IBDP Maths SL 1.9 The binomial theorem AA HL Paper 1- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
Binomial expansion of \( (1 + kx)^n \): \( \sum_{r=0}^n \binom{n}{r} k^r x^r \).
Given: \( 1 + 12x + 28k^2 x^2 + \ldots + k^n x^n \).
Coefficient of \( x \): \( \binom{n}{1} k = n k = 12 \quad (1) \).
Coefficient of \( x^2 \): \( \binom{n}{2} k^2 = \frac{n(n-1)}{2} k^2 = 28 k^2 \quad (2) \).
From (2): \( \frac{n(n-1)}{2} = 28 \implies n(n-1) = 56 \implies n^2 – n – 56 = 0 \).
Solve: \( n^2 – 8n + 7n – 56 = 0 \implies (n-8)(n+7) = 0 \implies n = 8 \) or \( n = -7 \).
Since \( n \in \mathbb{Z}^+ \), \( n = 8 \).
From (1): \( 8k = 12 \implies k = \frac{12}{8} = \frac{3}{2} \).
Thus, \( n = 8 \), \( k = \frac{3}{2} \).
Markscheme
Coefficient equations: \( n k = 12 \quad \mathbf{M1} \), \( \frac{n(n-1)}{2} k^2 = 28 k^2 \quad \mathbf{M1} \).
Solve: \( n(n-1) = 56 \implies n = 8 \quad \mathbf{A1} \).
\( k = \frac{12}{8} = \frac{3}{2} \quad \mathbf{A1} \).
Final answer: \( n = 8 \), \( k = \frac{3}{2} \quad \mathbf{A1} \).
▶️ Answer/Explanation
Binomial expansion: \( \left( x^2 – \frac{2}{x} \right)^4 = \sum_{r=0}^4 \binom{4}{r} (x^2)^{4-r} \left( -\frac{2}{x} \right)^r \).
Terms:
\( r=0 \): \( \binom{4}{0} (x^2)^4 = x^8 \).
\( r=1 \): \( \binom{4}{1} (x^2)^3 \left( -\frac{2}{x} \right) = 4 \cdot x^6 \cdot \left( -\frac{2}{x} \right) = -8 x^5 \).
\( r=2 \): \( \binom{4}{2} (x^2)^2 \left( -\frac{2}{x} \right)^2 = 6 \cdot x^4 \cdot \frac{4}{x^2} = 24 x^2 \).
\( r=3 \): \( \binom{4}{3} (x^2)^1 \left( -\frac{2}{x} \right)^3 = 4 \cdot x^2 \cdot \left( -\frac{8}{x^3} \right) = -\frac{32}{x} \).
\( r=4 \): \( \binom{4}{4} \left( -\frac{2}{x} \right)^4 = 1 \cdot \frac{16}{x^4} = \frac{16}{x^4} \).
Simplified: \( x^8 – 8x^5 + 24x^2 – \frac{32}{x} + \frac{16}{x^4} \).
Markscheme
Apply binomial theorem: \( \left( x^2 – \frac{2}{x} \right)^4 \quad \mathbf{M1} \).
Correct expansion: \( x^8 – 8x^5 + 24x^2 – \frac{32}{x} + \frac{16}{x^4} \quad \mathbf{A3} \).
Note: Deduct one A mark for each incorrect or omitted term.
[4 marks]
▶️ Answer/Explanation
Binomial expansion: \( (2 – 3x)^5 = \sum_{r=0}^5 \binom{5}{r} 2^{5-r} (-3x)^r \).
Terms:
\( r=0 \): \( \binom{5}{0} 2^5 = 1 \cdot 32 = 32 \).
\( r=1 \): \( \binom{5}{1} 2^4 (-3x) = 5 \cdot 16 \cdot (-3x) = -240x \).
\( r=2 \): \( \binom{5}{2} 2^3 (-3x)^2 = 10 \cdot 8 \cdot 9x^2 = 720x^2 \).
\( r=3 \): \( \binom{5}{3} 2^2 (-3x)^3 = 10 \cdot 4 \cdot (-27x^3) = -1080x^3 \).
\( r=4 \): \( \binom{5}{4} 2^1 (-3x)^4 = 5 \cdot 2 \cdot 81x^4 = 810x^4 \).
\( r=5 \): \( \binom{5}{5} (-3x)^5 = 1 \cdot (-243x^5) = -243x^5 \).
Result: \( 32 – 240x + 720x^2 – 1080x^3 + 810x^4 – 243x^5 \).
Markscheme
Binomial expansion with coefficients \(\quad \mathbf{M1} \).
Correct terms with moduli and powers \(\quad \mathbf{A1} \).
Correct signs \(\quad \mathbf{A1} \).
Final expansion: \( 32 – 240x + 720x^2 – 1080x^3 + 810x^4 – 243x^5 \quad \mathbf{A2} \).
Note: Award M1 only if binomial coefficients are seen. Award A1 for correct moduli and powers, A1 for correct signs.
[4 marks]