Question

If $f_1\left(x\right)=\frac{x}{2}+10,\forall xϵR$, and define $f_n\left(x\right)=f_1\left(f_{n-1}\right(x\left)\right),\forall n\ge 2$, and $\underset{n\to \infty }{lim}{f}_n\left(x\right)=g\left(x\right)$, and ${\int }_{\frac{1}{2}g\left(x\right)}^{g\left(x\right)-1}\left(\frac{sinx}{1+x^a}\right)<\frac{2}{g\left(x\right)-2}$, then minimum odd value of a(a > 1) is

• A. 3
• B. 5
• C. 17
• D. 19

Answer 1

The correct option is A 3

Let $\underset{n\to \infty }{lim}{f}_n\left(x\right)=g\left(x\right)$,
$\Rightarrow g\left(x\right).f_1\left(x\right)=\frac{g\left(x\right)}{\alpha }+10$
$g\left(x\right)={\int }_{10}^9\frac{sinx}{1+x^a}dx<\frac{1}{9}$
$\Rightarrow {\int }_{10}^9\frac{dx}{1+{10}^a}<\frac{1}{9}\Rightarrow {10}^a>80\Rightarrow a\ge 2$