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Show that the Function F : R − {3} → R − {2} Given By F(X) = `(X-2)/(X-3)` Is a Bijection.

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प्रश्न

Show that the function f : R − {3} → R − {2} given by f(x) = `(x-2)/(x-3)` is a bijection.

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उत्तर

f : R − {3} → R − {2} given by

`f (x) = (x-2)/(x-3)`

Injectivity:
Let x and y be any two elements in the domain (R − {3}), such that f(x) = f(y).

f(x) = f(y)

⇒ `(x-2)/(x-3) = (y-2)/(y-3)`

⇒ ( x-2 ) (y - 3) =  ( y-2 )  ( x-3 )

⇒ xy - 3x - 2y +  = xy = 3y - 2x + 6

⇒ x = y

So, f is one-one.

Surjectivity :

Let y be any element in the co-domain (R − {2}), such that f(x) = y for some element xin R − {3} (domain).

f(x) = y

⇒ `(x-2)/(x-3) =y`

⇒ x - 2 = xy - 3y

⇒ xy - x = 3y - 2

⇒ x ( y-1 ) = 3y - 2

⇒  x = `(3y - 2)/(y-1),`which is in R -{3}

So, for every element in the co-domain, there exists some pre-image in the domain. ⇒ f  is onto.

Since, f  is both one-one and onto, it is a bijection.

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अध्याय 2: Functions - Exercise 2.1 [पृष्ठ ३१]

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आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
अध्याय 2 Functions
Exercise 2.1 | Q 7 | पृष्ठ ३१

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