Let f be a non-negative function defined on the interval .[0,1].If `int_0^x sqrt[1-(f'(t))^2].dt`=`int_0^x f(t).dt`, `0<=x<=1` and f(0)=0,then

Let f be a non-negative function defined on the interval [0,1] . If int_0^xsqrt(1-(f\'(t))^2)dt=int_0^xf(t)dt, 0lexle1 and f(0)=0 , then (A) f(1/2)lt1/2 and f(1/3)gt1/3 (B) f(1/2)gt1/2 and f(1/3)gt1/3 (C) f(1/2)lt1/2 and f(1/3)lt1/3 (D) f(1/2)gt1/2 and f(1/3)lt1/3

Let f be a non-negative function defined on the interval [0,1]dot If int_0^xsqrt(1-(f^(prime)(t))^2)dt=int_0^xf(t)dt ,0lt=xlt=1,a n df(0)=0,t h e n (A)f(1/2) 1/3 (B)f(1/2)>1/2a n df(1/3)>1/3 (C)f(1/2) 1/2a n df(1/3)<1/3

Let f be a non-negative function in [0, 1] and twice differentiate in (0, 1). If int_0^x sqrt(1-(f'(t))^(2))dt=int_0^x f(t)dt , 0 lexle1 and f(0)=0 then the value of lim_(x to0)int_0^xf(t)/x^2 dt is

Let f be a function defined on the interval [0,2pi] such that int_(0)^(x)(f^(')(t)-sin2t)dt=int_(x)^(0)f(t)tantdt and f(0)=1 . Then the maximum value of f(x) is…………………..

Let f(x) be a non constant function such that int _(0) ^(x) (f (t))^(3) dt = (1)/(x ^(2))(int _(0) ^(x) (f(t)dt ) ) ^(3) AA x in R- {0} If f(1)=1 then f((1)/(2))=

Let f(x) be a continuous function which takes positive values for xge0 and satisfy int_(0)^(x)f(t)dt=x sqrt(f(x)) with f(1)=1/2 . Then

If int_(0)^(1) f(t)dt=x^2+int_(0)^(1) t^2f(t)dt , then f'(1/2)is