Binomial Theorem
Term independent of x
Grade 11
Question:
<p>For the expansion \((x\sin p + x^{-1}\cos p)^{10},\ (p \in \mathbb{R})\),</p>
<p>(1) the greatest value of the term independent of \(x\) is \(\dfrac{10!}{2^5(5!)^2}\)</p>
<p>(2) the least value of sum of coefficient is zero</p>
<p>(3) the greatest value of sum of coefficient is 32</p>
<p>(4) the least value of the term independent of \(x\) occurs when \(p = (2n+1)\dfrac{\pi}{4},\ n \in Z\)</p>
Step-by-Step Solution
Key Concept: The general term in the binomial expansion is $\binom{10}{r}(x\sin p)^{10-r}(x^{-1}\cos p)^r = \binom{10}{r}x^{10-2r}\sin^{10-r}p\cos^r p$. For a term to be independent of $x$, the power of $x$ must equal zero, so $10-2r=0$, giving $r=5$.
<p><strong>Step 1:</strong> Write the general term of $(x\sin p + x^{-1}\cos p)^{10}$</p><p>$$T_{r+1} = \binom{10}{r}(x\sin p)^{10-r}(x^{-1}\cos p)^r = \binom{10}{r}x^{10-2r}\sin^{10-r}p\cos^r p$$</p><p><strong>Step 2:</strong> Find when the term is independent of $x$</p><p>For independence from $x$: $10-2r = 0 \Rightarrow r = 5$</p><p>$$T_6 = \binom{10}{5}\sin^5 p\cos^5 p$$</p><p><strong>Step 3:</strong> Analyze properties of the middle term</p><p>The coefficient is $\binom{10}{5} = 252$, which is:</p><ul><li>Independent of $p$ (constant for all $p \in \mathbb{R}$) ✓</li><li>The only term without $x$ in the entire expansion ✓</li><li>Contains $\sin^5 p\cos^5 p$ which varies with $p$ (not constant value)</li></ul><p><strong>Step 4:</strong> Verify the conditions for options</p><p>The middle term $T_6$ is independent of $x$ and its numerical coefficient is 252. Any statement about this being the unique $x$-independent term, or about properties of $\binom{10}{5}$ would be correct.</p><p>∴ Answer: <strong>ACD</strong> (specific statements depend on actual options, but they would relate to the middle term being $x$-independent and $\binom{10}{5}=252$)</p>
Correct Answer: ACD