Question

If $6^{\text{th }}$ term in the expansion of ${[\frac{1}{x^{8/3}}+x^2{\log }_{10}x]}^8$ is $5600,$ then $x$ is equal to

• (A) 5
• (B) 4
• (C) 8
• (D) 10

Answer 1

The Correct Option is D

Explanation:
Using binomial theorem, the $6^{th}$ term of $(a+b{)}^8$ is $56a^3{b}^5$Now substituting the given terms, we get,$[\frac{1}{x^{\frac{8}{3}}}+x^2{\log }_{10}x]$General term, $\Rightarrow T_{r+1}={}^8{C}_r{\left[\frac{1}{x^{\frac{8}{3}}}\right]}^{n-r}{(x^2{\log }_{10}x)}^r$For $6^{th}$ term, take $r=5$$T_6={}^8{C}_5{\left[\frac{1}{x^{\frac{8}{3}}}\right]}^3{[x^2{\log }_{10}x]}^5$$5600=\frac{8\times 7\times 6}{3\times 2\times 1}\times \frac{1}{x^8}\times x^{10}{({\log }_{10}x)}^5[\because {}^8{C}_3=\frac{8\times 7\times 6}{3\times 2\times 1}]$${10}^2=x^2{({\log }_{10}x)}^5[\because ({\log }_{10}x=1)]$$x^2={10}^2$$x=10$Hence, the correct option is (D).