Question

If $4^{th}$ term in the expansion of ${(2+\frac{3x}{8})}^{10}$ has the greatest numerical value, then the range of values of $x$ is

• A. $\left|x\right|\ge 2$ or $\left|x\right|\le \frac{64}{21}$
• B. $-\frac{64}{21}<x<0$
• C. $2<x<\frac{64}{21}$
• D. $x\ge 2$ and $x\le \frac{64}{21}$

Answer 1

The correct option is D $x\ge 2$ and $x\le \frac{64}{21}$

${(2+\frac{3x}{8})}^{10}$

$4th$ term, $T_4={{}^{10}C}_3.2^7\times {\left(\frac{3x}{8}\right)}^3$

Now, $T_4$ is greatest numerical value, implies $T_4\ge T_5$ $\&$ $T_4\ge T_3$

$\therefore {{}^{10}C}_3.2^7\times {\left(\frac{3x}{8}\right)}^3\ge {{}^{10}C}_4.2^6\times {\left(\frac{3x}{8}\right)}^4$

$\Rightarrow \frac{3x}{8}\le \frac{{{}^{10}C}_3{2}^7}{{{}^{10}C}_4{2}^6}=\frac{8}{7}$

$\Rightarrow x\le \frac{64}{21}$

Also, $T_4\ge T_3$

${{}^{10}C}_3.2^7\times {\left(\frac{3x}{8}\right)}^3\ge {{}^{10}C}_2.2^8\times {\left(\frac{3x}{8}\right)}^2$

$=\left(\frac{3x}{8}\right)\frac{{{}^{10}C}_2{2}^8}{{{}^{10}C}_3{2}^7}$

$=x\ge 2$

$\therefore x\ge 2$ $\&$ $x\le \frac{64}{21}.$

Hence, the answer is $x\ge 2$ $\&$ $x\le \frac{64}{21}.$