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 $\quad$ (2019 Main, 11 Jan I)
(a) 4
(b) 0
(c) 6
(d) 8
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Answer:
Correct Answer: 11. (c)
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]
$$ \begin{array}{r} \therefore 5670={ }^{8} C _4 \frac{x^{3}}{3} \quad \frac{3}{x}=\frac{8 \cdot 7 \cdot 6 \cdot 5}{1 \cdot 2 \cdot 3 \cdot 4} x^{8} \\ \because T _{r+1}={ }^{8} C _r \quad \frac{x^{3}}{3} \quad \frac{3}{x}{ }^{r} \end{array} $$
$\Rightarrow \quad x^{8}=3^{4} \Rightarrow x= \pm \sqrt{3}$
So, sum of all values of $x$ i.e $+\sqrt{3}$ and $-\sqrt{3}=0$
