Question

Find the positive value of $\lambda $ for which the coefficient of ${{x}^{2}}$ in the expression ${{x}^{2}}{{\left( \sqrt{x}+\dfrac{\lambda }{{{x}^{2}}} \right)}^{10}}$ is $720$:
$\begin{align}
  & \text{A}\text{. }\sqrt{5} \\
 & \text{B}\text{. 4} \\
 & \text{C}\text{. 2}\sqrt{2} \\
 & \text{D}\text{. 3} \\
\end{align}$

Answer 1

Hint: To solve this question we will use binomial theorem to expand the given expression. The general formula used in the expansion of binomial raised to the power $n$, \[{{\left( x+a \right)}^{n}}\] is given as \[{{\left( x+a \right)}^{n}}=\sum\limits_{r=0}^{n}{\left( {}^{n}{{C}_{r}} \right){{x}^{n-r}}{{a}^{r}}}\]. In the expansion we put the coefficient of ${{x}^{2}}$ equal to $720$ and find the value of $\lambda $.

Complete step by step answer:
We have been given the expression ${{x}^{2}}{{\left( \sqrt{x}+\dfrac{\lambda }{{{x}^{2}}} \right)}^{10}}$.
We have to find the positive value of $\lambda $.
Now, we will use the binomial theorem to expand the given expression ${{x}^{2}}{{\left( \sqrt{x}+\dfrac{\lambda }{{{x}^{2}}} \right)}^{10}}$.
We know that binomial theorem states that for any positive integer $n$, the ${{n}^{th}}$ power of the sum of two real numbers $x$ and $a$ may be expressed as
\[{{\left( x+a \right)}^{n}}=\sum\limits_{r=0}^{n}{\left( {}^{n}{{C}_{r}} \right){{x}^{n-r}}{{a}^{r}}}\]
Now, when we compare the general term with the given expression, we have
$\begin{align}
  & x=\sqrt{x} \\
 & a=\dfrac{\lambda }{{{x}^{2}}} \\
 & n=10 \\
\end{align}$
So, ${{\left( \sqrt{x}+\dfrac{\lambda }{{{x}^{2}}} \right)}^{10}}=\sum\limits_{r=0}^{10}{\left( {}^{10}{{C}_{r}} \right){{\left( \sqrt{x} \right)}^{10-r}}{{\left( \dfrac{\lambda }{{{x}^{2}}} \right)}^{r}}}$
We know that we can write $\sqrt{x}={{x}^{\dfrac{1}{2}}}$ , we get
${{\left( \sqrt{x}+\dfrac{\lambda }{{{x}^{2}}} \right)}^{10}}=\sum\limits_{r=0}^{10}{\left( {}^{10}{{C}_{r}} \right){{\left( x \right)}^{\dfrac{10-r}{2}}}\left( \dfrac{{{\lambda }^{r}}}{{{x}^{2r}}} \right)}$
Now, we know that $\dfrac{{{x}^{m}}}{{{x}^{n}}}={{x}^{m-n}}$
So, we get
${{\left( \sqrt{x}+\dfrac{\lambda }{{{x}^{2}}} \right)}^{10}}=\sum\limits_{r=0}^{10}{\left( {}^{10}{{C}_{r}} \right){{\left( x \right)}^{\left( \dfrac{10-r}{2}-2r \right)}}{{\lambda }^{r}}}$
$\begin{align}
  & {{\left( \sqrt{x}+\dfrac{\lambda }{{{x}^{2}}} \right)}^{10}}=\sum\limits_{r=0}^{10}{\left( {}^{10}{{C}_{r}} \right){{\left( x \right)}^{\left( \dfrac{10-r-4r}{2} \right)}}{{\lambda }^{r}}} \\
 & {{\left( \sqrt{x}+\dfrac{\lambda }{{{x}^{2}}} \right)}^{10}}=\sum\limits_{r=0}^{10}{\left( {}^{10}{{C}_{r}} \right){{\left( x \right)}^{\left( \dfrac{10-5r}{2} \right)}}{{\lambda }^{r}}} \\
\end{align}$
Now, for the coefficient of ${{x}^{2}}$ in the above expression the value of $\left( \dfrac{10-5r}{2} \right) $ but we have $x^2$ in multiplication so value of $\dfrac{10-5r}{2}=0 $
$\begin{align}
  & \Rightarrow \left( \dfrac{10-5r}{2} \right)=0 \\
 & \Rightarrow 10-5r=0 \\
 & \Rightarrow 5r=10 \\
 & \Rightarrow r=\dfrac{10}{5} \\
 & r=2 \\
\end{align}$
Now, we have been given that the value of coefficient of ${{x}^{2}}$ is $720$.
So, when we compare the coefficient of ${{x}^{2}}$ we have
$\left( {}^{10}{{C}_{r}} \right){{\lambda }^{r}}=720$
Put $r=2$, we get
$\left( {}^{10}{{C}_{2}} \right){{\lambda }^{2}}=720$
Now, we know that ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$
So, we have
 \[\begin{align}
  & \dfrac{10!}{2!\left( 10-2 \right)!}{{\lambda }^{2}}=720 \\
 & \dfrac{10\times 9\times 8!}{2\times 1\left( 8 \right)!}{{\lambda }^{2}}=720 \\
 & \dfrac{10\times 9}{2}{{\lambda }^{2}}=720 \\
 & \dfrac{90}{2}{{\lambda }^{2}}=720 \\
 & 45{{\lambda }^{2}}=720 \\
 & {{\lambda }^{2}}=\dfrac{720}{45} \\
 & {{\lambda }^{2}}=16 \\
 & \lambda =\pm 4 \\
\end{align}\]
As we have asked to find the positive value of $\lambda $, so we take $\lambda =4$.

So, the correct answer is “Option B”.

Note: If binomial theorem were not there, then it would be a very complex and time-consuming task to calculate the binomial terms raised to the power more than $5$. In this question we only need the coefficient of ${{x}^{2}}$ that’s why we can’t expand the whole expression with all coefficients. Full expansion of the given expression is not needed in this question.