Let f(x)=Min`{x^(2)+1,x+1}` for `0 <= x <= 2` .If `A` is the area bounded by `y=f(x)`, the `x`-axis in the interval `[0,2]` then find the integral part of `A`

Let f(x)=min{x,(x-2)^(2)} for x>=0 then

Let f(x)=min{x,x^(2)}, for every x in R Then

Let f(x) = {{:(min{x, x^(2)}, x >= 0),(max{2x, x^(2)-1}, x (A) f(x) is continuous at x=0 (B) f(x) is differentiable at x=1 (C) f(x) is not differentiable at exactly three point (D) f(x) is every where continuous

Let f(x) =(x+1)^(2) - 1, x ge -1 Statement 1: The set {x : f(x) =f^(-1)(x)}= {0,-1} Statement-2: f is a bijection.

" If "f(x)=min{|x^(2)-1|,|x|}," then number of extremums of "f(x)" in "(-0.9,1.1)" is "

Let f(x)={[(x)/(2x^(2)+|x|) , If x!=0] , [1, If x=0] then

underset then Let f(x)=|min(x+1,3-x|AA x in R