`("lim")_(xvecoo)((n^2)/(n^2))^(n(n-1)i se q u a lto` e (b) `e^2` (c) `e^(-1)` (d) 1

Options

A. `f(1/2)gef(1)`
B. `f(1/3)lef(2/3)`
C. `f'(2)le0`
D. `(f'(3))/(f(3))ge(f'(2))/(f'(2))`

Answer

B, C

Explanation

`f(x)=lim_(n to oo) {(n^(2n)(x/n+1)……….(x/n+1/n))/(n!n^(2n)((x^(2))/(n^(2))+1)((x^(2))/(n^(2))+1/(2^(2)))…………((x^(2))/(n^(2))+1/(n^(2))))}^(x)`
`implies In f(x)=lim_(n to oo) x/n{sum_(r=1)^(n)In(x/n+1/r)-sum_(r=1)^(n)In((rx^(2))/(n^(2))+1/r)}`
`=lim_(n to oo) x/n{sum(r=1)^(n)In (1+(rx)/n)-sum_(r=1)^(n)In (1+(r^(2)x^(2))/(n^(2)))}`
`=x int_(0)^(1)In(In (1+xy)dy-x int_(0)^(1)In(1+x^(2)y^(2))dy)`
Let `xy=1`
`impliesIn(f(x))=int_(0)^(x)In(1+t)dt-int_(0)^(x)In(1+t^(2))dt`
`implies(f'(x))/(f(x))=In((1+x)/(1+x^(2)))`
`implies (f'(2))/(f(2))=In 3/5 (lt0)`
`impliesf'(2)lt0`
`(f'(3))/(f(3))=In(4/10)=In (2/5)`
`implies(f'(2))/(f(2))ge(f'(3))/(f(3))`
Now `(f'(x))/(f(x))gt0` in `(0,1)` and `(f'(x))/(f(x))lt0` in `(1,oo) `
`f(x)` in increasing in `(0,1)` and decreasing in `[1,oo)` (as `f(x)` is positive).
Hence `f(1)ge(1/2)` and `f(1/3)lef(2/3)`