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A-Level Maths Mark Scheme.pdf

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Preview of A-Level Maths Mark Scheme
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A(2x +5)+ B(5x -1) º 6x + 42, equating coefficients and constant terms yields 2A+5B = 6 and 5A- B = 42, solving gives A = 8 and B = −2.
Alternative method uses substitution, x = -5/2 gives -27/2B = 27, B = -2, x = 1/5 gives 27/5A = 43.2, A = 8.
Proof by contradiction assumes 25a+15b =1, dividing by 5 gives 5a + 3b = 1/5, but 5a + 3b must be an integer, contradiction.
Parametric differentiation, d(2sin2t)/dt = -4sin(t)cos(t), d(cos(4sin(t)))/dt = -4cosec(4sin(t)), dy/dx = -1/4cosec(4sin(t)) at t = -π/6, gradient of tangent is 1/2, normal is -2.
Equation of normal, y + 1/2 = -2(x + 1/2), simplifies to 4x + 2y + 1 = 0.
Integration using trigonometric identities, ∫(cos^3(x))/sin^3(x) dx, attempt to find ∫(cos^3(x))/sin^3(x) dx, writing ∫(f'(x)/f(x)) dx or ∫(ln(f(x)))' dx constitutes an attempt, correct answer is (1/3)ln|sin(x)| + C.
Vectors in 3 dimensions, finds AB = (0,4,2), AC = (5,4,8), BC = (5,0,6), |AB| = √20, |AC| = √105, |BC| = √125, Pythagoras' theorem |AB|^2 + |AC|^2 = |BC|^2.

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A(2x +5)+ B(5x -1) º 6x + 42, equating coefficients and constant terms yields 2A+5B = 6 and 5A- B = 42, solving gives A = 8 and B = −2.

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  • File Format: PDF
  • File Size: 853 KB
  • Pages: 17
  • Language: EN
  • Total Downloads: 200
  • Last Updated: 2 days ago

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The mark scheme provides notes on types of marks, including method marks (M), accuracy marks (A), and B marks for correct statements or steps.
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