Let A = R – {3} and B = R – {1}. Consider the function f : A → B defined by f(x) = (x-2)/(x-3) Is f one-one and onto? Justify your answer.

Class 12, NCERT Chapter 1, Exercise 1.2, Q10

A = R - {3}, B = R - {1}

f: A → B is defined as

Let x,y∈A such that f(x)=f(y).

⇒ (x-2)/(x-3)=(y-2)/(y-3)
⇒ (x-2)(y-3)=(y-2)(x-3)
⇒ xy-3x-2y+6=xy-3y-2x+6
⇒ -3x-2y=-3y-2x
⇒ 3x-2x=3y-2y
⇒ x=y
∴ f is one-one.

Let y ∈B = R - {1}. Then, y ≠ 1.

The function f is onto if there exists x ∈A such that f(x) = y.

Now,

f(x)=y
⇒(x-2)(x-3)=y
⇒x-2=xy-3y
⇒x(1-y)=-3y+2
⇒x=2-3y/1-y ∈A

Thus, for any y ∈ B, there exists x=2-3y/1-y ∈A such that

Hence, function f is one-one and onto.

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 a one-one function f : {1, 2, 3} → {1, 2, 3} must be onto.

Class 12, NCERT Chapter 1, Example14 Since f is one-one, three elements of {1, 2, 3} must be taken to 3 different elements of the co-domain {1, 2, 3} under f. Hence, f has to be onto .

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