Question

The term independent of x in the product $\left( 4+x+7{{x}^{2}} \right){{\left( x-\dfrac{3}{x} \right)}^{11}}$ is
(a) $7{{\cdot }^{11}}{{C}_{6}}$
(b) ${{3}^{6}}{{\cdot }^{11}}{{C}_{6}}$
(c) ${{3}^{5}}{{\cdot }^{11}}{{C}_{5}}$
(d) $-12\cdot {{2}^{11}}$

Answer 1

Hint: To find the term independent of x in the product $\left( 4+x+7{{x}^{2}} \right){{\left( x-\dfrac{3}{x} \right)}^{11}}$ , we will first apply distributive property in the given product. Then we will use binomial theorem to expand ${{\left( x-\dfrac{3}{x} \right)}^{11}}$ which will yield $4\left( ^{11}{{C}_{r}}{{x}^{11-r}}{{\left( -\dfrac{3}{x} \right)}^{r}} \right)+x\left( ^{11}{{C}_{r}}{{x}^{11-r}}{{\left( -\dfrac{3}{x} \right)}^{r}} \right)+7{{x}^{2}}\left( ^{11}{{C}_{r}}{{x}^{11-r}}{{\left( -\dfrac{3}{x} \right)}^{r}} \right)$ . Then, we will check each term by equating the powers of x to 0. From this, we will get the value of r. If r is an integer, the term that we considered will be independent of x and we will simplify that by substituting the value of r in the coefficient of x. If r is not an integer, then the term will not be independent of x. Finally, we have to sum all the independent terms.

Complete step by step solution:
We have to find the term independent of x in the product $\left( 4+x+7{{x}^{2}} \right){{\left( x-\dfrac{3}{x} \right)}^{11}}$ .Let us apply distributive property in the given product.
$4{{\left( x-\dfrac{3}{x} \right)}^{11}}+x{{\left( x-\dfrac{3}{x} \right)}^{11}}+7{{x}^{2}}{{\left( x-\dfrac{3}{x} \right)}^{11}}...\left( i \right)$
Let us expand the binomial term ${{\left( x-\dfrac{3}{x} \right)}^{11}}$ . We know that the binomial term ${{\left( x+a \right)}^{n}}$ can be expanded using binomial theorem as
\[{{T}_{r+1}}{{=}^{n}}{{C}_{r}}{{x}^{n-r}}{{a}^{r}}\] , where r is an integer.
Let us apply the above general form of expansion in (i). We can see that $n=11$ and $a=\dfrac{-3}{x}$ .
 $\Rightarrow 4\left( ^{11}{{C}_{r}}{{x}^{11-r}}{{\left( -\dfrac{3}{x} \right)}^{r}} \right)+x\left( ^{11}{{C}_{r}}{{x}^{11-r}}{{\left( -\dfrac{3}{x} \right)}^{r}} \right)+7{{x}^{2}}\left( ^{11}{{C}_{r}}{{x}^{11-r}}{{\left( -\dfrac{3}{x} \right)}^{r}} \right)$
The term independent of x can be found as
For the term independent of x, let us evaluate each term. First, we have to consider $4\left( ^{11}{{C}_{r}}{{x}^{11-r}}{{\left( -\dfrac{3}{x} \right)}^{r}} \right)$
This term will be independent of x when the power of x in $^{11}{{C}_{r}}{{x}^{11-r}}{{\left( -\dfrac{3}{x} \right)}^{r}}$ is 0. To find the coefficient of ${{x}^{0}}$ , we have to equate the power of x in $^{11}{{C}_{r}}{{x}^{11-r}}{{\left( -\dfrac{3}{x} \right)}^{r}}$ to 0.
$\begin{align}
  & \Rightarrow 11-r-r=0 \\
 & \Rightarrow 2r=11 \\
 & \Rightarrow r=\dfrac{11}{2} \\
\end{align}$
We can see that r is not an integer ( $r\notin I$ ). Therefore, the term $4\left( ^{11}{{C}_{r}}{{x}^{11-r}}{{\left( -\dfrac{3}{x} \right)}^{r}} \right)$ will never be independent of x.
Next, we have to consider the second term $x\left( ^{11}{{C}_{r}}{{x}^{11-r}}{{\left( -\dfrac{3}{x} \right)}^{r}} \right)...\left( ii \right)$ .
Here also, we have to equate the powers of x to 0.
$\begin{align}
  & \Rightarrow 11-r-r+1=0 \\
 & \Rightarrow 2r=12 \\
 & \Rightarrow r=\dfrac{12}{2}=6 \\
\end{align}$
We can see that r is an integer. Thus the second term will be independent of x. Let us substitute $r=6$ in (ii).
$x\left( ^{11}{{C}_{6}}{{x}^{11-6}}{{\left( -\dfrac{3}{x} \right)}^{6}} \right)$
We can calculate the value of the above term at $r=6$ by taking the terms other than x.
${{\Rightarrow }^{11}}{{C}_{6}}\cdot {{\left( -3 \right)}^{6}}$
We can write the above term by taking -1 from -3 as
${{\Rightarrow }^{11}}{{C}_{6}}\cdot {{\left( -1\times 3 \right)}^{6}}$
We know that ${{a}^{m}}\times {{b}^{m}}={{\left( ab \right)}^{m}}$ . Hence, the above term becomes
$\begin{align}
  & {{\Rightarrow }^{11}}{{C}_{6}}\cdot {{\left( 3 \right)}^{6}}{{\left( -1 \right)}^{6}} \\
 & {{=}^{11}}{{C}_{6}}\cdot {{\left( 3 \right)}^{6}} \\
\end{align}$
Now, let us consider the third term $7{{x}^{2}}\left( ^{11}{{C}_{r}}{{x}^{11-r}}{{\left( -\dfrac{3}{x} \right)}^{r}} \right)$ . Let us equate the powers of x to 0.
$\begin{align}
  & \Rightarrow 2+11-r-r=0 \\
 & \Rightarrow 13-2r=0 \\
 & \Rightarrow 2r=13 \\
 & \Rightarrow r=\dfrac{13}{2} \\
\end{align}$
We can see that r is not an integer. Hence, the third term will not be independent of x.
Hence, the term independent of x in the product $\left( 4+x+7{{x}^{2}} \right){{\left( x-\dfrac{3}{x} \right)}^{11}}$ is ${{3}^{6}}{{\cdot }^{11}}{{C}_{6}}$ . Hence, the correct option is b.

Note: Students must know the binomial theorem thoroughly to do the expansion. They must know that r in the binomial expansion must be an integer always. If there are more than one independent terms, we have to add all these.