt 3 \) A1
\(x = \frac{\pi }{3}\) , \(\frac{{4\pi }}{3}\) A1
attempt to substitute their x into \(f(x)\) M1
e.g. \(\cos \left( {\frac{{4\pi }}{3}} \right) + \sqrt 3 \sin \left( {\frac{{4\pi }}{3}} \right)\)
correct substitution A1
e.g. \( – \frac{1}{2} + \sqrt 3 \left( { – \frac{{\sqrt 3 }}{2}} \right)\)
correct working that clearly leads to \( – 2\) A1
e.g. \( – \frac{1}{2} – \frac{3}{2}\)
\(q = – 2\) AG N0
(ii) correct calculations to find \(f'(x)\) either side of \(x = \frac{{4\pi }}{3}\) A1A1
e.g. \(f'(\pi ) = 0 – \sqrt 3 \) , \(f'(2\pi ) = 0 + \sqrt 3 \)
\(f'(x)\) changes sign from negative to positive R1
so A is a minimum AG N0
[10 marks]
max when \(x = \frac{\pi }{3}\) R1
correctly substituting \(x = \frac{\pi }{3}\) into \(f(x)\) A1
e.g. \(\frac{1}{2} + \sqrt 3 \left( {\frac{{\sqrt 3 }}{2}} \right)\)
max value is 2 A1 N1
[3 marks]
\(r = 2\) , \(a = \frac{\pi }{3}\) A1A1 N2
[2 marks]