This problem is a fantastic blend of two essential JEE Mathematics topics: the Binomial Theorem for identifying specific terms in an expansion, and Differential Calculus for optimizing (finding the maximum value of) a function. Let's break it down methodically.

---

1. Understanding the General Term of a Binomial Expansion

The first step in tackling any binomial expansion problem is to recall the formula for the general term. For an expansion of $(a+b)^n$, the general term, denoted as $T_{r+1}$, is given by: $T_{r+1} = {}^{n}{C_r} a^{n-r} b^r$ where:

• $n$ is the power of the binomial.
• $r$ is the index of the term (starting from $r=0$ for the first term). It must be a non-negative integer $0 \le r \le n$.
• $a$ is the first term of the binomial.
• $b$ is the second term of the binomial.

Let's identify these components from our given expression: ${\left( {t{x^{{1 \over 5}}} + {{{{(1 - x)}^{{1 \over {10}}}}} \over t}} \right)^{10}}$

By comparing this with $(a+b)^n$:

• $n = 10$
• $a = t{x^{{1 \over 5}}}$
• $b = {{{{(1 - x)}^{{1 \over {10}}}}} \over t} = t^{-1} {(1-x)^{{1 \over {10}}}}$ (We rewrite $b$ with a negative exponent for $t$ to simplify combining 't' terms later.)

Now, substitute these into the general term formula: $T_{r+1} = {}^{10}{C_r} {\left(t{x^{{1 \over 5}}}\right)^{10 - r}} {\left(t^{-1} {(1-x)^{{1 \over {10}}}}\right)^r}$

2. Finding the Term Independent of 't'

The problem asks for the "term independent of 't'". This means we need to find the term where the variable 't' is not present, which translates to its exponent being zero. To do this, we collect all terms involving 't' in our $T_{r+1}$ expression and set their combined exponent to zero.

Let's expand the powers in the expression for $T_{r+1}$: $T_{r+1} = {}^{10}{C_r} (t^{10-r}) (x^{{1 \over 5}(10-r)}) (t^{-r}) ((1-x)^{{1 \over {10}}r})$

Now, combine the terms involving 't' using the exponent rule $p^m \cdot p^n = p^{m+n}$: $T_{r+1} = {}^{10}{C_r} t^{(10-r-r)} x^{{{10-r} \over 5}} (1-x)^{{r \over {10}}}$ $T_{r+1} = {}^{10}{C_r} t^{10-2r} x^{{{10-r} \over 5}} (1-x)^{{r \over {10}}}$

For the term to be independent of 't', the exponent of 't' must be zero: $10 - 2r = 0$ Solving for $r$: $2r = 10$ $r = 5$

Since $r=5$ is a non-negative integer and falls within the range of 0 to $n=10$, it is a valid value. This means the 6th term ($T_{5+1}$) of the expansion will be independent of 't'.

Now, substitute $r=5$ back into our simplified general term expression to find this specific term: $T_6 = {}^{10}{C_5} x^{{{10-5} \over 5}} (1-x)^{{5 \over {10}}}$ $T_6 = {}^{10}{C_5} x^{{5 \over 5}} (1-x)^{{1 \over 2}}$ $T_6 = {}^{10}{C_5} x^1 (1-x)^{1/2}$ $T_6 = {}^{10}{C_5} x\sqrt{1-x}$

This is the term independent of 't', and it is now a function of 'x'. Let's denote this function as $f(x)$.

Tip: Always double-check that the value of $r$ you find is an integer and within the valid range ($0 \le r \le n$). If not, there might be no term independent of the variable, or you might have made a calculation error.

3. Maximizing the Term with Respect to 'x'

The problem asks for the maximum value of this term, given that $x \in (0, 1)$. This is an optimization problem that we solve using differential calculus. To find the maximum value of a function $f(x)$, we follow these steps:

1. Find the first derivative of $f(x)$, i.e., $f'(x)$.
2. Set $f'(x) = 0$ to find the critical points.
3. Solve for $x$.
4. Verify that this critical point corresponds to a maximum (often, in JEE problems, a single critical point within the domain is the required extremum).

Let $f(x) = {}^{10}{C_5} x\sqrt{1-x}$. Since ${}^{10}{C_5}$ is a positive constant, maximizing $f(x)$ is equivalent to maximizing the part $g(x) = x\sqrt{1-x}$.

We will differentiate $g(x)$ using the product rule: $(uv)' = u'v + uv'$. Let $u = x \implies u' = {d \over {dx}}(x) = 1$ Let $v = \sqrt{1-x} = (1-x)^{1/2} \implies v' = {d \over {dx}}((1-x)^{1/2})$ Using the chain rule for $v'$, we get: $v' = {1 \over 2}(1-x)^{{1 \over 2} - 1} \cdot {d \over {dx}}(1-x)$ $v' = {1 \over 2}(1-x)^{-{1 \over 2}} \cdot (-1)$ $v' = -{1 \over {2\sqrt{1-x}}}$

Now, apply the product rule to find $g'(x)$: $g'(x) = (1)\sqrt{1-x} + (x)\left(-{1 \over {2\sqrt{1-x}}}\right)$ $g'(x) = \sqrt{1-x} - {x \over {2\sqrt{1-x}}}$

To find the critical points, set $g'(x) = 0$: $\sqrt{1-x} - {x \over {2\sqrt{1-x}}} = 0$ Add the second term to both sides: $\sqrt{1-x} = {x \over {2\sqrt{1-x}}}$ To eliminate the square root from the denominator, multiply both sides by $2\sqrt{1-x}$: $2(\sqrt{1-x})(\sqrt{1-x}) = x$ $2(1-x) = x$ Distribute on the left side: $2 - 2x = x$ Add $2x$ to both sides: $2 = 3x$ $x = {2 \over 3}$

This critical value $x = 2/3$ lies within the given domain $(0, 1)$. This value of $x$ will yield the maximum value of $f(x)$. (A quick check using the second derivative or analyzing the sign of $g'(x)$ around $x=2/3$ would confirm this is a maximum).

Alternative for Maximization: Instead of differentiating $g(x) = x\sqrt{1-x}$, we can maximize $g(x)^2 = x^2(1-x)$ for $x \in (0,1)$, since $g(x)$ is always positive. Let $h(x) = x^2 - x^3$. $h'(x) = 2x - 3x^2$ Set $h'(x) = 0$: $x(2 - 3x) = 0$ This gives $x=0$ or $x=2/3$. Since $x \in (0,1)$, we choose $x=2/3$. Substitute $x=2/3$ into $h(x)$: $h(2/3) = \left({2 \over 3}\right)^2 \left(1 - {2 \over 3}\right) = {4 \over 9} \cdot {1 \over 3} = {4 \over {27}}$ So, the maximum value of $g(x)^2$ is $4/27$. Therefore, the maximum value of $g(x) = \sqrt{{4 \over {27}}} = {2 \over {\sqrt{27}}} = {2 \over {3\sqrt 3 }}$. This confirms our previous result.

4. Calculating the Maximum Value

Finally, substitute the optimal value $x = 2/3$ back into the expression for $T_6$ to find its maximum value: $T_{6 \text{ max}} = {}^{10}{C_5} \left( {2 \over 3} \sqrt{1 - {2 \over 3}} \right)$ Simplify the term inside the square root: $T_{6 \text{ max}} = {}^{10}{C_5} \left( {2 \over 3} \sqrt{{1 \over 3}} \right)$ Evaluate the square root: $T_{6 \text{ max}} = {}^{10}{C_5} \left( {2 \over 3} \cdot {1 \over {\sqrt 3 }} \right)$ Combine the terms: $T_{6 \text{ max}} = {}^{10}{C_5} \left( {2 \over {3\sqrt 3 }} \right)$

Now, let's calculate the binomial coefficient ${}^{10}{C_5}$: ${}^{10}{C_5} = {{10!} \over {5!(10-5)!}} = {{10!} \over {5!5!}}$ ${}^{10}{C_5} = {{10 \times 9 \times 8 \times 7 \times 6} \over {5 \times 4 \times 3 \times 2 \times 1}} = 2 \times 9 \times 2 \times 7 = 252$

Substitute this back into the expression for $T_{6 \text{ max}}$: $T_{6 \text{ max}} = {{10!} \over {5!5!}} \cdot {2 \over {3\sqrt 3 }}$ $T_{6 \text{ max}} = {{2 \cdot 10!} \over {3\sqrt 3 {{(5!)}^2}}}$

This result matches option (B).

Summary and Key Takeaways:

1. Binomial Theorem Foundation: Always start by writing down the general term $(T_{r+1})$ and carefully identifying $a$, $b$, and $n$. Pay attention to signs and exponents.
2. Term Independent of a Variable: To find a term independent of a specific variable (like 't' here), collect all instances of that variable, combine their exponents, and set the total exponent to zero to solve for $r$.
3. Optimization with Calculus: Once you have the term as a function of another variable (like 'x' here), use differential calculus to find its maximum or minimum value.
   • Find the first derivative.
   • Set it to zero to find critical points.
   • (Optional but good practice) Use the second derivative test or check the sign of the first derivative around the critical point to confirm if it's a maximum or minimum.
   • Substitute the optimal value back into the function to find the maximum/minimum value.
4. Algebraic Precision: Be careful with exponent rules, square roots, and simplifying fractions, especially with factorials.
5. Alternative Maximization: For functions of the form $x^a(1-x)^b$ (or its square root), maximizing the squared term can sometimes simplify differentiation. The maximum for $x^a(1-x)^b$ occurs at $x = \frac{a}{a+b}$. Here, for $x^1(1-x)^{1/2}$, this formula gives $x = \frac{1}{1+1/2} = \frac{1}{3/2} = \frac{2}{3}$, which confirms our critical point.