Binomial Theorem
Grade 11
Question:
<p>If the value of x is so small that x<sup>2 </sup>and greater powers can be neglected, then <span class="math-tex">\(\frac{\sqrt{1+x}+\sqrt[3]{(1-x)^{2}}}{1+x+\sqrt{1+x}}\)</span> is equal to</p>
<p style="display:inline">1 + <span class="math-tex">\(\frac{5}{6}\)</span>x</p>
<p style="display:inline">1 - <span class="math-tex">\(\frac{2}{3}\)</span>x</p>
<p style="display:inline">1 - <span class="math-tex">\(\frac{5}{6}\)</span>x</p>
<p style="display:inline">1 + <span class="math-tex">\(\frac{2}{3}\)</span>x</p>
Step-by-Step Solution
Key Concept: Apply the binomial approximation (1+x)^n ≈ 1+nx for small x to all terms, then simplify the resulting fraction by converting the denominator into a negative power expansion.
<p>Given expression can be written as<br />
<span class="math-tex">\(=\frac{(1+x)^{\frac{1}{2}}+(1-x)^{\frac{2}{3}}}{1+x+(1+x)^{\frac{1}{2}}}\)</span><br />
<span class="math-tex">\(=\frac{\left[1+\frac{1}{2} x-\frac{1}{8} x^{2}+\ldots\right]+\left[1-\frac{2}{3} x-\frac{1}{9} x^{2}-\ldots\right]}{1+x+\left[1+\frac{1}{2} x-\frac{1}{8} x^{2}+\ldots\right]}\)</span><br />
<span class="math-tex">\(=\frac{2-\frac{1}{6} x}{2+\frac{3}{2} x}\)</span>, neglecting other terms<br />
<span class="math-tex">\(=\frac{1-\frac{1}{12} x}{1+\frac{3}{4} x}=\left(1-\frac{1}{12} x\right)\left(1+\frac{3}{4} x\right)^{-1}\)</span><br />
= <span class="math-tex">\(\left(1-\frac{1}{12} x\right)\left(1-\frac{3}{4} x\right)\)</span>, neglecting terms of higher degree<br />
= 1 - <span class="math-tex">\(\frac{3}{4}\)</span>x - <span class="math-tex">\(\frac{1}{12}\)</span>x, neglecting terms of higher degree<br />
= 1 - <span class="math-tex">\(\frac{5}{6}\)</span>x</p>
Correct Answer: C