Let `f:[-1.2]->R` be differentiable such that `0 le f'(t) le 1` for `t in [-1,0]` and `-1 le f'(t) le 0` for `t in [0,2]` then

Let g (x) int_(0)^(x) f(t) dy=t , where f is such that (1)/(2) le f (t) le 1 for tin [0,1] and 0 le f (t) le (1)/(2) for t in [1 ,2] . then . g (2) satis fies the inequality

Let g(x)=int_(0)^(x) f(t) dt , where f is continuous function in [0,3] such that 1/3 le f(t) le 1 for all t in [0,1] and 0 le f(t) le 1/2 for all tin (1,3] . The largest possible interval in which g(3) lies is:

Let g(x) =int_0^x f(t) dt where f is such that 1//2lef(t)le1 for tin[0, 1] and 0 lef(t) le1//2 for tin[1, 2] Then the interval in which g(2) lies.

Let g(x) = int_0^(x) f(t) dt , where 1/2 le f(t) < 1, t in [0,1] and 0 < f(t) le 1/2 "for" t in[1,2] . Then :

Let g(x)=int_(0)^(x)f(t)dt , where f is such that 1/2lef(t)le1 , for tepsilon[0,1] and 0lef(t)le1/2 , for tepsilon[1,2] . Then prove that 1/2leg(2)le3/2 .

Let g(x)=int_(0)^(x)f(t)dt , where f is such that 1/2lef(t)le1 , for tepsilon[0,1] and 0lef(t)le1/2 , for tepsilon[1,2] . Then prove that 1/2leg(2)le3/2 .