Question

What is the maximum value of the term independent of $t$ in the expansion of ${\left[ {t{x^{\frac{1}{5}}} + \dfrac{{{{\left( {1 - x} \right)}^{\frac{1}{{10}}}}}}{t}} \right]^{10}}$, where $x \in \left( {0,1} \right)$?
A. $\dfrac{{10!}}{{\sqrt 3 {{\left( {5!} \right)}^2}}}$
B. $\dfrac{{2 \cdot 10!}}{{3{{\left( {5!} \right)}^2}}}$
C. $\dfrac{{10!}}{{3{{\left( {5!} \right)}^2}}}$
D. $\dfrac{{2 \cdot 10!}}{{3\sqrt 3 {{\left( {5!} \right)}^2}}}$

Answer 1

Hint: Find the general term of the given binomial expression. Let the general term ${t_{r + 1}}$ be the term independent of $t$. Make an equation which shows the exponent of $t$ in ${t_{r + 1}}$ is zero. Solving it find the value of $r$. Substituting the value of $r$ in ${t_{r + 1}}$, you’ll get a function of $x$. Use second derivative test to find the maximum value of the term which is independent of $t$.

Formula Used:
General term of a binomial expression ${\left( {a + b} \right)^n}$ is the $\left( {r + 1} \right)$th term ${t_{r + 1}} = {}^n{C_r}{a^{n - r}}{b^r}$
Product rule of differentiation is $\dfrac{d}{{dx}}\left\{ {f\left( x \right)g\left( x \right)} \right\} = f\left( x \right)\dfrac{d}{{dx}}\left\{ {g\left( x \right)} \right\} + g\left( x \right)\dfrac{d}{{dx}}\left\{ {f\left( x \right)} \right\}$
$\dfrac{d}{{dx}}{\left\{ {f\left( x \right)} \right\}^n} = n{\left\{ {f\left( x \right)} \right\}^{n - 1}}\dfrac{d}{{dx}}\left\{ {f\left( x \right)} \right\}$
${}^n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!r!}}$

Complete step by step solution:
A binomial expression ${\left[ {t{x^{\frac{1}{5}}} + \dfrac{{{{\left( {1 - x} \right)}^{\frac{1}{{10}}}}}}{t}} \right]^{10}}$ is given.
Find the general term i.e. the $\left( {r + 1} \right)$th term of its expansion.
Let it be ${t_{r + 1}}$
Then ${t_{r + 1}} = {}^{10}{C_r}{\left( {t{x^{\frac{1}{5}}}} \right)^{10 - r}}{\left\{ {\dfrac{{{{\left( {1 - x} \right)}^{\frac{1}{{10}}}}}}{t}} \right\}^r}$
$ = {}^{10}{C_r}{t^{10 - r}}{x^{\frac{{10 - r}}{5}}}\dfrac{{{{\left( {1 - x} \right)}^{\frac{r}{{10}}}}}}{{{t^r}}}$, applying the rule ${\left( {{a^m}} \right)^n} = {a^{mn}}$
$ = {}^{10}{C_r}{t^{10 - 2r}}{x^{\frac{{10 - r}}{5}}}{\left( {1 - x} \right)^{\frac{r}{{10}}}}$, applying the rule $\dfrac{{{a^m}}}{{{a^n}}} = {a^{m - n}}$
Let us find the term which is independent of $t$.
Let ${t_{r + 1}}$ be independent of $t$.
Then the exponent of ${t^{10 - 2r}}$ will be zero.
$10 - 2r = 0 \Rightarrow r = 5$
$\therefore $ The sixth term is independent of $t$.
Putting $r = 5$ in the expression of ${t_{r + 1}}$, we get
${t_6} = {}^{10}{C_5}{x^{\frac{{10 - 5}}{5}}}{\left( {1 - x} \right)^{\frac{5}{{10}}}}$
$ = {}^{10}{C_5}{x^1}{\left( {1 - x} \right)^{\frac{1}{2}}}$
$ = {}^{10}{C_5}x\sqrt {1 - x} $
Now, we have to find its maximum value.
Let $f\left( x \right) = {}^{10}{C_5}x\sqrt {1 - x} $
Apply the second derivative test to find its maximum value.
Differentiating with respect to $x$, we get
$f'\left( x \right) = {}^{10}{C_5}\dfrac{d}{{dx}}\left( {x\sqrt {1 - x} } \right)$
Apply the product rule $\dfrac{d}{{dx}}\left\{ {f\left( x \right)g\left( x \right)} \right\} = f\left( x \right)\dfrac{d}{{dx}}\left\{ {g\left( x \right)} \right\} + g\left( x \right)\dfrac{d}{{dx}}\left\{ {f\left( x \right)} \right\}$
Substituting $f\left( x \right) = x$ and $g\left( x \right) = \sqrt {1 - x} $, we get
$\dfrac{d}{{dx}}\left( {x\sqrt {1 - x} } \right) = x\dfrac{d}{{dx}}\left( {\sqrt {1 - x} } \right) + \sqrt {1 - x} \dfrac{d}{{dx}}\left( x \right)$
Use the formula $\dfrac{d}{{dx}}{\left\{ {f\left( x \right)} \right\}^n} = n{\left\{ {f\left( x \right)} \right\}^{n - 1}}\dfrac{d}{{dx}}\left\{ {f\left( x \right)} \right\}$
Substituting $f\left( x \right) = 1 - x$ and $n = \dfrac{1}{2}$, we get
$\dfrac{d}{{dx}}{\left( {1 - x} \right)^{\frac{1}{2}}} = \dfrac{1}{2}{\left( {1 - x} \right)^{\frac{1}{2} - 1}}\dfrac{d}{{dx}}\left( {1 - x} \right)$
$ \Rightarrow \dfrac{d}{{dx}}\left( {\sqrt {1 - x} } \right) = \dfrac{1}{2}{\left( {1 - x} \right)^{ - \dfrac{1}{2}}}\left( { - 1} \right)$
$ \Rightarrow \dfrac{d}{{dx}}\left( {\sqrt {1 - x} } \right) = - \dfrac{1}{{2\sqrt {1 - x} }}$
Substituting $f\left( x \right) = x$ and $n = 1$, we get
$\dfrac{d}{{dx}}\left( x \right) = 1$
$\therefore \dfrac{d}{{dx}}\left( {x\sqrt {1 - x} } \right) = x\left( { - \dfrac{1}{{2\sqrt {1 - x} }}} \right) + \sqrt {1 - x} \left( 1 \right)$
$ \Rightarrow \dfrac{d}{{dx}}\left( {x\sqrt {1 - x} } \right) = - \dfrac{x}{{2\sqrt {1 - x} }} + \sqrt {1 - x} $
$ \Rightarrow \dfrac{d}{{dx}}\left( {x\sqrt {1 - x} } \right) = \dfrac{{ - x + 2\left( {1 - x} \right)}}{{2\sqrt {1 - x} }} = \dfrac{{ - x + 2 - 2x}}{{2\sqrt {1 - x} }} = \dfrac{{2 - 3x}}{{2\sqrt {1 - x} }}$
$\therefore f'\left( x \right) = {}^{10}{C_5}\left( {\dfrac{{2 - 3x}}{{2\sqrt {1 - x} }}} \right)$
Solve the equation $f'\left( x \right) = 0$ to find the critical points.
$ \Rightarrow {}^{10}{C_5}\left( {\dfrac{{2 - 3x}}{{2\sqrt {1 - x} }}} \right) = 0$
$ \Rightarrow 2 - 3x = 0$
$ \Rightarrow x = \dfrac{2}{3}$
So, the critical point is $x = \dfrac{2}{3}$
Again differentiating $f'\left( x \right)$ with respect to $x$, we get
$f''\left( x \right) = {}^{10}{C_5}\dfrac{d}{{dx}}\left( {\dfrac{{2 - 3x}}{{2\sqrt {1 - x} }}} \right)$
$ = {}^{10}{C_5}\left\{ {\dfrac{{2\sqrt {1 - x} \dfrac{d}{{dx}}\left( {2 - 3x} \right) - \left( {2 - 3x} \right)\dfrac{d}{{dx}}\left( {2\sqrt {1 - x} } \right)}}{{{{\left( {2\sqrt {1 - x} } \right)}^2}}}} \right\}$
$ = {}^{10}{C_5}\left\{ {\dfrac{{2\sqrt {1 - x} \left( { - 3} \right) - \left( {2 - 3x} \right) \times 2 \times \dfrac{1}{{2\sqrt {1 - x} }} \times \left( { - 1} \right)}}{{4\left( {1 - x} \right)}}} \right\}$
$ = {}^{10}{C_5}\left\{ {\dfrac{{ - 6\left( {1 - x} \right) + \left( {2 - 3x} \right)}}{{4\left( {1 - x} \right)\sqrt {1 - x} }}} \right\}$
$ = {}^{10}{C_5}\left\{ {\dfrac{{ - 6 + 6x + 2 - 3x}}{{4{{\left( {1 - x} \right)}^{\frac{3}{2}}}}}} \right\}$
$ = {}^{10}{C_5}\left\{ {\dfrac{{ - 4 + 3x}}{{4{{\left( {1 - x} \right)}^{\frac{3}{2}}}}}} \right\}$
Putting $x = \dfrac{2}{3}$, we get
$f''\left( {\dfrac{2}{3}} \right) = {}^{10}{C_5}\left\{ {\dfrac{{ - 4 + 3\left( {\dfrac{2}{3}} \right)}}{{4{{\left( {1 - \dfrac{2}{3}} \right)}^{\frac{3}{2}}}}}} \right\} = {}^{10}{C_5}\left\{ {\dfrac{{ - 4 + 2}}{{4{{\left( {\dfrac{1}{3}} \right)}^{\frac{3}{2}}}}}} \right\} = {}^{10}{C_5}\left\{ {\dfrac{{ - 2}}{{4{{\left( {\dfrac{1}{3}} \right)}^{\frac{3}{2}}}}}} \right\} < 0$
$\therefore $ The value of $f\left( x \right)$ is maximum at $x = \dfrac{2}{3}$ and the maximum value is given by $f\left( {\dfrac{2}{3}} \right)$
Now, $f\left( {\dfrac{2}{3}} \right) = {}^{10}{C_5}\left( {\dfrac{2}{3}} \right)\sqrt {1 - \dfrac{2}{3}} = \dfrac{{10!}}{{5!5!}} \times \dfrac{2}{3} \times \sqrt {\dfrac{1}{3}} = \dfrac{{10!}}{{5!5!}} \times \dfrac{2}{{3\sqrt 3 }} = \dfrac{{2 \cdot 10!}}{{3\sqrt 3 {{\left( {5!} \right)}^2}}}$

Option ‘D’ is correct

Note: Always remember that to solve such types of problems, at first you need to find out the general term and then proceed as per the question. Here at first, the general term has been found and then found the maximum value of that term using the second order derivative test.