A-Level Mathematics (9MA0) Paper 3: Statistics
& Mechanics - 2020 Marking Scheme & Examiner
Report
Section 1: Algebra & Functions (Questions 1–3)
Question 1:
Solve the equation 2e2x−5ex+2=02e^{2x} - 5e^x + 2 = 02e2x−5ex+2=0.
Solution:
1. Let y=exy = e^xy=ex. The equation becomes: 2y2−5y+2=02y^2 - 5y + 2 = 02y2−5y+2=0.
(1 mark for substitution)
2. Factorise: (2y−1)(y−2)=0(2y - 1)(y - 2) = 0(2y−1)(y−2)=0. (1 mark for correct
factorisation)
3. Solve for yyy: y=12y = \frac{1}{2}y=21 or y=2y = 2y=2. (1 mark for solutions)
4. Convert back to xxx: ex=12⇒x=ln(12)=−ln2e^x = \frac{1}{2} \Rightarrow x =
\ln\left(\frac{1}{2}\right) = -\ln 2ex=21⇒x=ln(21)=−ln2, ex=2⇒x=ln2e^x = 2 \Rightarrow x =
\ln 2ex=2⇒x=ln2. (1 mark for each logarithmic solution)
Final Answer: x=−ln2x = -\ln 2x=−ln2 or x=ln2x = \ln 2x=ln2.
Question 2:
Given f(x)=3x+1x−2f(x) = \frac{3x + 1}{x - 2}f(x)=x−23x+1, find f−1(x)f^{-1}(x)f−1(x).
Solution:
1. Set y=3x+1x−2y = \frac{3x + 1}{x - 2}y=x−23x+1. (1 mark for substitution)
2. Cross-multiply: y(x−2)=3x+1y(x - 2) = 3x + 1y(x−2)=3x+1. (1 mark for rearrangement)
3. Solve for xxx: yx−2y=3x+1yx - 2y = 3x + 1yx−2y=3x+1, yx−3x=1+2yyx - 3x = 1 +
2yyx−3x=1+2y, x(y−3)=1+2yx(y - 3) = 1 + 2yx(y−3)=1+2y, x=1+2yy−3x = \frac{1 + 2y}{y -
3}x=y−31+2y. (2 marks for solving and inverse function)
4. Replace yyy with xxx: f−1(x)=1+2xx−3f^{-1}(x) = \frac{1 + 2x}{x - 3}f−1(x)=x−31+2x.
Final Answer: f−1(x)=1+2xx−3f^{-1}(x) = \frac{1 + 2x}{x - 3}f−1(x)=x−31+2x.
, Question 3:
Find the exact solutions to sinθ=−12\sin \theta = -\frac{1}{2}sinθ=−21 for 0≤θ≤2π0 \leq \theta \leq
2\pi0≤θ≤2π.
Solution:
1. Reference angle: sin−1(12)=π6\sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6}sin−1(21)=6π.
(1 mark for reference angle)
2. Sine is negative in 3rd and 4th quadrants: θ=π+π6=7π6\theta = \pi + \frac{\pi}{6} =
\frac{7\pi}{6}θ=π+6π=67π, θ=2π−π6=11π6\theta = 2\pi - \frac{\pi}{6} =
\frac{11\pi}{6}θ=2π−6π=611π. (2 marks for correct quadrants and solutions)
Final Answer: θ=7π6,11π6\theta = \frac{7\pi}{6}, \frac{11\pi}{6}θ=67π,611π.
Section 2: Coordinate Geometry (Questions 4–6)
Question 4:
Find the equation of the line perpendicular to 3x+4y=123x + 4y = 123x+4y=12 passing through
(2,−1)(2, -1)(2,−1).
Solution:
1. Rewrite in slope-intercept form: 4y=−3x+12⇒y=−34x+34y = -3x + 12 \Rightarrow y =
-\frac{3}{4}x + 34y=−3x+12⇒y=−43x+3. (1 mark for rearrangement)
2. Slope of perpendicular line: m=43m = \frac{4}{3}m=34. (1 mark for negative reciprocal)
3. Use point-slope form: y+1=43(x−2)y + 1 = \frac{4}{3}(x - 2)y+1=34(x−2). (1 mark for
substitution)
4. Simplify: y=43x−83−1=43x−113y = \frac{4}{3}x - \frac{8}{3} - 1 = \frac{4}{3}x -
\frac{11}{3}y=34x−38−1=34x−311. (1 mark for final equation)
Final Answer: y=43x−113y = \frac{4}{3}x - \frac{11}{3}y=34x−311.
Question 5:
Find the distance between the points A(3,−4)A(3, -4)A(3,−4) and B(−1,2)B(-1, 2)B(−1,2).
Solution:
& Mechanics - 2020 Marking Scheme & Examiner
Report
Section 1: Algebra & Functions (Questions 1–3)
Question 1:
Solve the equation 2e2x−5ex+2=02e^{2x} - 5e^x + 2 = 02e2x−5ex+2=0.
Solution:
1. Let y=exy = e^xy=ex. The equation becomes: 2y2−5y+2=02y^2 - 5y + 2 = 02y2−5y+2=0.
(1 mark for substitution)
2. Factorise: (2y−1)(y−2)=0(2y - 1)(y - 2) = 0(2y−1)(y−2)=0. (1 mark for correct
factorisation)
3. Solve for yyy: y=12y = \frac{1}{2}y=21 or y=2y = 2y=2. (1 mark for solutions)
4. Convert back to xxx: ex=12⇒x=ln(12)=−ln2e^x = \frac{1}{2} \Rightarrow x =
\ln\left(\frac{1}{2}\right) = -\ln 2ex=21⇒x=ln(21)=−ln2, ex=2⇒x=ln2e^x = 2 \Rightarrow x =
\ln 2ex=2⇒x=ln2. (1 mark for each logarithmic solution)
Final Answer: x=−ln2x = -\ln 2x=−ln2 or x=ln2x = \ln 2x=ln2.
Question 2:
Given f(x)=3x+1x−2f(x) = \frac{3x + 1}{x - 2}f(x)=x−23x+1, find f−1(x)f^{-1}(x)f−1(x).
Solution:
1. Set y=3x+1x−2y = \frac{3x + 1}{x - 2}y=x−23x+1. (1 mark for substitution)
2. Cross-multiply: y(x−2)=3x+1y(x - 2) = 3x + 1y(x−2)=3x+1. (1 mark for rearrangement)
3. Solve for xxx: yx−2y=3x+1yx - 2y = 3x + 1yx−2y=3x+1, yx−3x=1+2yyx - 3x = 1 +
2yyx−3x=1+2y, x(y−3)=1+2yx(y - 3) = 1 + 2yx(y−3)=1+2y, x=1+2yy−3x = \frac{1 + 2y}{y -
3}x=y−31+2y. (2 marks for solving and inverse function)
4. Replace yyy with xxx: f−1(x)=1+2xx−3f^{-1}(x) = \frac{1 + 2x}{x - 3}f−1(x)=x−31+2x.
Final Answer: f−1(x)=1+2xx−3f^{-1}(x) = \frac{1 + 2x}{x - 3}f−1(x)=x−31+2x.
, Question 3:
Find the exact solutions to sinθ=−12\sin \theta = -\frac{1}{2}sinθ=−21 for 0≤θ≤2π0 \leq \theta \leq
2\pi0≤θ≤2π.
Solution:
1. Reference angle: sin−1(12)=π6\sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6}sin−1(21)=6π.
(1 mark for reference angle)
2. Sine is negative in 3rd and 4th quadrants: θ=π+π6=7π6\theta = \pi + \frac{\pi}{6} =
\frac{7\pi}{6}θ=π+6π=67π, θ=2π−π6=11π6\theta = 2\pi - \frac{\pi}{6} =
\frac{11\pi}{6}θ=2π−6π=611π. (2 marks for correct quadrants and solutions)
Final Answer: θ=7π6,11π6\theta = \frac{7\pi}{6}, \frac{11\pi}{6}θ=67π,611π.
Section 2: Coordinate Geometry (Questions 4–6)
Question 4:
Find the equation of the line perpendicular to 3x+4y=123x + 4y = 123x+4y=12 passing through
(2,−1)(2, -1)(2,−1).
Solution:
1. Rewrite in slope-intercept form: 4y=−3x+12⇒y=−34x+34y = -3x + 12 \Rightarrow y =
-\frac{3}{4}x + 34y=−3x+12⇒y=−43x+3. (1 mark for rearrangement)
2. Slope of perpendicular line: m=43m = \frac{4}{3}m=34. (1 mark for negative reciprocal)
3. Use point-slope form: y+1=43(x−2)y + 1 = \frac{4}{3}(x - 2)y+1=34(x−2). (1 mark for
substitution)
4. Simplify: y=43x−83−1=43x−113y = \frac{4}{3}x - \frac{8}{3} - 1 = \frac{4}{3}x -
\frac{11}{3}y=34x−38−1=34x−311. (1 mark for final equation)
Final Answer: y=43x−113y = \frac{4}{3}x - \frac{11}{3}y=34x−311.
Question 5:
Find the distance between the points A(3,−4)A(3, -4)A(3,−4) and B(−1,2)B(-1, 2)B(−1,2).
Solution:
