SATHEE: Binomial Theorem 1 Question 10

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

(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 $T_{4 + 1}$ .

$[∵$ Here, $n = 8$ , which is even, therefore middle term $= \frac{n + 2}{2}$ th term]

$\begin{array}{r} ∴ 5670 =^{8} C_{4} \frac{x^{3}}{3} \frac{3}{x} = \frac{8 \cdot 7 \cdot 6 \cdot 5}{1 \cdot 2 \cdot 3 \cdot 4} x^{8} \\ ∵ T_{r + 1} =^{8} C_{r} \frac{x^{3}}{3} \frac{3}{x}^{r} \end{array}$

$\Rightarrow x^{8} = 3^{4} \Rightarrow x = \pm \sqrt{3}$

So, sum of all values of $x$ i.e $+ \sqrt{3}$ and $- \sqrt{3} = 0$

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Binomial Theorem 1 Question 10

11. The sum of the real values of x for which the middle term in the binomial expansion of x33+3x8 equals 5670 is (2019 Main, 11 Jan I)

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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)