IB Mathematics SL 1.9 The binomial theorem AA SL Paper 2- Exam Style Questions- New Syllabus
Question
Most-appropriate topic codes (IB Mathematics AA SL 2021):
▶️ Answer/Explanation
Method:
The general term in the expansion of \( (2x + k)^{10} \) is: \( T_{r+1} = ^{10}C_r (2x)^{10-r} k^r \).
For the term containing \( x^6 \), we need \( 10 – r = 6 \), so \( r = 4 \).
The term is: \( T_5 = ^{10}C_4 (2x)^6 k^4 \).
The coefficient of \( x^6 \) is: \( ^{10}C_4 \cdot 2^6 \cdot k^4 = 210 \times 64 \times k^4 = 13440 k^4 \).
Set this equal to \( 8.4 \times 10^6 \): \( 13440 k^4 = 8400000 \) \( k^4 = \frac{8400000}{13440} = 625 \) \( k^4 = 5^4 \) \( k = \pm 5 \)
Since \( k \in \mathbb{Z} \), both values are valid, but the question may accept only the positive integer in context.
\( \boxed{k = 5} \) (or \( k = \pm 5 \)).
▶️ Answer/Explanation
First, write \(\sqrt{x}\) as \(x^{\frac12}\).
The general term in the expansion of \(\left(x^{\frac12}+k\right)^{12}\) is:
\(\binom{12}{r}\left(x^{\frac12}\right)^r k^{12-r}\)
The power of \(x\) in this term is \(x^{\frac{r}{2}}\).
For the term in \(x^{\frac32}\):
\(\dfrac{r}{2}=\dfrac{3}{2}\)
\(r=3\)
Therefore, the required term is:
\(\binom{12}{3}\left(x^{\frac12}\right)^3k^9\)
\(\binom{12}{3}k^9x^{\frac32}\)
Hence, its coefficient is \(\binom{12}{3}k^9\).
\(\binom{12}{3}k^9=-112640\)
\(220k^9=-112640\)
\(k^9=-512\)
Since \((-2)^9=-512\):
✅ Answer: \(k=-2\)