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

Solution:

According to the given problem:

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

and f : A → B defined by f (x) = (x - 2) / (x - 3)

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 - 2 y + 6 = xy - 3y - 2x + 6

⇒ - 3x - 2 y = -3y - 2x

⇒ 3x - 2x = 3y - 2 y

⇒ x = y

Therefore,

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 [y ≠ 1]

Thus,

for any y ∈ B, there exists (2 - 3y)/(1 - y) ∈ A such that

f((2 - 3y)/(1 - y)) = [(2 - 3y)/(1 - y)]/[(2 - 3y)/(1 - (- y))]

= (2 - 3y - 2 + 2 y)/(2 - 3y - 3 + 3y) = - y/- 1 = y

Therefore,

f is onto.

Hence, the function is one-one and onto

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NCERT Solutions for Class 12 Maths - Chapter 1 Exercise 1.2 Question 10

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

Summary:

For the given function A = R - {3}, B = R - {1} and f : A → B defined by f (x) = (x - 2)/(x - 3), we have shown that the given function is one - one and onto