Q. 205.0( 1 Vote )

# Let A = R – {3}, B = R – {1}. Let f : A → B be defined by ∀ x ∈ A . Then show that f is bijective.

Answer :

Given that,

In order to prove that f is one-one, it is sufficient to prove that f(x_{1})=f(x_{2}) ⇒ x_{1}=x_{2}∀ x_{1}, x_{2} ∈ A .

Let f(x_{1})=f(x_{2})

⇒

⇒ (x_{1}-2)(x_{2}-3) = (x_{2}-2)(x_{1}-3)

⇒ x_{1}x_{2}-3x_{1}-2x_{2}+6 = x_{1}x_{2}-3x_{2}-2x_{1}+6

⇒ -3x_{1}-2x_{2} = -3x_{2}-2x_{1}

⇒ -3x_{1}+2x_{1} = -3x_{2}+2x_{2}

⇒ (-3+2) x_{1} = (-3+2)x_{2}

⇒ x_{1} = x_{2}

∴ f is one-one.

f is onto if every element of B is the f-image of some element of A.

let f(x) = y

⇒

⇒ x-2 = y(x-3)

⇒ x-2 = xy-3y

⇒ x-xy = -3y+2

⇒ x(1-y) = -3y+2

⇒

⇒

Thus, for each y ∈ B there exists such that f(x) = y.

Hence, f is onto.

Since, f is one-one and onto therefore f is bijective.

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