The minmumu value of the fucntion
`f(x)=a^2/x+b^2/(a-x),a gt 0, b gt 0` , in (0,a) is

Options

A. a+b
B. `1/(a+b)`
C. `((a+b)^2)/(a)`
D. `(a+b)/(a^2)`

Answer

C

Explanation

We have
`f(x)=a^2/x+b^2/(a-x)`
`rArr f(x)=-a^2/x^2+b^2/(a-x)^2`
`therefore f(x)=0`
`rArr a^2/x^2=(b^2)/(a-x)^2`
`rArr a^2(a-x)^2=b^2 x^2`
`a(a-x)= pm bx`
`rArr x=(a^2)/(a+b) [therefore a^2/(a-b) ne (0,a)]`
given by
Now `f'' (x)=(2ab^2)/(x^3)+(2b^2)/(a-x)^2 gt 0` for all `in (0,a)`
Hence f(x) minimum at `x=a^2/(a+b)` with the minimum value `f(a^2/(a+b))=(a+b)+(b^2)/(a-a^2/(a+b))=(a+b)+(b^2(a+b))/(ab)=(a+b)^2/a`