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 $720$:
$\begin{aligned} & \text{A}\text{. }\sqrt{5} \\\ & \text{B}\text{. 4} \\\ & \text{C}\text{. 2}\sqrt{2} \\\ & \text{D}\text{. 3} \\\ \end{aligned}$

Answer 1

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
$\begin{aligned} & x=\sqrt{x} \\\ & a=\dfrac{\lambda }{{{x}^{2}}} \\\ & n=10 \\\ \end{aligned}$
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
${{\left( \sqrt{x}+\dfrac{\lambda }{{{x}^{2}}} \right)}^{10}}=\sum\limits_{r=0}^{10}{\left( {}^{10}{{C}_{r}} \right){{\left( x \right)}^{\left( \dfrac{10-r}{2}-2r \right)}}{{\lambda }^{r}}}$
$\begin{aligned} & {{\left( \sqrt{x}+\dfrac{\lambda }{{{x}^{2}}} \right)}^{10}}=\sum\limits_{r=0}^{10}{\left( {}^{10}{{C}_{r}} \right){{\left( x \right)}^{\left( \dfrac{10-r-4r}{2} \right)}}{{\lambda }^{r}}} \\\ & {{\left( \sqrt{x}+\dfrac{\lambda }{{{x}^{2}}} \right)}^{10}}=\sum\limits_{r=0}^{10}{\left( {}^{10}{{C}_{r}} \right){{\left( x \right)}^{\left( \dfrac{10-5r}{2} \right)}}{{\lambda }^{r}}} \\\ \end{aligned}$
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 $\dfrac{10-5r}{2}=0$
$\begin{aligned} & \Rightarrow \left( \dfrac{10-5r}{2} \right)=0 \\\ & \Rightarrow 10-5r=0 \\\ & \Rightarrow 5r=10 \\\ & \Rightarrow r=\dfrac{10}{5} \\\ & r=2 \\\ \end{aligned}$
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

& \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{aligned}$$ 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.