`f(x)>0AAx in R` and is bounded. If `lim_(n->oo)[int_0^a(f(x)dx)/(f(x)+f(a-x))+aint_a^(2a)(f(x)dx)/(f(x)+f(3a-x))``+a^2int_(2a)^(3a)(f(x)dx)/(f(x)+f(5a-x))+...+a^(n-1)int_((n-1)a)^(n a)(f(x)dx)/(f(x)+f[(2n-1)a-x]]]``=7//5` (where `a<1),` then `a` is equal to

Options

A. `2/7`
B. `1/7`
C. `14/19`
D. `9/14`

Answer

C

Explanation

`int_(0)^(a) (f(x))(f(x)+f(a-x))=int_(a)^(2a)(f(x)dx)/(f(x)+f(3a-x))=……………=a/2`
So from given sum,
`lim_(n to oo) [a/2+(a^(2))/2+(a^(3))/2+………..+(a^(n))/2]=7/5`
or `a/(1-a)=14/5` ( sumof infinite G.P.)
or `5a=14-14a`
or `a=14/9`