Binomial Theorem 1 Question 10
11.
The sum of the real values of $x$ for which the middle term in the binomial expansion of $(\frac{x^{3}}{3}+\frac{3}{x})^{8}$ equals $5670$ is
(2019 Main, 11 Jan I)
(a) 4
(b) 0
(c) 6
(d) 8
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Answer:
Correct Answer: 11. (b)
Solution:
- In the expansion of $(\frac{x^{3}}{3}+\frac{3}{x})^{8}$, the middle term is $T_{4+1}$.
$[\because$ Here, $n=8$, which is even, therefore middle term $=(\frac{n+2}{2})$ th term]
$ \therefore 5670={ }^{8} C_{4} (\frac{x^{3}}{3} )^4(\frac{3}{x})^4=\frac{8 \cdot 7 \cdot 6 \cdot 5}{1 \cdot 2 \cdot 3 \cdot 4} x^{8} $
$\because [T_{r+1}={ }^{8} C_{r} \quad (\frac{x^{3}} {3} )^{8-r}\quad (\frac{3}{x}){ }^{r}] $
$\Rightarrow \quad x^{8}=3^{4} \Rightarrow x= \pm \sqrt{3}$
So, sum of all values of $x$ i.e $+\sqrt{3}$ and $-\sqrt{3}=0$
