1.

If f(x) is differentiable function in the interval (0,oo) such that f(1) = 1 and lim_(trarrx) (t^(2)f(x)-x^(2)(t))/(t-x)=1 for each x gt 0, then f((3)/(2)) is equal tv

Answer»

`(13)/(6)`
`(23)/(18)`
`(25)/(9)`
`(31)/(18)`

Solution :We have,
`underset(trarrx)(lim)(t^(2)f(x)-x^(2)(t))/(t-x)=1`
`rArr""underset(trarrx)(lim)(t^(2){f(x)-f(t)}-(x^(2)-t^(2))f(t))/(t-x)=1`
`rArr""underset(trarrx)(lim){(f(x)-f(t))/(t-x)}t^(2)+underset(trarrx)(lim)(x+t)f(t)=1`
`rArr""-x^(2)f'(x)+2xf(x)=1""[{:(because" f is differentiable and"),("so continuous also"):}]`
`rArr""f'(x)-(2)/(x)f(x)=(1)/(x^(2))"...(i)"`
This is a linear differential equation with I.F. `e^(-int(2)/(x)DX)=(1)/(x^(2))`
MULTIPLYING (i) by I.F.`=(1)/(x^(2))` integrating, we get
`(f(x))/(x^(2))=(1)/(3x^(3))+C`
It is given that f(1) = 1 Putting x = 1 in (ii), we get
`=(1)/(3)+CrArr C=(2)/(3)`
`therefore""(f(x))/(x^(2))=(1)/(3x^(3))+(2)/(3)`
`rArr""f(x)=(1)/(3x)+(2x^(2))/(3)rArrf((3)/(2))=(2)/(9)+(3)/(2)=(31)/(18)`


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