Binomial Theorem 1 Question 11
12. The positive value of $\lambda$ for which the coefficient of $x^{2}$ in the expression $x^{2} \sqrt{x}+\frac{\lambda}{x^{2}}$ is 720 , is
(a) 3
(b) $\sqrt{5}$
(c) $2 \sqrt{2}$
(d) 4
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Answer:
Correct Answer: 12. (a)
Solution:
- The general term in the expansion of binomial expression $(a+b)^{n}$ is $T _{r+1}={ }^{n} C _r a^{n \rightarrow r} b^{r}$, so the general term in the expansion of binomial expression $x^{2} \sqrt{x}+\frac{\lambda}{x^{2}}$ is
$$ \begin{aligned} T _{r+1} & =x^{2} \quad{ }^{10} C _r(\sqrt{x})^{10 \rightarrow r} \frac{\lambda}{x^{2}}{ }^{r}{ }^{10} C _r x^{2} \cdot x^{\frac{10 \rightarrow r}{2}} \lambda^{r} x^{-2 r} \\ & { }^{10} C _r \lambda^{r} x^{2+\frac{10 \rightarrow}{2}-2 r} \end{aligned} $$
Now, for the coefficient of $x^{2}$, put $2+\frac{10-r}{2}-2 r=2$
$$ \begin{array}{rlrl} \Rightarrow & & \frac{10-r}{2}-2 r & =0 \\ \Rightarrow & 10-r & =4 r \Rightarrow r=2 \end{array} $$
So, the coefficient of $x^{2}$ is ${ }^{10} C _2 \lambda^{2}=720$ [given]
$$ \begin{array}{lc} \Rightarrow & \frac{10 !}{2 ! 8 !} \lambda^{2}=720 \Rightarrow \frac{10 \cdot 9 \cdot 8 !}{2 \cdot 8 !} \lambda^{2}=720 \\ \Rightarrow & 45 \lambda^{2}=720 \\ \Rightarrow & \lambda^{2}=16 \Rightarrow \lambda= \pm 4 \\ \therefore & \lambda=4 \end{array} $$
