In each of the following cases, state whether the function is one-one, onto or bijective. Justify your answer. (i) f : R → R defined by f(x) = 3 – 4x (ii) f : R → R defined by f(x) = 1 + x^2

Class 12, NCERT Chapter 1, Exercise 1.2, Q7

(i)

Here f(x) = 3 – 4x

Then f(x1) = 3 – 4x1

and f (x2) = 3 – 4x2

Now f(x1) = f (x2)

⇒ 3 – 4x1 = 3 – 4x2

⇒ – 4x1 = –4x2

⇒ x1 = x2

∴ f(x1) = f(x2)

& x1 = x2

So, f is one-one function.

Let f(x) = y ∈ R

Then y=3 - 4x

⇒ x= 3-y/4

(ii)

Let x1 = 2 and x2 = –2 ∈ R

Here f(x) = 1 + x2

Then f(x1) = f (2) = 1 + (2) 2 = 5

and f (x2) = f(–2) = 1 + (–2) 2 = 5

∴ f (x1) = f(x2) but x1 ≠ x2

So, f is not one-one function.

Let f(x) = – 2 ⇒R

Then 1 + x2 = –2

⇒ x2 = –3 = ± √–3 ∈R

So, f is not onto function.

Comments

. Let f : R → R be defined as f(x) = 3x. Choose the correct answer. (A) f is one-one onto (B) f is many-one onto (C) f is one-one but not onto (D) f is neither one-one nor onto.

Class 12, NCERT Chapter 1, Exercise 1.2, Q12 f : R → R defined as f ( x ) = 3 x . Let x , y ∈ R such that f ( x ) = f ( y ). ⇒ 3 x = 3 y ⇒ x = y ∴ f is one-one. Also, for any real number ( y) in co-domain R , there exists in R such that . ∴ f is onto. Hence, function f is one-one and onto. The correct answer is A.

Let f : N → N be defined by f (n) ={ (n+1)/2, if n is odd and (n-1)/2, if n is even, for all n ∈ N.State whether the function f is bijective. Justify your answer

Class 12, NCERT Chapter 1, Exercise 1.2, Q9 Thus it is bijective.

Show that the relation R in the set {1, 2, 3} given by R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3)} is reflexive but neither symmetric nor transitive.

RELATIONS AND FUNCTIONS Class 12, NCERT Chapter 1, Example4 Solution R is reflexive since (1, 1), (2, 2) and (3, 3) lie in R. Also, R is not symmetric, as (1, 2) ∈ R but (2, 1) ∉ R. Similarly, R is not transitive, as (1, 2) ∈ R and (2, 3) ∈ R but (1, 3) ∉ R.

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