Prove that for `a =3, b = 1`, the function `f(x) =4; x leq 1 , f(x)= ax^2+b; -1 lt x lt 0, f(x)=cosx; x geq 0` is continuous

Show that the function f(x) = {{:(2 + x"," , "if " x ge 0),(1",", "if " x lt 0):} is not continuous at x = 0

Let f(x)=1+x , 0 leq x leq 2 and f(x)=3−x , 2 lt x leq 3 . Find f(f(x)) .

If the function f(x) {:(=5x-4", ... "0 lt x le 1 ), (=4x^(2)+3bx", ... "1 lt x lt 2 ) :} is continuous at every point of its domain , then : b =

If the function f(x) ={(ax-b, x le1),(3x, 1lt x lt 2),(bx^(2)-a, x ge 2):} is continuous at x =1 and discontinuous at x =2, then

At x =1, the function given by f(x) = {(sinx,x lt 0),(2x,x ge 0):} is (a) Continuous (b) Not continuous ( c) Not defined

If f(x)=3x-1 when x geq1 and f(x) = -x when x lt 1 then value of f(-2) is

If f(x) = 3x-1 when x geq 1 and f(x) = -x when x lt 1 then value of f(-2) is

If f(x) = 3x-1 when x geq 1 and f(x) = -x when x lt 1 then value of f(-2) is