The functions defined by `f(x)="max" {x^(2),(x-1)^(2),2x(1-x)}, 0lexle1`

Examine continuity at the given point of the function f defined by f(x)={(1-x,0 lexle1),(x-1,xgt1,):} at x=1

The minimum value of the function defined by f(x)=max {x,x+1,2-x} is

Discuss the continmuity of function f defined by f(x)={{:((1)/(2)-x", if "0lexle(1)/(2)),("1, if "x=(1)/(2)),((3)/(2)-x", if "(1)/(2)ltxle1):}

Consider the function f(x)=max. [(x^(2) , (1-x)^(2) , 2x(1-x))] , x in [0,1] The interval in which f(x) is increasing, is

Consider the function f(x)=max x^(2), [((1-x)^(2),2x(1-x))],x in [0,1] The interval in which f(x) is decreasing is

Consider the function f(x)=max. [(x^(2) , (1-x)^(2) , 2x(1-x))] , x in [0,1] The interval in which f(x) is increasing, is

The minimum value of the function defined by f(x)=max(x,x+1,2-x) is

The function f defined by f(x)=(4sinx-2x-xcosx)/(2+cosx),0lexle2pi is :

Examine continuity at the given point of the function f defined by f(x)={(-x,xlt0),(x,0lexle1,),(2-x,xgt1,):} at x=0,1

The relation 'f' is defined by f(x)={(x^2,0lexle3), (3x, 3lexle10):} The relation 'g' is defined by g(x)= {(x^2,0lexle2), (3x, 2lexle10):} Show that 'f' is a function and 'g' is not a function.