Let
`f(x)=lim_(n to oo ) {(n^(n)(x+n)(x+(n)/(2))....(x+(n)/(2)))/(n!(x^(2)+n^(2))(x^(2)+(n^(2))/(4))...(x^(2)+(n^(2))/(n^(2))))}^(x//n)`for all `x gt0`. Then,

Options

A. `f((1)/(2))gef(1)`
B. `f((1)/(3))lef((2)/(3))`
C. `f'(2) le 0`
D. `(f'(3))/(f(3))ge(f'(2))/(f(2))`

Answer

B, C

Explanation

We have,
`f(x)=underset(n tooo)lim{(n^(n)(x+n)(x+(n)/(2))....(x+(n)/(2)))/(n!(x^(2)+n^(2))(x^(2)+(n^(2))/(4))...(x^(2)+(n^(2))/(n^(2))))}^(x//n)`
`rArr logf(x)=underset(n to oo)lim(x)/(n)"log"{underset(r=1)overset(n)prod((x+(n)/(r )))/((x^(2)+(n^(2))/(r^(2)))).(n)/(r)}`
`rArr log f(x)=underset(n to oo)lim(x)/(2)underset(r=1)overset(n)sum log{((x+(n)/(r ))/(x^(2)+(n^(2))/(r^(2)))).(1)/((r)/(n))}`
`rArr log f(x)=underset(n to oo)lim(x)/(n)underset(r=1)overset(n)sumlog{(1+(r )/(n)x)/(1+((r )/(n)x)^(2))}`
`rArr log f(x)=xunderset(n to oo)lim underset(r=1)overset(n)sum log{(1+(r)/(n)x)/(1+((r )/(n)x)^(2))}(1)/(n)`
`rArr log f(x)=x underset(0)overset(1)int log{(1+tx)/(1+(tx)^(2))}dt`
`rArr log f(x)=underset(0)overset(x)int log((1+u)/(1+u^(2)))du`, where u=tx
`rArr(f'(x))/(f(x))=log((1+x)/(1+x^(2)))" "`.......(i)
We observe that `f(x)gt 0` for all`a gt 0`.
Also, `log((1+x)/(1+x^(2)))gt0hArr(1+x)/(1+x^(2))gt1hArrxgtx^(2)hArr0ltxlt1`
`:.(f'(x))/(f(x))gt0" for "0ltxlt1" and "(f'(x))/(f(x))lt0" for "x gt1" "`...(ii)
`rArr f(x)` is increasing on (0,1) and decreasing on `(1,oo)`.
`rArr f((1)/(2))lef(1)" and "f((1)/(3))lef((2)/(3))`
So option(b) is correct and option (a) is incorrect.
From(ii),we obtain
`f'(2)lt 0`. So,option (c ) is correct.
From(i), we obtain
`(f'(3))/(f(3))-(f'(2))/(f(2))=log(4)/(10)-log(3)/(5)=log((2)/(3))lt 0`
`rArr (f'(3))/(f(3))lt(f'(2))/(f(2))`. So, option (d) is not correct.