Let `f` be a non-negative function defined on the interval `[0,1]`. If `int_0^xsqrt(1-(f^(prime)(t))^2)dt=int_0^xf(t)dt ,0lt=xlt=1,a n d \ f(0)=0`, then

Options

A. `f((1)/(2))lt(1)/(2)andf((1)/(3))gt(1)/(3)`
B. `f((1)/(2))gt(1)/(2)andf((1)/(3))gt(1)/(3)`
C. `f((1)/(2))lt(1)/(2)andf((1)/(3))lt(1)/(3)`
D. `f((1)/(2))gt(1)/(2)andf((1)/(3))lt(1)/(3)`

Answer

C

Explanation

We have,
`int_(0)^(x)sqrt(1-(f'(t)}^(2))dt=int_(0)^(x)f(t)dt`
Differentiating w.r.to x, we get
`sqrt(1-{f'(x)}^(2))=f(x)`
`rArr" "{f'(x)}=1-{f(x)}^(2)`
`rArr" "f'(x)=pmsqrt(1-{f(x)}^(2))`
`rArr" "(f'(x))/(sqrt(1-{f(x)}^(2)))=pm1`
`rArr" "int(f'(X))/(sqrt(1-{f(x)}^(2)))dx=pmint1.dxrArr sin^(-1){f(x)}= pm x+C`
It is given that f(0) = 0. Therefore, C = 0.
`therefore" "sin^(-1){f(x)}=pm x`
But, f(x) gt 0 all `x in [0,1]`
`therefore" "sin^(-1){f(x)}=x rArrf(x)=sin x`
We know that
`sinx lt x " for all "x gt0`
`rArr" "f(x) lt x " for all "x gt 0`
`rArr" "f((1)/(2))lt(1)/(2) and f((1)/(3))lt (1)/(3)`