AQA C2 Unofficial Markscheme 24th May 2017 Watch

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AQA MPC2 June 2017 Unofficial Markscheme

1. A sector has a perimeter of 22 cm and an angle of \theta radians. The radius is 8 cm.

(a) Find the value of \theta [2 marks]
16 + 8\theta = 22 \Rightarrow \theta = \frac{3}{4}

(b) Find the area of the sector. [2 marks]
Using A = \frac{1}{2}r^2\theta,

A = \frac{1}{2}\times 8^2 \times \frac{3}{4} = 24

2. A triangle ABC has an angle BAC of 120^{\circ}. There are sides of 6 cm and 16 cm.

(a) Show that the angle ACB is 19^{\circ} to the nearest degree [3 marks]
Using the sine rule:
\frac{\sin (ACB)}{6} = \frac{\sin 120}{16} \Rightarrow ACB = \arcsin \left(\frac{6 \sin 120}{16}\right) = 18.95\dots, which is 19^{\circ} to the nearest degree.

(b) Find the area of the triangle. [3 marks]
Other angle is 180 - (120+19) = 41^{\circ}. Hence the area is \frac{1}{2}\times 16 \times 6 \times \sin 41 = 31.5cm2

3.
(a) Express \frac{\sqrt{27^x}}{3^{2x-1}} in the form 3^p where p is an expression in x. [3 marks]

\frac{\sqrt{27^x}}{3^{2x-1}} = \frac{3^{\frac{3x}{2}}}{3^{2x-1}} = 3^{1 - \frac{x}{2}}

(b) Hence solve the equation \frac{\sqrt{27^x}}{3^{2x-1}} = \sqrt[3]{81}. [2 marks]

3^{1 - \frac{x}{2}} = 3^{\frac{4}{3}} \Rightarrow 1 - \frac{x}{2} = \frac{4}{3} \Rightarrow x = -\frac{2}{3}

4. A geometric series has the nth term u_n = 162\left(\frac{2}{3}\right)^n.
(a) Find u_1 and u_2. [2 marks]
u_1 = 108, \;u_2 = 72.

(b) Find the sum to infinity of the series [3 marks]
First term a = 108, and common ratio r = \frac{2}{3}.
Hence sum to infinity is \frac{a}{1-r}=\frac{108}{1 - \frac{2}{3}} = 324

(c) Find the smallest value of k such that \sum_{n=k}^{\infty} u_n < 2.5. [3 marks]
\sum_{n=k}^{\infty} u_n = 486\left(\frac{2}{3}\right)^k

\left(\frac{2}{3}\right)^k < \frac{2.5}{486} \Rightarrow k > \log_{\frac{2}{3}}\frac{2.5}{486  }. The signs change here as \log_{\frac{2}{3}}x is a decreasing function so if a<b, then \log_{\frac{2}{3}}a>\log_{\frac{  2}{3}}b.

Hence k>12.997 so k is at least 13.

5. A curve satisfies \frac{\mathrm{d}y}{\mathrm{d}x} = x^{\frac{3}{2}} - 2x, where x>0.
(a) Show that there is only one value of x for which there is a stationary point. [2 marks]
At stationary points, \frac{\mathrm{d}y}{\mathrm{d}x} = 0 \Rightarrow x(x^\frac{1}{2} - 2) = 0 \Rightarrow x^{1/2} = 2 \Rightarrow x=  4 (discarding x=0 as x>0). Hence there is only one value of x for which there is a stationary point.

(b) Find \frac{\mathrm{d}^2y}{\mathrm{d}x  ^2} and show that this is a minimum point. [3 marks]
\frac{\mathrm{d}^2y}{\mathrm{d}x  ^2} = \frac{3}{2}\sqrt{x} - 2.
When x = 4, this is equal to 1. Hence as the second derivative is positive, the point is a minimum.

(c) The line y=2 is a tangent to the curve. Find the equation of the curve. [4 marks]
Must be a tangent at the stationary point. Hence the point (4, 2) lies on the curve (the stationary point). Integrating gives y = \frac{2}{5}x^{\frac{5}{2}} - x^2 + c. As y(4) = 2, c = \frac{26}{5} so y = \frac{2}{5}x^{\frac{5}{2}} - x^2 + \frac{26}{5}.

6. There is a curve y = 2^{3x}.
(a) (i) Use the trapezium rule with five ordinates (four strips) to approximate \int^{1}_0 2^{3x}\;\mathrm{d}x to two decimal places. [4 marks]
Call the integral I. Then I \approx \frac{1}{2}h(y_0 + y_4 + 2(y_1+y_2+y_3)).

Now, h = \frac{1}{4} and y_r = 2^{\frac{3r}{4}}.

Hence I \approx \frac{1}{8}(2^0 + 2^3 + 2(2^{\frac{3}{4}} + 2^{\frac{3}{2}} + 2^{\frac{9}{4}})) = 3.44, to two decimal places.

(a) (ii) State how the approximation may be improved. [1 mark]
Use more strips/ordinates.

(a) (iii) The point (1, k) lies on the curve. Use your answer to part (a)(i) to find the area bounded by the curve, the line x=0 and the line y=k, to two decimal places. [3 marks]
k=8 so, by a sketch, the area is a rectangle with part (a)(i) subtracted. The area is 8 - 3.44 = 4.56 to two decimal places.

(b) The curve y=2^{3x} may be mapped onto the curve of y = 2^{3x-4} by a translation and a stretch. Describe:
(i) the translation [2 marks]

Translation in the vector \binom{\frac{4}{3}}{0}

(ii) the stretch [2 marks]

Stretch parallel to the y-axis scale factor \frac{1}{16}.

(c) Use logarithms to solve the equation 2^{3x-4} = 7, giving your answer to three significant figures. [2 marks]

x = \frac{1}{3}(4 + \log_2 7) = 2.27 to three significant figures.

7. A curve has the equation y = 7x + 6  - \frac{1}{x^2}.
(a) The region bounded by the curve and the lines x=1 and x=2 is above the x-axis. Show that the area of this region is 16. [5 marks]

Area (as curve is above the x-axis) is:
\int^{2}_{1} 7x + 6 - \frac{1}{x^2}\;\mathrm{d}x = \left[\frac{7x^2}{2} + 6x + \frac{1}{x}\right]^2_1
=\left(\frac{7}{2}(2^2) + 6(2) + \frac{1}{2}\right) - \left(\frac{7}{2}(1^2) + 6(1) + 1\right)
= \left(14 + 12 + \frac{1}{2}\right) - \left(\frac{7}{2} + 6 + 1\right)
= \frac{53}{2} - \frac{21}{2}
= \frac{32}{2} = 16

(b) The normal to the curve is parallel to the line 2y + 8x = 3. Find the equation of the normal to the curve. [6 marks]
Differentiating gives \frac{\mathrm{d}y}{\mathrm{d}x} = 7 + \frac{2}{x^3}. Now, the normal has a gradient of -4. Hence the gradient of the tangent to the curve is \frac{1}{4}. So

7 + \frac{2}{x^3} = \frac{1}{4} \Rightarrow \frac{8}{x^3} = -27 \Rightarrow x = -\frac{2}{3}. Hence y = -\frac{11}{12}.

So the equation of the normal is y + \frac{11}{12} = -4(x + \frac{2}{3}).

8.
(a) Solve the equation \cos \theta = \frac{2}{3} in the range 0^{\circ}\leqslant \theta \leqslant 360^{\circ}. Give your answers to the nearest degree. [2 marks]

\theta = \left\lbrace \arccos \frac{2}{3},\;360 - \arccos \frac{2}{3}\right\rbrace = \left\lbrace 48^{\circ},\;312^{\circ} \right\rbrace, to the nearest degree.

(b) (i) Given that 4\sin \theta \tan \theta = 4 - \cos \theta, show that 3\cos^2\theta + 4\cos \theta - 4 = 0. [3 marks]

4\sin\theta \frac{\sin\theta}{\cos\theta}= 4 - \cos \theta

4\frac{\sin^2\theta}{\cos\theta} = 4 - \cos \theta

4\sin^2\theta = 4\cos\theta - \cos^2 \theta

4 - 4\cos^2\theta = 4\cos\theta - \cos^2 \theta

3\cos^2\theta + 4\cos\theta - 4 = 0

(ii) Show that there is only one possible value of \cos \theta. [2 marks]
Factorising gives (3 \cos \theta - 2)(\cos \theta + 2) = 0. As |\cos\theta| \leqslant 1, the only possible value of \cos \theta is \frac{2}{3}.

(c) Hence solve the equation 4\sin 4x \tan 4x = 4 - \cos 4x in the range 0^{\circ}\leqslant x \leqslant 180^{\circ}, giving your answer to the nearest degree. [4 marks]

Solving \cos 4x = \frac{2}{3} in the range 0^{\circ}\leqslant 4x \leqslant 720^{\circ}.

Use part (a) and then add on 360^{\circ} to both solutions to give
4x = \left\lbrace 48^{\circ},\; 312^{\circ},\;408^{\circ},\;672^  {\circ}\right\rbrace, to the nearest degree.

So
x = \left\lbrace 12^{\circ},\; 78^{\circ},\;102^{\circ},\;168^{  \circ}\right\rbrace, to the nearest degree.

9. Given that 3\log_2\left(c+2\right) - \log_2\left(\frac{c^3}{2} + k\right) = 1, express (c+1)^2 in terms of k. [7 marks]

\log_2\left((c+2)^3\right)- \log_2\left( \frac{c^3+2k}{2} \right) = \log_2 2

\log_2\left(\frac{2(c+2)^3}{c^3+  2k}\right) = \log_2 2

\frac{2(c+2)^3}{2k+c^3} = 2

2(c+2)^3 = 4k + 2c^3

2c^3 + 12c^2 + 24c + 16 = 4k + 2c^3

12c^2 + 24c + 16 = 4k

12(c+1)^2 = 4k - 4

(c+1)^2 = \frac{k - 1}{3}
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MajorFader
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One of the trig equations were

48, 312
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MajorFader
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Equation of curve is 2/5x^5/2-x^2+ 26/5

If I am not wrong
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Integer123
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(Original post by MajorFader)
Equation of curve is 2/5x^5/2-x^2+ 26/5

If I am not wrong
(Original post by MajorFader)
One of the trig equations were

48, 312
I agree I'm writing the markscheme question by question and updating it.
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alextttt
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(a) (iii) The point (1, ) lies on the curve. Use your answer to part (a)(i) to find the area bounded by the curve, the line and the line , to two decimal places. [3 marks]
so, by a sketch, the area is a rectangle with part (a)(i) subtracted. The area is to two decimal places.

k=8 not 2?
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Integer123
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(Original post by alextttt)
(a) (iii) The point (1, ) lies on the curve. Use your answer to part (a)(i) to find the area bounded by the curve, the line and the line , to two decimal places. [3 marks]
so, by a sketch, the area is a rectangle with part (a)(i) subtracted. The area is to two decimal places.

k=8 not 2?
Sorry that was a typo
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niffty
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all the singles in here say "Resit!"
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remate91
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(b) The curve may be mapped onto the curve of by a translation and a stretch.
(i) the translation [2 marks]

Translation in the vector (, 0)

Translation should be by vector (4/3, 0) as a positive integer as
y = 2^3(x- 4/3)
x - 4 /3 => x = 4/3
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Impressive getting this up so quick!
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all the people who failed say "Mc Donald!"
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niffty
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i guess im not going to uni and living off your guys taxes
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SRowe
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Do you think slightly different methods for 7 will still be fully credited? I did the binomial expansion while it was still in log form and found an equation equal to k-1, then factored out 3
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Integer123
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(Original post by solrowe)
Do you think slightly different methods for 7 will still be fully credited? I did the binomial expansion while it was still in log form and found an equation equal to k-1, then factored out 3
I think that's fine, whether it is in log form or not
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Why is the translation -4/3 and not +4/3?
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Niallll99
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(Original post by Integer123)
AQA MPC2 June 2017 Unofficial Markscheme

1. A sector has a perimeter of 22cm and an angle of \theta radians. The radius is 8cm.
(a) Find the value of \theta [2 marks]
16 + 8\theta = 22
\theta = \frac{3}{4}

(b) Find the area of the sector [2 marks]
\frac{1}{2}\times 8^2 \times \frac{3}{4} = 24

2. A triangle ABC has an angle BAC of 120\textdegree. There are sides of 6cm and 16cm.

(a) Show that the angle ACB is 19\textdegree to the nearest degree [3 marks]
\frac{\sin (ACB)}{6} = \frac{\sin 120}{16} \Rightarrow ACB = \arcsin \left(\frac{6 \sin 120}{16}\right) = 18.95\dots, which is 19 degrees to the nearest degree.

(b) Find the area of the triangle. [3 marks]
Other angle is 41 degrees. Hence the area is \frac{1}{2}\times 16 \times 6 \times \sin 41 = 31.5cm2

3.
(a) Express \frac{\sqrt{27^x}}{3^{2x-1}} in the form 3^p where p is an expression in x. [3 marks]

\frac{\sqrt{27^x}}{3^{2x-1}} = \frac{3^{\frac{3x}{2}}}{3^{2x-1}} = 3^{1 - \frac{x}{2}}

(b) Hence solve the equation \frac{\sqrt{27^x}}{3^{2x-1}} = \sqrt[3]{81}. [2 marks]

3^{1 - \frac{x}{2}} = 3^{\frac{4}{3}} \Rightarrow 1 - \frac{x}{2} = \frac{4}{3} \Rightarrow x = -\frac{2}{3}

4. A geometric series has the nth term u_n = 162\left(\frac{2}{3}\right)^n.
(a) Find u_1 and u_2. [2 marks]
u_1 = 108, \;u_2 = 72.

(b) Find the sum to infinity of the series [3 marks]
First term is 108, common ratio is 2/3.
Hence sum to infinity is \frac{108}{1 - \frac{2}{3}} = 324

(c) Find the smallest value of k such that \sum_{n=k}^{\infty} u_n < 2.5. [3 marks]
\sum_{n=k}^{\infty} u_n = 486\left(\frac{2}{3}\right)^k

486\left(\frac{2}{3}\right)^k < 2.5 \Rightarrow k > \log_{\frac{2}{3}}\frac{2.5}{486  }.

Hence k>12.997 so k is at least 13.

5. A curve satisfies \frac{\mathrm{d}y}{\mathrm{d}x} = x^{\frac{3}{2}} - 2x, where x>0.
(a) Show that there is only one value of x for which there is a stationary point. [2 marks]
\frac{\mathrm{d}y}{\mathrm{d}x} = 0 \Rightarrow x(x^\frac{1}{2} - 2) = 0 \Rightarrow x^{1/2} = 2 \Rightarrow x=  4. Hence there is only one value of x for which there is a stationary point.

(b) Find \frac{\mathrm{d}^2y}{\mathrm{d}x  ^2} and show that this is a minimum point. [3 marks]
\frac{\mathrm{d}^2y}{\mathrm{d}x  ^2} = \frac{3}{2}\sqrt{x} - 2.
When x = 4, this is equal to 1. Hence it is positive, so the point is a minimum.

(c) The line y=2 is a tangent to the curve. Find the equation of the curve. [4 marks]
Must be a tangent at the stationary point. Hence the point (4, 2) lies on the curve (the stationary point). Integrating gives y = \frac{2}{5}x^{\frac{5}{2}} - x^2 + c. As y(4) = 2, c = \frac{26}{5} so y = \frac{2}{5}x^{\frac{5}{2}} - x^2 + \frac{26}{5}.

6. There is a curve y = 2^{3x}.
(a) (i) Use the trapezium rule with five ordinates (four strips) to approximate \int^{1}_0 2^{3x}\;\mathrm{d}x to two decimal places. [4 marks]
Use the trapezium rule formula to find that it is 3.44 to two decimal places.

(a) (ii) State how the approximation may be improved. [1 mark]
Use more strips/ordinates.

(a) (iii) The point (1, k) lies on the curve. Use your answer to part (a)(i) to find the area bounded by the curve, the line x=0 and the line y=k, to two decimal places. [3 marks]
k=8 so, by a sketch, the area is a rectangle with part (a)(i) subtracted. The area is 8 - 3.44 = 4.56 to two decimal places.

(b) The curve y=2^{3x} may be mapped onto the curve of y = 2^{3x-4} by a translation and a stretch.
(i) the translation [2 marks]

Translation in the vector (-\frac{4}{3}, 0)

(ii) stretch [2 marks]

Stretch parallel to the y-axis scale factor \frac{1}{16}.

(c) Use logarithms to solve the equation 2^{3x-4} = 7, using your answer to three significant figures. [2 marks]

x = \frac{1}{3}(4 + \log_2 7) = 2.27 to three significant figures.

7. A curve has the equation y = 7x + 6  - \frac{1}{x^2}.
(a) The region bounded by the curve and the lines x=1 and x=2 is above the x-axis. Show that the area of this region is 16. [5 marks]

Area (as curve is above the x-axis) is:
\int^{2}_{1} 7x + 6 - \frac{1}{x^2}\;\mathrm{d}x = \left[\frac{7x^2}{2} + 6x + \frac{1}{x}\right]^2_1. Substitute in the limits to show that the area is 16.

(b) The normal to the curve is parallel to the line 2y + 8x = 3. Find the equation of the normal to the curve. [6 marks]
Differentiating gives \frac{\mathrm{d}y}{\mathrm{d}x} = 7 - \frac{2}{x^3}. Now, the normal has a gradient of -4. Hence the gradient of the tangent to the curve is \frac{1}{4}. So

7 - \frac{2}{x^3} = \frac{1}{4} \Rightarrow \frac{8}{x^3} = -27 \Rightarrow x = -\frac{2}{3}. Hence y = -\frac{11}{12}.

So the equation of the normal is y + \frac{11}{12} = -4(x + \frac{2}{3}).

8.
(a) Solve the equation \cos \theta = \frac{2}{3} in the range 0\leqslant \theta \leqslant 360. Give your answers to the nearest degree. [2 marks]

The values of \theta are 48 and 312.

(b) (i) Given that 4\sin \theta \tan \theta = 4 - \cos \theta, show that 3\cos^2\theta + 4\cos \theta - 4 = 0. [3 marks]

4\frac{\sin\theta}{\cos\theta}\s  in\theta = 4 - \cos \theta

4\frac{\sin^2\theta}{\cos\theta} = 4 - \cos \theta

4\sin^2\theta = 4\cos\theta - \cos^2 \theta

4 - 4\cos^2\theta = 4\cos\theta - \cos^2 \theta

3\cos^2\theta + 4\cos\theta - 4 = 0

(ii) Show that there is only one possible value of \cos \theta. [2 marks]
Factorising gives (3 \cos \theta - 2)(\cos \theta + 2) = 0. As |\cos\theta| \leqslant 1, the only possible value of \cos \theta is \frac{2}{3}.

(c) Hence solve the equation 4\sin 4x \tan 4x = 4 - \cos 4x in the range 0\leqslant x \leqslant 180. [4 marks]

Solving \cos 4x = \frac{2}{3} in the range 0 \leqslant 4x \leqslant 720.

Use part (a) and then add on 360 to both solutions to give
4x = \left\lbrace 48,\; 312,\;408,\;672\right\rbrace

So
x = \left\lbrace 12,\; 78,\;102,\;168\right\rbrace

9. Given that 3\log_2\left(c+2\right) - \log_2\left(\frac{k}{2} + c^3\right) = 1, express (c+1)^2 in terms of k. [7 marks]

\log_2((c+2)^3) - \log_2\left( \frac{2k+c^3}{2} \right) = \log_2 2

\log_2(\frac{2(c+2)^3}{2k+c^3}) = \log_2 2

\frac{2(c+2)^3}{2k+c^3} = 2

2(c+2)^3 = 4k + 2c^3

2c^3 + 12c^2 + 24c + 16 = 4k + 2c^3

12c^2 + 24c + 16 = 4k

12(c+1)^2 = 4k - 4

(c+1)^2 = \frac{k - 1}{3}
For 6)b)I) I'm pretty sure the translation was with the vector (4/3,0) and for the last question it asked to fine k in the form (c+1)^2 therefore k=3(c+1)^(2) +1
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Andrew Dainty
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Yeah, the translation should be +4/3
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mancmed
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I used log laws for the first bit and ended up crossing out correct working so 5 marks down drain ?
Also how much do you lose for putting gradient of Q as 1/4. I got x and y fine and used equation right just will it be one accuracy mark gone?
And for trapezium rule for some reason I have no idea I halved the rectangle coz I'm used to doing triangles hahahahha, will I lose 2 or 3 as I still got 8 but then subtracted 3.44 from 4.
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Integer123
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(Original post by Niallll99)
For 6)b)I) I'm pretty sure the translation was with the vector (4/3,0) and for the last question it asked to fine k in the form (c+1)^2 therefore k=3(c+1)^(2) +1
(Original post by Andrew Dainty)
Yeah, the translation should be +4/3
Ooops, another typo, I put 4/3 on my exam paper
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(Original post by Niallll99)
For 6)b)I) I'm pretty sure the translation was with the vector (4/3,0) and for the last question it asked to fine k in the form (c+1)^2 therefore k=3(c+1)^(2) +1
I believe it asked to find (c+1)^2 in terms of k.
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