But (f + g) (0) = sin 0 + cos 0 = 1 and
(f + g)
π2





 = sin
cos
1
2
2
π
π
+
= Therefore, f + g is not one-one Miscellaneous Exercise on Chapter 1
1 Show that the function f : R → {x ∈ R : – 1 < x < 1} defined by 
( )
1 |
|
x
f x
= +x
,
x ∈ R is one one and onto function