If the greatest value of the term independent of x in the expansion of (x+ax)^10 is 10!(5!…

Question

If the greatest value of the term independent of $x$ in the expansion of $\left(x\sin\alpha+\dfrac{a\cos\alpha}{x}\right)^{10}$ is $\dfrac{10!}{(5!)^2}$, then the value of $a$ is equal to

Accepted Answer

$${T_{r + 1}} = {}^{10}{C_r}{(x\sin \alpha )^{10 - r}}{\left( {{{a\cos \alpha } \over x}} \right)^r}$$

r = 0, 1, 2, ......., 10

Tr + 1 will be independent of x when 10 - 2r = 0 $$\Rightarrow$$ r = 5

$${T_6} = {}^{10}{C_5}{(x\sin \alpha )^5} \times {\left( {{{a\cos \alpha } \over x}} \right)^5}$$

$$ = {}^{10}{C_5} \times {a^5} \times {1 \over {{2^5}}}{(\sin 2\alpha )^5}$$

will be greatest when sin2$$\alpha$$ = 1

$$ \Rightarrow {}^{10}{C_5}{{{a^5}} \over {{2^5}}} = {}^{10}{C_5} \Rightarrow a = 2$$

Solutions · 1