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Show that the Function F : Q → Q, Defined by F(X) = 3x + 5, is Invertible. Also, Find F−1

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Question

Show that the function f : Q → Q, defined by f(x) = 3x + 5, is invertible. Also, find f−1

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Solution

Injectivity of f:
Let x and y be two elements of the domain (Q), such that
f(x)=f(y)

⇒ 3x + 5 =3y + 5

⇒ 3x = 3y

⇒ x = y
So, f is one-one.

Surjectivity of f:
Let y be in the co-domain (Q), such that f(x) = y

⇒ 3x +5 = y 

⇒ 3x = y - 5

⇒ `x = (y -5)/3 in` (domain)

⇒ f is onto.
So, f is a bijection and, hence, it is invertible.

Finding f-1:

Let f−1(x)=y                    ...(1)

⇒ x=f(y)

⇒ x=3y+5

⇒ x −5 = 3y

⇒ `y = (x -5)/3`

So, `f^-1 (x) = (x-5)/3`      [from (1)]

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Chapter 2: Functions - Exercise 2.4 [Page 68]

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R.D. Sharma Mathematics Volume 1 and 2 [English] Class 12
Chapter 2 Functions
Exercise 2.4 | Q 5 | Page 68

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