`f(x)=min {x-[x],-x-[-x]}` and `x=0` and `x=4` then the area of `f(x)` in sq. units

f(x)= minimum of {x-[x],-x-[-x]} and x=0 and x=4 then the area of f(x) in sq.units

f(x)=minimum of {|x-1|,|x|,|x+1|} and x=+-1 the area of f(x) is in sq.units

The function f(x)=min{x-[x],-x-|-x|} is a

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

Let f(x)=min(x+1,sqrt(1-x))AA x le 1 . Then, the area (in sq. units( bounded by y=f(x), y=0 and x=0 from y=0 to x=1 is equal to

Let f(x)=min{|x+2|,|x|,|x-2|} then the value of the integral int_-2^2 f(x)dx is (i) 4 sq.unit (ii)10 sq.unit (iii)2 sq.unit (iv) 7 sq.unit

Let f(x)={{:(2^([x]),, x lt0),(2^([-x]),,x ge0):} , then the area bounded by y=f(x) and y=0 is

Let f(x)=min{4x+1,x+2,-2x+4}. then the maximum value of f(x) is

If f(x)= min(x^(2),x,sgn(x^(2)+4x-12)) then the value of f(2) is equal to