Let `f(x)=(2x +1)/x` , `x != 0`, where

If f(x) is continuous at x=0 , where f(x)=(1+2x)^(1/x)", for " x!=0 , then f(0)=

Let f(x) =ax^(2) -b|x| , where a and b are constants. Then at x = 0, f (x) is

Let f(x)=f_1(x)-2f_2 (x), where ,where f_1(x)={((min{x^2,|x|},|x|le 1),(max{x^2,|x|},|x| le 1)) and f_2(x)={((min{x^2,|x|},|x| lt 1),({x^2,|x|},|x| le 1)) and let g(x)={ ((min{f(t):-3letlex,-3 le x le 0}),(max{f(t):0 le t le x,0 le x le 3})) for -3 le x le -1 the range of g(x) is

Let F(x)=f(x)+f((1)/(x)), where f(x)=int_(t)^(x)(log t)/(1+t)dt (1) (1)/(2)(2)0(3)1(4)2

If f(x)=(1)/(x)-1 where x ne 0, then: f(x)-2.f(2x)=

Let f(x) = ||x|-1| , then points where, f(x) is not differentiable is/are

Let f(x)=ax^(2)-b|x| , where a and b are constant . Then at x=0 , f(x) has