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Step-by-Step Solution
Step 1: Write the general term of the expansion
Consider the expansion of
$ (t x^{1/5} + \frac{(1 - x)^{1/10}}{t})^{10} $.
The general term (the $(r+1)$th term) in the binomial expansion is:
$ T_{r+1} \;=\; \binom{10}{r}\,\bigl(t\,x^{1/5}\bigr)^{\,10-r}\,\bigl(\frac{(1 - x)^{1/10}}{t}\bigr)^{\,r}. $
Step 2: Simplify to identify the power of $t$
Simplifying, we get:
$ T_{r+1} \;=\; \binom{10}{r}\,t^{\,10-r}\,x^{\frac{10-r}{5}}\;\times\;t^{-r}\,(1 - x)^{\frac{r}{10}}. $
Combine the powers of $t$:
$ T_{r+1} \;=\; \binom{10}{r}\;t^{\,10 - r - r}\;x^{\frac{10-r}{5}}\;(1 - x)^{\frac{r}{10}}. $
Hence, the exponent of $t$ is $ (10 - 2r) $.
Step 3: Determine the term independent of $t$
For the term to be independent of $t$, set the exponent of $t$ to zero:
$ 10 - 2r \;=\; 0 \;\;\Rightarrow\;\; r = 5. $
Thus, the term independent of $t$ appears when $r = 5$.
Step 4: Evaluate the term for $r = 5$
Substituting $ r = 5 $ in $ T_{r+1} $:
$ T_{5+1} \;=\; \binom{10}{5}\,x^{\frac{10-5}{5}}\,(1 - x)^{\frac{5}{10}}
\;=\;\binom{10}{5}\,x^{1}\,(1 - x)^{\frac{1}{2}}. $
Therefore,
$ T_{6} \;=\;\binom{10}{5}\,x\,(1 - x)^{1/2}. $
Step 5: Find $x$ that maximizes this expression
We want to maximize
$ T_{6}(x) = \binom{10}{5}\;x\,(1 - x)^{1/2} $
for $ x \in (0,1) $. Let us denote
$ f(x) = x\,(1 - x)^{1/2}. $
The binomial coefficient $ \binom{10}{5} $ is a constant factor and does not affect the location of the maximum.
Differentiate $ f(x) $ w.r.t. $ x $ and set it to zero:
$ f'(x) = (1 - x)^{1/2} + x \cdot \frac{1}{2}(1 - x)^{-1/2}\,(-1)
= \sqrt{1 - x} \;-\; \frac{x}{2\sqrt{1 - x}}. $
Setting $ f'(x) = 0 $:
$ \sqrt{1 - x} \;-\; \frac{x}{2\sqrt{1 - x}} = 0. $
Multiply through by $ 2\,\sqrt{1 - x} $:
$ 2(1 - x) - x \;=\; 0 \;\;\Rightarrow\;\; 2 - 2x - x = 0 \;\;\Rightarrow\;\; 2 = 3x \;\;\Rightarrow\;\; x = \frac{2}{3}. $
Step 6: Substitute $ x = \tfrac{2}{3} $ back into $T_{6}$
So the maximum value of the $t$-independent term is:
$ T_{6}\bigl(\frac{2}{3}\bigr)
= \binom{10}{5}\,\left(\frac{2}{3}\right)\,\left(1 - \frac{2}{3}\right)^{1/2}
= \binom{10}{5}\,\left(\frac{2}{3}\right)\,\left(\frac{1}{3}\right)^{1/2}
= \binom{10}{5}\,\frac{2}{3}\,\frac{1}{\sqrt{3}}
= \binom{10}{5}\,\frac{2}{3\sqrt{3}}.
$
Since $ \binom{10}{5} = \frac{10!}{5!\,5!}, $
the maximum value becomes:
$
\frac{10!}{5!\,5!} \cdot \frac{2}{3\sqrt{3}}
= \frac{2\,10!}{3\,\sqrt{3}\,(5!)^{2}}.
$
Thus, the maximum value of the term independent of $t$ is
$ \displaystyle \frac{2\,10!}{3\,\sqrt{3}\,(5!)^2}\,.
