IBDP Maths SL 5.10 Indefinite integral of xn , sinx, cosx, and ex AA HL Paper 2- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
[6 marks]
(a) \( \int (6x^2 – 3x) \, dx = 2x^3 – \frac{3x^2}{2} + c \) (A1, A1).
(b) \( \int_1^2 (6x^2 – 3x) \, dx \) (A1).
\( \left[ 2x^3 – \frac{3x^2}{2} \right]_1^2 = (2 \cdot 2^3 – \frac{3 \cdot 2^2}{2}) – (2 \cdot 1^3 – \frac{3 \cdot 1^2}{2}) \) (M1).
\( = (16 – 6) – (2 – 1.5) = 10 – 0.5 = \frac{19}{2} \) (A1, A1).
Markscheme Answers:
(a) \( 2x^3 – \frac{3x^2}{2} + c \) or \( \frac{6x^3}{3} – \frac{3x^2}{2} + c \) (A1, A1)
(b) Substitution of limits or function \( \int_1^2 f(x) \, dx \) or \( \left[ 2x^3 – \frac{3x^2}{2} \right]_1^2 \) (A1), substituting limits (M1), correct working (A1), \( \frac{19}{2} \) (A1)
[Total 6 marks]
▶️ Answer/Explanation
[15 marks]
(a) \( \frac{1}{x(k – x)} = \frac{a}{x} + \frac{b}{k – x} \) (M1).
\( a(k – x) + bx = 1 \) (A1).
Comparing coefficients or substituting \( x = 0 \) and \( x = k \): \( a = \frac{1}{k} \), \( b = \frac{1}{k} \) (A1).
(b) \( f(x) = \frac{1}{k} \int \left( \frac{1}{x} + \frac{1}{k – x} \right) dx \) (M1).
\( = \frac{1}{k} \left( \ln |x| – \ln |k – x| \right) + c = \frac{1}{k} \ln \left| \frac{x}{k – x} \right| + c \) (A1, A1).
(c) \( 5k \int \frac{1}{P(k – P)} dP = \int 1 dt \) (M1).
\( 5 (\ln P – \ln (k – P)) = t + c \) (A1).
EITHER \( c = 5 \ln \frac{1200}{k – 1200} \) at \( t = 0 \), \( P = 1200 \) (M1).
\( 5 (\ln P – \ln (k – P)) = t + 5 \ln \frac{1200}{k – 1200} \), \( P = \frac{1200k}{(k – 1200)e^{\frac{t}{5}} + 1200} \) (A1).
OR \( \frac{P}{k – P} = A e^{\frac{t}{5}} \), \( A = \frac{1200}{k – 1200} \) at \( t = 0 \), \( P = 1200 \) (M1).
\( P = \frac{1200k}{(k – 1200)e^{\frac{t}{5}} + 1200} \) (A1).
(d) \( 2400 = \frac{1200k}{(k – 1200)e^{-2} + 1200} \) at \( t = 10 \) (M1).
Solving gives \( k = 2845.34\ldots \), so \( k = 2845 \) (A1, A1).
(e) Maximum rate at \( P = \frac{k}{2} = 1422.67\ldots \) (M1).
\( 1422.67 = \frac{1200 \cdot 2845}{(2845 – 1200)e^{\frac{t}{5}} + 1200} \), \( t = 1.57814\ldots \), so \( t = 1.58 \) days (A1, A1).
Markscheme Answers:
(a) \( a(k – x) + bx = 1 \) (M1), coefficients or substitutions (A1), \( a = \frac{1}{k} \), \( b = \frac{1}{k} \) (A1)
(b) \( \int \left( \frac{1}{x} + \frac{1}{k – x} \right) dx \) (M1), \( f(x) = \frac{1}{k} \ln \left| \frac{x}{k – x} \right| + c \) (A1, A1)
(c) \( 5k \int \frac{1}{P(k – P)} dP = \int 1 dt \) (M1), \( 5 (\ln P – \ln (k – P)) = t + c \) (A1)
EITHER \( c = 5 \ln \frac{1200}{k – 1200} \) (M1), \( P = \frac{1200k}{(k – 1200)e^{\frac{t}{5}} + 1200} \) (A1)
OR \( \frac{P}{k – P} = A e^{\frac{t}{5}} \), \( A = \frac{1200}{k – 1200} \) (M1), \( P = \frac{1200k}{(k – 1200)e^{\frac{t}{5}} + 1200} \) (A1)
(d) \( 2400 = \frac{1200k}{(k – 1200)e^{-2} + 1200} \) (M1), \( k = 2845.34\ldots \) (A1), \( k = 2845 \) (A1)
(e) \( P = \frac{k}{2} \) (M1), \( t = 1.57814\ldots \) (A1), \( t = 1.58 \) (A1)
[Total 15 marks]