Show that the function f : R → R defined by f ( x ) = cos x is neither one-one nor onto.

One-one - Since 0 , 2 π ∈ R f ( 0 ) = cos 0 = 1 ; f ( 2 π ) = cos 2 π = 1 So different elements in R D f have same images in R (codomain). ∴ f is not 1 - 1 . Also, 2 ∈ R (codomain of f ). Then f ( x ) = 2 ⇒ co s x = 2 which does not gives any solution since - 1 ≤ cos ≤ 1 ∴ f is not onto. Hence f is neither one-one nor onto.

Show that the function f : R → R defined by f ( x ) = sin 2 x + cos 2 x is neither one-one nor onto.

Given function f : R → R defined by f ( x ) = s i n 2 x + c o s 2 x ∀ x ∈ R Since, f ( 0 ) = sin 2 0 + cos 2 0 = 1 f ( π ) = sin 2 π + cos 2 π = 0 2 + ( - 1 ) 2 = 1 Thus, different elements of R (in domain of f ) have same image in R (codomain). ∴ f is many one function i . e . not a one-one function. Since, f ( x ) = s in 2 x + cos 2 x = 1 ∀ x ∈ R i . e . f ( x ) be a constant function and all elements in R ( D f ) have same image in R . Since, 2 ∈ R (codomain) there does not exists any arbitrary element x ∈ R (domain), f ( x ) = 2 . ∴ f is not onto. Hence, f is neither one-one nor onto.

Show that the function f : I → I defined by f ( x ) = x 2 + x is neither one-one nor onto.

Given that, f : z → z defined by f ( x ) = x 2 + x Clearly 0 , - 1 ∈ Z (domain), here f ( 0 ) = 0 f ( - 1 ) = ( - 1 ) 2 - 1 = 1 - 1 = 0 ∴ Different elements have same image in Z (codomain). Thus, f is not injective. also - 1 ∈ Z (codomain) have no pre-image in Z (domain). [Let if possible - 1 ∈ Z has pre image x in Z (domain). then f ( x ) = - 1 ⇒ x 2 + x + 1 = 0 ⇒ x = - 1 ± - 3 2 ∉ Z ∴ - 1 has no pre-image in Z ] Thus, f is not onto.

If the function f : 1 → 1 is defined by f ( x ) = 2 x + 1 , if x is odd 2 x - 1 , if x is even , find f ( 3 ) .

f : 1 → 1 defined by f ( x ) = 2 x + 1 ; x is odd 2 x - 1 ; x is even We know that, the range is the resulting y -values we get after substituting all the possible x -values. Here, x = 3 f ( 3 ) = 2 × 3 + 1 = 6 + 1 = 7 [ ∵ 3 be an odd number ∴ f ( x ) = 2 x + 1 ]

If the function f : 1 → 1 is defined by f ( x ) = 2 x + 1 , if x is odd 2 x - 1 , if x is even , find f ( 4 )

Given that, f : 1 → 1 defined by f ( x ) = 2 x + 1 ; x is odd 2 x - 1 ; x is even We know that, the range is the resulting y -values we get after substituting all the possible x -values. Here, x = 4 f ( 4 ) = 2 × 4 - 1 = 8 - 1 = 7 [ ∵ 4 be an even number ∴ f ( x ) = 2 x - 1 ]

If the function f : I → I is defined by f ( x ) = 2 x + 1 , if x is odd 2 x - 1 , if x is even find x , if f ( x ) = - 9 .

Given that, f ( x ) = - 9 f ( x ) = 2 x + 1 , if x is odd 2 x - 1 , if x is even When x is odd then f ( x ) = 2 x + 1 ⇒ 2 x + 1 = - 9 ⇒ x = - 5 When x is even then f ( x ) = 2 x - 1 ⇒ 2 x - 1 = - 9 ⇒ x = - 4 Clearly 3 , 4 ∈ I and have same image 7 . ∴ Different elements of I have same image in I . ∴ f is not one-one.