Binomial Theorem
Grade 11
Question:
<p>The term independent of x in the expansion of <span class="math-tex">\(\left(\sqrt{\frac{x}{3}}+\frac{3}{2 x^{2}}\right)^{10}\)</span> will be</p>
<p style="display:inline"><span class="math-tex">\(\frac{5}{2}\)</span></p>
<p style="display:inline"><span class="math-tex">\(\frac{5}{4}\)</span></p>
<p style="display:inline"><span class="math-tex">\(\frac{3}{2}\)</span></p>
<p style="display:inline"><span class="math-tex">\(\frac{7}{4}\)</span></p>
Step-by-Step Solution
Key Concept: To find the term independent of x, identify the general term formula T_{r+1} = nCr (a)^{n-r} (b)^r and solve for r by setting the net exponent of x to zero.
<p>Term independent of x in <span class="math-tex">$\left(\sqrt{\frac{x}{3}}+\frac{3}{2 x^{2}}\right)^{10}$</span> = T<sub>r+1</sub>,<br />
where r = <span class="math-tex">$\frac{10 \times \frac{1}{2}}{\frac{1}{2}+2}$</span> = 2<br />
The required term = T<sub>3</sub> = <sup>10</sup>C<sub>2</sub> <span class="math-tex">$\times\left(\frac{1}{3}\right)^{4}\left(\frac{3}{2}\right)^{2}$</span><br />
<span class="math-tex">$=\frac{10 \times 9}{2} \times \frac{1}{36}$</span><br />
= <span class="math-tex">$\frac{5}{4}$</span></p>
Correct Answer: B