f:R rarr R,quad f(x)=x|x|" is "

Function f:R rarr R,f(x)=x|x| is

f: R rarr R,f(x)=x|x| is (A) one-one but not onto (B) onto but not one-one (C) Both one-one and onto (D) neither one-one nor onto

Function f : R rarr R , f(x) = x + |x| , is

Function f : R rarr R , f(x) = x + |x| , is

Function is said to be onto if range is same as co - domain otherwise it is into. Function is said to be one - one if for all x_(1) ne x_(2) rArr f(x_(1)) ne f(x_(2)) otherwise it is many one. Function f:R rarr R, f(x)=x^(2)+x , is

If R is the set of real numbers prove that a function f:R rarr R,f(x)=e^(x),x in R is one to one mapping.

Examine f:R_+ rarr R +, f(x) =x +1/x "where"R_+={x in R :x gt 0} functions if it is (i) injective (ii) surjective, (iii) bijective and (iv) none of the three.

Let f:R rarr R, then f(x)=2x+|cos x| is

Find the range of f: R rarr R, f(x)= x^(2)-6x+ 7

If f:R^(+) rarr R , f(x) = x/(x+1) is .......