Follow Us


Question 2

Check the injectivity and surjectivity of the following functions:

(i) f : N → N given by f(x) = x2

(ii) f : Z → Z given by f(x) = x2

(iii) f : R → R given by f(x) = x2

(iv) f : N → N given by f(x) = x3

(v) f : Z → Z given by f(x) = x

Answer

(i) fN → N is given by,

f(x) = x2

It is seen that for xy ∈Nf(x) = f(y) ⇒ x2 = y2 ⇒ x = y.

f is injective.

Now, 2 ∈ N. But, there does not exist any x in N such that f(x) = x2 = 2.

∴ f is not surjective.

Hence, function f is injective but not surjective.

 

(ii) fZ → Z is given by,

f(x) = x2

It is seen that f(-1) = f(1) = 1, but -1 ≠ 1.

∴ f is not injective.

Now,-2 ∈ Z. But, there does not exist any element x ∈Z such that f(x) = x2 = -2.

∴ f is not surjective.

Hence, function f is neither injective nor surjective.

 

(iii) fR → R is given by,

f(x) = x2

It is seen that f(-1) = f(1) = 1, but -1 ≠ 1.

∴ f is not injective.

Now,-2 ∈ R. But, there does not exist any element x ∈ R such that f(x) = x2 = -2.

∴ f is not surjective.

Hence, function f is neither injective nor surjective.

 

(iv) fN → N given by,

f(x) = x3

It is seen that for xy ∈Nf(x) = f(y) ⇒ x3 = y3 ⇒ x = y.

f is injective.

Now, 2 ∈ N. But, there does not exist any element x in domain N such that f(x) = x3 = 2.

∴ f is not surjective

Hence, function f is injective but not surjective.

 

(v) fZ → Z is given by,

f(x) = x3

It is seen that for xy ∈ Zf(x) = f(y) ⇒ x3 = y3 ⇒ x = y.

∴ f is injective.

Now, 2 ∈ Z. But, there does not exist any element x in domain Z such that f(x) = x3 = 2.

∴ f is not surjective.

Hence, function f is injective but not surjective.

Popular Questions of Class 12th mathematics

 

">

Let A = R – {3} and B = R – {1}. Consider the function  f : A → B defined by

. Is f one-one and onto? Justify your answer. 

 

  • Q:-

     Prove that the Greatest Integer Function f : R → R, given by f(x) = [x], is neither one-one nor onto, where [x] denotes the greatest integer less than or equal to x.

  • Q:- Show that the relation R in the set R of real numbers, defined as R = {(a, b): a ≤ b2} is neither reflexive nor symmetric nor transitive.
  • Q:-

     Show that the Modulus Function f : R → R, given by f(x) = |x|, is neither oneone nor onto, where | x | is x, if x is positive or 0 and |x| is – x, if x is negative.

  • Q:- Show that the relation R in the set A = {1, 2, 3, 4, 5} given by R = { (a,b) ; |a - b| is even}, is an equivalence relation. Show that all the elements of {1, 3, 5} are related to each other and all the elements of {2, 4} are related to each other. But no element of {1, 3, 5} is related to any element of {2, 4}.
  • Q:- Show that the relation R in R defined as R = {(a, b): a ≤ b}, is reflexive and transitive but not symmetric.
  • Recently Viewed Questions of Class 12th mathematics

     

    ">

    Let A = R – {3} and B = R – {1}. Consider the function  f : A → B defined by

    . Is f one-one and onto? Justify your answer. 

     

  • Q:-

    Show that the function f : R* → R* defined by f(x) = 1/x is one-one and onto,where R* is the set of all non-zero real numbers. Is the result true, if the domain R* is replaced by N with co-domain being same as R

  • Q:-

    Answer the following as true or false.

     \begin{align}(i) \overrightarrow{a}\;  and\; \overrightarrow{-a}\; are\; collinear.\end{align}

    (ii) Two collinear vectors are always equal in magnitude.

    (iii) Two vectors having same magnitude are collinear.

    (iv) Two collinear vectors having the same magnitude are equal.

  • Q:- Let L be the set of all lines in XY plane and R be the relation in L defined as R = {(L1, L2): L1 is parallel to L2}. Show that R is an equivalence relation. Find the set of all lines related to the line y = 2x + 4.
  • Write a Comment: