Let `A={-1 le x le 1} and f:A to A` such that `f(x)=x|x|` then f is:

Let A={x-1 le x le 1} and f:A to A such that f(x)=x|x| then f is:

A={x:-1 le x le 1}, f: A to A defined by f(x) =x|x| . Then f is:

Let A ={x:-1 le x le 1} and f: A rarr A be defined by f(x) ="sin" (pix)/(2) . Show that f is a one- one onto mapping and hence find a formula that defines f^(-1)

If A= {x:pi/6 le x le pi/3} and f(x) = cosx - x(1 + x) then f(A) is equal to-

Let f(x) = {{:(-1,-2 le x lt 0),(x^2-1,0 le x le 2):} and g(x)=|f(x)|+f(|x|) . Then , in the interval (-2,2),g is

Let f:R to R be such that f(0)=0 and |f'(x)| le 5 for all x. Then f(1) is in

Given A={x:(pi)/(6) le x le (pi)/(3)} and f(x)=cosx-x(1+x) , then f(A)=

Let f(x) be defined on [-2,2] and is given by f(x)={{:(,-1,-2 le x le 0),(,x-1,0 lt x le 2):} and g(x) =f(|x|)+|f(x)| . Then g(x) is equal to