find the coefficient of the term of \[{{x}^{-5}}\]in the binomial expansion of \[{{\left( \dfrac{x+1}{{{x}^{\dfrac{2}{3}}}-{{x}^{\dfrac{1}{3}}}+1}-\dfrac{x-1}{x-{{x}^{\dfrac{1}{2}}}} \right)}^{10}}\] , where \[x\ne 0,1\].
(a) 1
(b) 4
(c) \[-4\]
(d) \[-1\]
Answer
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Hint: In this question, we have to first find the binomial expansion of \[{{\left( \dfrac{x+1}{{{x}^{\dfrac{2}{3}}}-{{x}^{\dfrac{1}{3}}}+1}-\dfrac{x-1}{x-{{x}^{\dfrac{1}{2}}}} \right)}^{10}}\].
Using the formula of binomial expansion of elements say \[a\] and \[b\] raised to the power \[n\] which is given by \[\begin{align}
& {{\left( a-b \right)}^{n}}{{=}^{n}}{{C}_{0}}{{\left( a \right)}^{n}}{{\left( b \right)}^{0}}{{\left( -1 \right)}^{0}}{{+}^{n}}{{C}_{1}}{{\left( a \right)}^{n+1}}{{\left( b \right)}^{1}}{{\left( -1 \right)}^{1}}+...{{+}^{n}}{{C}_{r}}{{\left( a \right)}^{n-r}}{{\left( b \right)}^{r}}{{\left( -1 \right)}^{r}}+... \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{+}^{n}}{{C}_{n-1}}{{\left( a \right)}^{1}}{{\left( b \right)}^{n-1}}{{\left( -1 \right)}^{n-1}}{{+}^{n}}{{C}_{n}}{{\left( a \right)}^{0}}{{\left( b \right)}^{n}}{{\left( -1 \right)}^{n}} \\
\end{align}\]
Where we have \[^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\]. Also since the number of terms in the binomial expansion of \[{{\left( a+b \right)}^{n}}\] is equal to \[n+1\]. Using this we will have that the number of terms in the binomial expansion of \[{{\left( \dfrac{x+1}{{{x}^{\dfrac{2}{3}}}-{{x}^{\dfrac{1}{3}}}+1}-\dfrac{x-1}{x-{{x}^{\dfrac{1}{2}}}} \right)}^{10}}\] is equals to 11. After finding the binomial expansion of \[{{\left( \dfrac{x+1}{{{x}^{\dfrac{2}{3}}}-{{x}^{\dfrac{1}{3}}}+1}-\dfrac{x-1}{x-{{x}^{\dfrac{1}{2}}}} \right)}^{10}}\]we will have to determine the coefficient of the term of \[{{x}^{-5}}\]in the binomial expansion.
Complete step by step answer:
Let us first determine the binomial expansion of \[{{\left( \dfrac{x+1}{{{x}^{\dfrac{2}{3}}}-{{x}^{\dfrac{1}{3}}}+1}-\dfrac{x-1}{x-{{x}^{\dfrac{1}{2}}}} \right)}^{10}}\].
Since we know that the binomial expansion of \[{{\left( a-b \right)}^{n}}\] raised to the power \[n\] which is given by \[\begin{align}
& {{\left( a-b \right)}^{n}}{{=}^{n}}{{C}_{0}}{{\left( a \right)}^{n}}{{\left( b \right)}^{0}}{{\left( -1 \right)}^{0}}{{+}^{n}}{{C}_{1}}{{\left( a \right)}^{n+1}}{{\left( b \right)}^{1}}{{\left( -1 \right)}^{1}}+...{{+}^{n}}{{C}_{r}}{{\left( a \right)}^{n-r}}{{\left( b \right)}^{r}}{{\left( -1 \right)}^{r}}+... \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{+}^{n}}{{C}_{n-1}}{{\left( a \right)}^{1}}{{\left( b \right)}^{n-1}}{{\left( -1 \right)}^{n-1}}{{+}^{n}}{{C}_{n}}{{\left( a \right)}^{0}}{{\left( b \right)}^{n}}{{\left( -1 \right)}^{n}} \\
\end{align}\]Where we have \[^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\].
On comparing the expression \[{{\left( \dfrac{x+1}{{{x}^{\dfrac{2}{3}}}-{{x}^{\dfrac{1}{3}}}+1}-\dfrac{x-1}{x-{{x}^{\dfrac{1}{2}}}} \right)}^{10}}\] with \[{{\left( a-b \right)}^{n}}\], we get that
\[a=\dfrac{x+1}{{{x}^{\dfrac{2}{3}}}-{{x}^{\dfrac{1}{3}}}+1}\], \[b=\dfrac{x-1}{x-{{x}^{\dfrac{1}{2}}}}\] and \[n=10\].
We will now simplify the value of \[a=\dfrac{x+1}{{{x}^{\dfrac{2}{3}}}-{{x}^{\dfrac{1}{3}}}+1}\] using the identity that \[{{x}^{3}}+{{y}^{3}}=\left( x+1 \right)\left( {{x}^{2}}-xy+{{y}^{2}} \right)\].
Then we have
\[\begin{align}
& a=\dfrac{x+1}{{{x}^{\dfrac{2}{3}}}-{{x}^{\dfrac{1}{3}}}+1} \\
& =\dfrac{{{\left( {{x}^{\dfrac{1}{3}}} \right)}^{3}}+1}{{{x}^{\dfrac{2}{3}}}-{{x}^{\dfrac{1}{3}}}+1} \\
& =\dfrac{\left( {{x}^{\dfrac{1}{3}}}+1 \right)\left( {{x}^{\dfrac{2}{3}}}-{{x}^{\dfrac{1}{3}}}+1 \right)}{{{x}^{\dfrac{2}{3}}}-{{x}^{\dfrac{1}{3}}}+1} \\
& ={{x}^{\dfrac{1}{3}}}+1
\end{align}\]
That is we have \[a={{x}^{\dfrac{1}{3}}}+1\].
We will now simplify the value of \[b=\dfrac{x-1}{x-{{x}^{\dfrac{1}{2}}}}\] using the identity that \[{{x}^{2}}-{{y}^{2}}=\left( x+y \right)\left( x-y \right)\].
Then we have
\[\begin{align}
& b=\dfrac{x-1}{x-{{x}^{\dfrac{1}{2}}}} \\
& =\dfrac{{{\left( {{x}^{2}} \right)}^{\dfrac{1}{2}}}-1}{x-{{x}^{\dfrac{1}{2}}}} \\
& =\dfrac{\left( {{x}^{\dfrac{1}{2}}}+1 \right)\left( {{x}^{\dfrac{1}{2}}}-1 \right)}{{{x}^{\dfrac{1}{2}}}\left( {{x}^{\dfrac{1}{2}}}-1 \right)} \\
& =\dfrac{{{x}^{\dfrac{1}{2}}}+1}{{{x}^{\dfrac{1}{2}}}} \\
& =1+{{x}^{-\dfrac{1}{2}}}
\end{align}\]
That is we have \[b=1+{{x}^{-\dfrac{1}{2}}}\].
Therefore the value of \[a-b\] is given by
\[\begin{align}
& a-b={{x}^{\dfrac{1}{3}}}+1-\left( 1+{{x}^{-\dfrac{1}{2}}} \right) \\
& ={{x}^{\dfrac{1}{3}}}+1-1-{{x}^{-\dfrac{1}{2}}} \\
& ={{x}^{\dfrac{1}{3}}}-{{x}^{-\dfrac{1}{2}}}
\end{align}\]
Therefore we have simplified the binomial expansion of \[{{\left( \dfrac{x+1}{{{x}^{\dfrac{2}{3}}}-{{x}^{\dfrac{1}{3}}}+1}-\dfrac{x-1}{x-{{x}^{\dfrac{1}{2}}}} \right)}^{10}}\] into
\[{{\left( {{x}^{\dfrac{1}{3}}}-{{x}^{-\dfrac{1}{2}}} \right)}^{10}}\]
We now have to expand the binomial expansion of \[{{\left( {{x}^{\dfrac{1}{3}}}-{{x}^{-\dfrac{1}{2}}} \right)}^{10}}\].
Now using \[\begin{align}
& {{\left( a-b \right)}^{n}}{{=}^{n}}{{C}_{0}}{{\left( a \right)}^{n}}{{\left( b \right)}^{0}}{{\left( -1 \right)}^{0}}{{+}^{n}}{{C}_{1}}{{\left( a \right)}^{n+1}}{{\left( b \right)}^{1}}{{\left( -1 \right)}^{1}}+...{{+}^{n}}{{C}_{r}}{{\left( a \right)}^{n-r}}{{\left( b \right)}^{r}}{{\left( -1 \right)}^{r}}+... \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{+}^{n}}{{C}_{n-1}}{{\left( a \right)}^{1}}{{\left( b \right)}^{n-1}}{{\left( -1 \right)}^{n-1}}{{+}^{n}}{{C}_{n}}{{\left( a \right)}^{0}}{{\left( b \right)}^{n}}{{\left( -1 \right)}^{n}} \\
\end{align}\]Where we have \[^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\].
We will have
\[\begin{align}
& {{\left( {{x}^{\dfrac{1}{3}}}-{{x}^{-\dfrac{1}{2}}} \right)}^{10}}{{=}^{10}}{{C}_{0}}{{\left( {{x}^{\dfrac{1}{3}}} \right)}^{10}}{{\left( {{x}^{-\dfrac{1}{2}}} \right)}^{0}}{{\left( -1 \right)}^{0}}{{+}^{10}}{{C}_{1}}{{\left( {{x}^{\dfrac{1}{3}}} \right)}^{10-1}}{{\left( {{x}^{-\dfrac{1}{2}}} \right)}^{1}}{{\left( -1 \right)}^{1}}+...\,+ \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{\,}^{10}}{{C}_{6}}{{\left( {{x}^{\dfrac{1}{3}}} \right)}^{10-6}}{{\left( {{x}^{-\dfrac{1}{2}}} \right)}^{6}}{{\left( -1 \right)}^{6}}+...{{+}^{10}}{{C}_{9}}{{\left( {{x}^{\dfrac{1}{3}}} \right)}^{1}}{{\left( {{x}^{-\dfrac{1}{2}}} \right)}^{9}}{{\left( -1 \right)}^{9}}{{+}^{10}}{{C}_{10}}{{\left( {{x}^{\dfrac{1}{3}}} \right)}^{0}}{{\left( {{x}^{-\dfrac{1}{2}}} \right)}^{10}}{{\left( -1 \right)}^{10}} \\
\end{align}\]
Now since we know that the number of terms in the binomial expansion of \[{{\left( a-b \right)}^{n}}\] is equals to \[n+1\].
Using this we will have that the number of terms in the binomial expansion of \[{{\left( {{x}^{\dfrac{1}{3}}}-{{x}^{-\dfrac{1}{2}}} \right)}^{10}}\] is equals to
\[10+1=11\]
Also since by seeing the above binomial expansion, we have that the general term is given by
\[^{10}{{C}_{r}}{{\left( {{x}^{\dfrac{1}{3}}} \right)}^{r}}{{\left( {{x}^{-\dfrac{1}{2}}} \right)}^{10-r}}{{\left( -1 \right)}^{10-r}}{{=}^{10}}{{C}_{r}}{{\left( x \right)}^{\dfrac{1}{3}\left( r \right)-\dfrac{1}{2}\left( 10-r \right)}}{{\left( -1 \right)}^{10-r}}..............(1)\]
Therefore in order to find coefficient of the term of \[{{x}^{-5}}\]in the binomial expansion, we must have
\[\dfrac{1}{3}\left( r \right)-\dfrac{1}{2}\left( 10-r \right)=-5\]
Solving the above equation, we get
\[\begin{align}
& \dfrac{r}{3}+\dfrac{r}{2}-\dfrac{10}{2}=-5 \\
& \Rightarrow \dfrac{r}{3}+\dfrac{r}{2}-5=-5 \\
& \Rightarrow \dfrac{r}{3}+\dfrac{r}{2}=0 \\
& \Rightarrow \dfrac{2r+3r}{6}=0 \\
& \Rightarrow \dfrac{5r}{6}=0 \\
& \Rightarrow r=0
\end{align}\]
Using substituting the value of \[r=0\] in equation (1), we get
\[^{10}{{C}_{r}}{{\left( x \right)}^{\dfrac{1}{3}\left( r \right)-\dfrac{1}{2}\left( 10-r \right)}}{{\left( -1 \right)}^{10-r}}{{=}^{10}}{{C}_{0}}{{\left( x \right)}^{\dfrac{1}{3}\left( 0 \right)-\dfrac{1}{2}\left( 10-0 \right)}}{{\left( -1 \right)}^{10-0}}\]
Now since
\[\begin{align}
& ^{10}{{C}_{0}}=\dfrac{10!}{10!} \\
& =1
\end{align}\]
Therefore we have
\[\begin{align}
& ^{10}{{C}_{r}}{{\left( x \right)}^{\dfrac{1}{3}\left( r \right)-\dfrac{1}{2}\left( 10-r \right)}}{{\left( -1 \right)}^{10-r}}{{=}^{10}}{{C}_{0}}{{\left( x \right)}^{\dfrac{1}{3}\left( 0 \right)-\dfrac{1}{2}\left( 10-0 \right)}}{{\left( -1 \right)}^{10-0}} \\
& =1\times {{x}^{-\dfrac{10}{2}}}\times 1 \\
& ={{x}^{-5}}
\end{align}\]
Therefore the coefficient of the term of \[{{x}^{-5}}\]in the binomial expansion of \[{{\left( \dfrac{x+1}{{{x}^{\dfrac{2}{3}}}-{{x}^{\dfrac{1}{3}}}+1}-\dfrac{x-1}{x-{{x}^{\dfrac{1}{2}}}} \right)}^{10}}\] is equals to 1.
So, the correct answer is “Option A”.
Note: In this problem, in order to determine the coefficient of the term of \[{{x}^{-5}}\]in the binomial expansion of \[{{\left( \dfrac{x+1}{{{x}^{\dfrac{2}{3}}}-{{x}^{\dfrac{1}{3}}}+1}-\dfrac{x-1}{x-{{x}^{\dfrac{1}{2}}}} \right)}^{10}}\] we have to find the binomial expansion of \[{{\left( \dfrac{x+1}{{{x}^{\dfrac{2}{3}}}-{{x}^{\dfrac{1}{3}}}+1}-\dfrac{x-1}{x-{{x}^{\dfrac{1}{2}}}} \right)}^{10}}\] . Then carefully take the general term to find the coefficient.
Using the formula of binomial expansion of elements say \[a\] and \[b\] raised to the power \[n\] which is given by \[\begin{align}
& {{\left( a-b \right)}^{n}}{{=}^{n}}{{C}_{0}}{{\left( a \right)}^{n}}{{\left( b \right)}^{0}}{{\left( -1 \right)}^{0}}{{+}^{n}}{{C}_{1}}{{\left( a \right)}^{n+1}}{{\left( b \right)}^{1}}{{\left( -1 \right)}^{1}}+...{{+}^{n}}{{C}_{r}}{{\left( a \right)}^{n-r}}{{\left( b \right)}^{r}}{{\left( -1 \right)}^{r}}+... \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{+}^{n}}{{C}_{n-1}}{{\left( a \right)}^{1}}{{\left( b \right)}^{n-1}}{{\left( -1 \right)}^{n-1}}{{+}^{n}}{{C}_{n}}{{\left( a \right)}^{0}}{{\left( b \right)}^{n}}{{\left( -1 \right)}^{n}} \\
\end{align}\]
Where we have \[^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\]. Also since the number of terms in the binomial expansion of \[{{\left( a+b \right)}^{n}}\] is equal to \[n+1\]. Using this we will have that the number of terms in the binomial expansion of \[{{\left( \dfrac{x+1}{{{x}^{\dfrac{2}{3}}}-{{x}^{\dfrac{1}{3}}}+1}-\dfrac{x-1}{x-{{x}^{\dfrac{1}{2}}}} \right)}^{10}}\] is equals to 11. After finding the binomial expansion of \[{{\left( \dfrac{x+1}{{{x}^{\dfrac{2}{3}}}-{{x}^{\dfrac{1}{3}}}+1}-\dfrac{x-1}{x-{{x}^{\dfrac{1}{2}}}} \right)}^{10}}\]we will have to determine the coefficient of the term of \[{{x}^{-5}}\]in the binomial expansion.
Complete step by step answer:
Let us first determine the binomial expansion of \[{{\left( \dfrac{x+1}{{{x}^{\dfrac{2}{3}}}-{{x}^{\dfrac{1}{3}}}+1}-\dfrac{x-1}{x-{{x}^{\dfrac{1}{2}}}} \right)}^{10}}\].
Since we know that the binomial expansion of \[{{\left( a-b \right)}^{n}}\] raised to the power \[n\] which is given by \[\begin{align}
& {{\left( a-b \right)}^{n}}{{=}^{n}}{{C}_{0}}{{\left( a \right)}^{n}}{{\left( b \right)}^{0}}{{\left( -1 \right)}^{0}}{{+}^{n}}{{C}_{1}}{{\left( a \right)}^{n+1}}{{\left( b \right)}^{1}}{{\left( -1 \right)}^{1}}+...{{+}^{n}}{{C}_{r}}{{\left( a \right)}^{n-r}}{{\left( b \right)}^{r}}{{\left( -1 \right)}^{r}}+... \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{+}^{n}}{{C}_{n-1}}{{\left( a \right)}^{1}}{{\left( b \right)}^{n-1}}{{\left( -1 \right)}^{n-1}}{{+}^{n}}{{C}_{n}}{{\left( a \right)}^{0}}{{\left( b \right)}^{n}}{{\left( -1 \right)}^{n}} \\
\end{align}\]Where we have \[^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\].
On comparing the expression \[{{\left( \dfrac{x+1}{{{x}^{\dfrac{2}{3}}}-{{x}^{\dfrac{1}{3}}}+1}-\dfrac{x-1}{x-{{x}^{\dfrac{1}{2}}}} \right)}^{10}}\] with \[{{\left( a-b \right)}^{n}}\], we get that
\[a=\dfrac{x+1}{{{x}^{\dfrac{2}{3}}}-{{x}^{\dfrac{1}{3}}}+1}\], \[b=\dfrac{x-1}{x-{{x}^{\dfrac{1}{2}}}}\] and \[n=10\].
We will now simplify the value of \[a=\dfrac{x+1}{{{x}^{\dfrac{2}{3}}}-{{x}^{\dfrac{1}{3}}}+1}\] using the identity that \[{{x}^{3}}+{{y}^{3}}=\left( x+1 \right)\left( {{x}^{2}}-xy+{{y}^{2}} \right)\].
Then we have
\[\begin{align}
& a=\dfrac{x+1}{{{x}^{\dfrac{2}{3}}}-{{x}^{\dfrac{1}{3}}}+1} \\
& =\dfrac{{{\left( {{x}^{\dfrac{1}{3}}} \right)}^{3}}+1}{{{x}^{\dfrac{2}{3}}}-{{x}^{\dfrac{1}{3}}}+1} \\
& =\dfrac{\left( {{x}^{\dfrac{1}{3}}}+1 \right)\left( {{x}^{\dfrac{2}{3}}}-{{x}^{\dfrac{1}{3}}}+1 \right)}{{{x}^{\dfrac{2}{3}}}-{{x}^{\dfrac{1}{3}}}+1} \\
& ={{x}^{\dfrac{1}{3}}}+1
\end{align}\]
That is we have \[a={{x}^{\dfrac{1}{3}}}+1\].
We will now simplify the value of \[b=\dfrac{x-1}{x-{{x}^{\dfrac{1}{2}}}}\] using the identity that \[{{x}^{2}}-{{y}^{2}}=\left( x+y \right)\left( x-y \right)\].
Then we have
\[\begin{align}
& b=\dfrac{x-1}{x-{{x}^{\dfrac{1}{2}}}} \\
& =\dfrac{{{\left( {{x}^{2}} \right)}^{\dfrac{1}{2}}}-1}{x-{{x}^{\dfrac{1}{2}}}} \\
& =\dfrac{\left( {{x}^{\dfrac{1}{2}}}+1 \right)\left( {{x}^{\dfrac{1}{2}}}-1 \right)}{{{x}^{\dfrac{1}{2}}}\left( {{x}^{\dfrac{1}{2}}}-1 \right)} \\
& =\dfrac{{{x}^{\dfrac{1}{2}}}+1}{{{x}^{\dfrac{1}{2}}}} \\
& =1+{{x}^{-\dfrac{1}{2}}}
\end{align}\]
That is we have \[b=1+{{x}^{-\dfrac{1}{2}}}\].
Therefore the value of \[a-b\] is given by
\[\begin{align}
& a-b={{x}^{\dfrac{1}{3}}}+1-\left( 1+{{x}^{-\dfrac{1}{2}}} \right) \\
& ={{x}^{\dfrac{1}{3}}}+1-1-{{x}^{-\dfrac{1}{2}}} \\
& ={{x}^{\dfrac{1}{3}}}-{{x}^{-\dfrac{1}{2}}}
\end{align}\]
Therefore we have simplified the binomial expansion of \[{{\left( \dfrac{x+1}{{{x}^{\dfrac{2}{3}}}-{{x}^{\dfrac{1}{3}}}+1}-\dfrac{x-1}{x-{{x}^{\dfrac{1}{2}}}} \right)}^{10}}\] into
\[{{\left( {{x}^{\dfrac{1}{3}}}-{{x}^{-\dfrac{1}{2}}} \right)}^{10}}\]
We now have to expand the binomial expansion of \[{{\left( {{x}^{\dfrac{1}{3}}}-{{x}^{-\dfrac{1}{2}}} \right)}^{10}}\].
Now using \[\begin{align}
& {{\left( a-b \right)}^{n}}{{=}^{n}}{{C}_{0}}{{\left( a \right)}^{n}}{{\left( b \right)}^{0}}{{\left( -1 \right)}^{0}}{{+}^{n}}{{C}_{1}}{{\left( a \right)}^{n+1}}{{\left( b \right)}^{1}}{{\left( -1 \right)}^{1}}+...{{+}^{n}}{{C}_{r}}{{\left( a \right)}^{n-r}}{{\left( b \right)}^{r}}{{\left( -1 \right)}^{r}}+... \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{+}^{n}}{{C}_{n-1}}{{\left( a \right)}^{1}}{{\left( b \right)}^{n-1}}{{\left( -1 \right)}^{n-1}}{{+}^{n}}{{C}_{n}}{{\left( a \right)}^{0}}{{\left( b \right)}^{n}}{{\left( -1 \right)}^{n}} \\
\end{align}\]Where we have \[^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\].
We will have
\[\begin{align}
& {{\left( {{x}^{\dfrac{1}{3}}}-{{x}^{-\dfrac{1}{2}}} \right)}^{10}}{{=}^{10}}{{C}_{0}}{{\left( {{x}^{\dfrac{1}{3}}} \right)}^{10}}{{\left( {{x}^{-\dfrac{1}{2}}} \right)}^{0}}{{\left( -1 \right)}^{0}}{{+}^{10}}{{C}_{1}}{{\left( {{x}^{\dfrac{1}{3}}} \right)}^{10-1}}{{\left( {{x}^{-\dfrac{1}{2}}} \right)}^{1}}{{\left( -1 \right)}^{1}}+...\,+ \\
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{\,}^{10}}{{C}_{6}}{{\left( {{x}^{\dfrac{1}{3}}} \right)}^{10-6}}{{\left( {{x}^{-\dfrac{1}{2}}} \right)}^{6}}{{\left( -1 \right)}^{6}}+...{{+}^{10}}{{C}_{9}}{{\left( {{x}^{\dfrac{1}{3}}} \right)}^{1}}{{\left( {{x}^{-\dfrac{1}{2}}} \right)}^{9}}{{\left( -1 \right)}^{9}}{{+}^{10}}{{C}_{10}}{{\left( {{x}^{\dfrac{1}{3}}} \right)}^{0}}{{\left( {{x}^{-\dfrac{1}{2}}} \right)}^{10}}{{\left( -1 \right)}^{10}} \\
\end{align}\]
Now since we know that the number of terms in the binomial expansion of \[{{\left( a-b \right)}^{n}}\] is equals to \[n+1\].
Using this we will have that the number of terms in the binomial expansion of \[{{\left( {{x}^{\dfrac{1}{3}}}-{{x}^{-\dfrac{1}{2}}} \right)}^{10}}\] is equals to
\[10+1=11\]
Also since by seeing the above binomial expansion, we have that the general term is given by
\[^{10}{{C}_{r}}{{\left( {{x}^{\dfrac{1}{3}}} \right)}^{r}}{{\left( {{x}^{-\dfrac{1}{2}}} \right)}^{10-r}}{{\left( -1 \right)}^{10-r}}{{=}^{10}}{{C}_{r}}{{\left( x \right)}^{\dfrac{1}{3}\left( r \right)-\dfrac{1}{2}\left( 10-r \right)}}{{\left( -1 \right)}^{10-r}}..............(1)\]
Therefore in order to find coefficient of the term of \[{{x}^{-5}}\]in the binomial expansion, we must have
\[\dfrac{1}{3}\left( r \right)-\dfrac{1}{2}\left( 10-r \right)=-5\]
Solving the above equation, we get
\[\begin{align}
& \dfrac{r}{3}+\dfrac{r}{2}-\dfrac{10}{2}=-5 \\
& \Rightarrow \dfrac{r}{3}+\dfrac{r}{2}-5=-5 \\
& \Rightarrow \dfrac{r}{3}+\dfrac{r}{2}=0 \\
& \Rightarrow \dfrac{2r+3r}{6}=0 \\
& \Rightarrow \dfrac{5r}{6}=0 \\
& \Rightarrow r=0
\end{align}\]
Using substituting the value of \[r=0\] in equation (1), we get
\[^{10}{{C}_{r}}{{\left( x \right)}^{\dfrac{1}{3}\left( r \right)-\dfrac{1}{2}\left( 10-r \right)}}{{\left( -1 \right)}^{10-r}}{{=}^{10}}{{C}_{0}}{{\left( x \right)}^{\dfrac{1}{3}\left( 0 \right)-\dfrac{1}{2}\left( 10-0 \right)}}{{\left( -1 \right)}^{10-0}}\]
Now since
\[\begin{align}
& ^{10}{{C}_{0}}=\dfrac{10!}{10!} \\
& =1
\end{align}\]
Therefore we have
\[\begin{align}
& ^{10}{{C}_{r}}{{\left( x \right)}^{\dfrac{1}{3}\left( r \right)-\dfrac{1}{2}\left( 10-r \right)}}{{\left( -1 \right)}^{10-r}}{{=}^{10}}{{C}_{0}}{{\left( x \right)}^{\dfrac{1}{3}\left( 0 \right)-\dfrac{1}{2}\left( 10-0 \right)}}{{\left( -1 \right)}^{10-0}} \\
& =1\times {{x}^{-\dfrac{10}{2}}}\times 1 \\
& ={{x}^{-5}}
\end{align}\]
Therefore the coefficient of the term of \[{{x}^{-5}}\]in the binomial expansion of \[{{\left( \dfrac{x+1}{{{x}^{\dfrac{2}{3}}}-{{x}^{\dfrac{1}{3}}}+1}-\dfrac{x-1}{x-{{x}^{\dfrac{1}{2}}}} \right)}^{10}}\] is equals to 1.
So, the correct answer is “Option A”.
Note: In this problem, in order to determine the coefficient of the term of \[{{x}^{-5}}\]in the binomial expansion of \[{{\left( \dfrac{x+1}{{{x}^{\dfrac{2}{3}}}-{{x}^{\dfrac{1}{3}}}+1}-\dfrac{x-1}{x-{{x}^{\dfrac{1}{2}}}} \right)}^{10}}\] we have to find the binomial expansion of \[{{\left( \dfrac{x+1}{{{x}^{\dfrac{2}{3}}}-{{x}^{\dfrac{1}{3}}}+1}-\dfrac{x-1}{x-{{x}^{\dfrac{1}{2}}}} \right)}^{10}}\] . Then carefully take the general term to find the coefficient.
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