First slide

Binomial theorem for positive integral Index

Question

$If the middle term of {(\frac{1}{x} + x \sin x)}^{10} is equal to 7 \frac{7}{8}, then the value of x is$

Moderate

A

$2 n π + \frac{π}{6}, n \in z$
B

$n π + \frac{π}{6}, n \in z$
C

$n π + {(- 1)}^{n} \frac{π}{6}, n \in z$
D

$n π + {(- 1)}^{n} \frac{π}{3}, n \in z$

Solution

$\begin{array}{l} {(\frac{1}{x} + x \sin x)}^{10} n = 10 is even \\ M. T = T_{\frac{10}{2} + 1} = T_{5 + 1} = 7 \frac{7}{8} \\ \Rightarrow 10 c_{5} {(\frac{1}{x})}^{5} (x \sin x)^{5} = \frac{63}{8} \\ \Rightarrow \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} \frac{1}{x^{5}} x^{5} \sin^{5} x = \frac{63}{8} \\ \Rightarrow 63 \times 4 \sin^{5} x = \frac{63}{8} \end{array}$
$\begin{array}{l} \Rightarrow \sin^{5} x = \frac{1}{32} \\ \Rightarrow \sin x = \frac{1}{2} \Rightarrow x = n π + (- 1)^{n} \frac{π}{6}, n \in z \end{array}$

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