# The maximum value of the term independent of

Question:

The maximum value of the term independent of ' $\mathrm{t}$ ' in the expansion of $\left(t x^{\frac{1}{5}}+\frac{(1-x)^{\frac{1}{10}}}{t}\right) \quad$ where $x \in(0,1)$ is:

1. (1) $\frac{10 !}{\sqrt{3}(5 !)^{2}}$

2. (2) $\frac{2.10 !}{3(5 !)^{2}}$

3. (3) $\frac{10 !}{3(5 !)^{2}}$

4. (4) $\frac{2.10 !}{3 \sqrt{3}(5 !)^{2}}$

Correct Option: , 4

Solution:

$T_{r+1}={ }^{10} C_{r}\left(t x^{1 / 5}\right)^{10-r}\left[\frac{(1-x)^{1 / 10}}{t}\right]^{r}$

$={ }^{10} C_{r} t^{(10-2 r)} \times x^{\frac{10-r}{5}} \times(1-x)^{\frac{t}{10}}$

$\Rightarrow 10-2 r=0 \Rightarrow r=5$

$T_{6}={ }^{10} C_{5} x \sqrt{1-x}$

$\frac{d T_{6}}{d x}={ }^{10} C_{5}\left[\sqrt{1-x}-\frac{x}{2 \sqrt{1-x}}\right]=0$

$=1-x=x / 2 \Rightarrow 3 x=2$

$\Rightarrow x=2 / 3$

$\left.T_{6}\right|_{\max }=\frac{10 !}{5 ! 5 !} \times \frac{2}{3 \sqrt{3}}$

Leave a comment