Question

In the expansion of $ \left( \frac{1}{x} + x \sin x \right)^{10} $, the coefficient of the 6th term is equal to $ 7^7g $, then the principal value of $ x $ is.

• A. \( 45^\circ \)
• B. \( 60^\circ \)
• C. \( 25^\circ \)
• D. \( 30^\circ \)

Hint:

In binomial expansions, identify the general term and solve for the required values. Pay attention to the powers of the terms involved.

Answer 1

The Correct Option is D

In this expansion, we are dealing with the binomial expansion. The given expression is \( \left( \frac{1}{x} + x \sin x \right)^{10} \). The general term in the expansion of \( (a + b)^n \) is given by: \[ T_r = \binom{n}{r} a^{n-r} b^r \] where \( n = 10 \), \( a = \frac{1}{x} \), and \( b = x \sin x \). The general term becomes: \[ T_r = \binom{10}{r} \left( \frac{1}{x} \right)^{10-r} (x \sin x)^r \] Simplifying: \[ T_r = \binom{10}{r} \cdot x^{r - (10 - r)} \cdot (\sin x)^r \] \[ T_r = \binom{10}{r} \cdot x^{2r - 10} \cdot (\sin x)^r \] For the 6th term, substitute \( r = 6 \): \[ T_6 = \binom{10}{6} \cdot x^{2(6) - 10} \cdot (\sin x)^6 = \binom{10}{6} \cdot x^2 \cdot (\sin x)^6 \] Now, using the fact that the coefficient of the 6th term is equal to \( 7^7g \), we can solve for the principal value of \( x \). Solving the equation yields: \[ x = 30^\circ \] Thus, the principal value of \( x \) is \( 30^\circ \).