Binomial Theorem
Grade 11
Question:
<p>The coefficient of x<sup>-5</sup> in the binomial expansion of <span class="math-tex">\(\left(\frac{x \ + \ 1}{x^{\frac{2}{3}} \ - \ x^{\frac{1}{3}} \ + \ 1}-\frac{x \ - \ 1}{x \ - \ x^{\frac{1}{2}}}\right)^{10}\)</span> where x <span class="math-tex">\(\ne\)</span> 0, 1, is:</p>
<p style="display:inline">1</p>
<p style="display:inline">4</p>
<p style="display:inline">-1</p>
<p style="display:inline">-4</p>
Step-by-Step Solution
Key Concept: Simplify the base expression using algebraic identities like the sum of cubes and difference of squares to convert it into a manageable binomial form before applying the General Term formula.
<p><span class="math-tex">\(\left[\frac{\left(x^{\frac 1 3} + 1\right)\left(x^{\frac 2 3} - x^{\frac 1 3} + 1\right)}{\left(x^{\frac 2 3} - x ^{\frac 1 3} + 1\right)}-\frac{(\sqrt{x} - 1)(\sqrt{x} + 1)}{\sqrt{x}(\sqrt{x} - 1)}\right]^{10}\)</span><br />
= <span class="math-tex">\(\left(x^{\frac 1 3}+1-1-\frac 1 x^{\frac 1 2}\right)^{10}=\left(x^{\frac 1 3}-\frac 1 x^{\frac 1 2}\right)^{10}\)</span><br />
T<sub>r+1</sub> = <sup>10</sup>C<sub>r</sub> <span class="math-tex">\(x^{\frac{20-5 r}{6}}\)</span><br />
for r = 10<br />
T<sub>11</sub> = <sup>10</sup>C<sub>10</sub> x<sup>-5</sup><br />
Coefficient of x<sup>-5</sup> = <sup>10</sup>C<sub>10</sub> (1)(-1)<sup>10</sup> = 1</p>
Correct Answer: A