Let `f:(0,1) in (0,1)` be a differenttiable function such that `f(x)ne 0` for all `x in (0,1)` and `f((1)/(2))=(sqrt(3))/(2)`. Suppose for all x,
`lim_(x to x)(int_(0)^(1) sqrt(1(f(s))^(2))dxint_(0)^(x) sqrt(1(f(s))^(2))ds)/(f(t)-f(x))=f(x)`
Then, the value of `f((1)/(4))` belongs to

Options

A. `{(sqrt(7),sqrt(15))}`
B. `{sqrt(7)/(2),sqrt(15)/(2)}`
C. `{sqrt(7)/(3),sqrt(15)/(3)}`
D. `{sqrt(7)/(4),sqrt(15)/(4)}`

Answer

D

Explanation

Appling L' Hospital's rule on LHS, we get
`underset(t inx)lim(sqrt(1-{f(t)}^(2)))/(f'(t))=f(x)`
`rArr (sqrt(1-(f(x))^(2)))/(f'(x))=f(x)`
`rArr (f(x)f'(x))/(sqrt(1-(f(x))^(2)))=1`
`rArr (-2f(x)f'(x))/(sqrt(1-(f(x))^(2)))=-2`
`rArr (1)/(sqrt(1-(f(x))^(2)))d{1-(f(x))^(2)}=-2`
`rArr 2sqrt(1-(f(x))^(2))=-2x+C` [On integrating]`" "`......(i)
Putting `x=(1)/(2)` on both sides and using `f((1)/(2))=(sqrt(3))/(2)`, we get C=2
Putting C=2 in (i), we get
`sqrt(1-(f(x))^(2))=-x+1`
`rArrf(x)=sqrt(2x-x^(2))`
`rArr f((1)/(4))=(sqrt(7))/(4)`
Hence,`f((1)/(4))` belongs to `{(sqrt(7))/(4),(sqrt(15))/(4)}`.