1.

Find whether the following functions are one-one or many-one?(i) f: R – {3} → R defined by f (x) = \(\frac{5x+7}{x-3}\) ,  x ∈ R − {3} (ii) Modulus function f : R → R defined by f (x) = \(\ f(n) =\begin{cases}x \,; & \quad x \geq 0\\-x \,; & \quad x <0\end{cases}\)(iii) Greatest integer function f : R → R defined by f (x) = [x] = n (an integer) for all n ≤ x ≤ n + 1.

Answer»

Given:

f (x) = \(\frac{5x+7}{x-3}\) , x ∈ R - {3}

Let a1, a2 be two arbitrary elements ∈R – {3} such that f (a1) = f (a2).

Then, f (a1) = f(a2) ⇒ \(\frac{5a_1+7}{a_1-3} = \frac{5a_2+7}{a_2-3}\)  

⇒ 5a1a2 + 7a2 – 15a1 – 21 = 5a2a1 – 21 + 7a1 – 15a2 

⇒ – 22a1 = – 22a

⇒ a1 = a2 

⇒ f is one-one.

(ii) Given: f (x) = | x | 

⇒ f (1) = | 1 | = 1 and f (– 1) = |– 1| = 1 

Thus, f (1)= f (– 1) = 1 but 1 ≠ – 1 

∴  f is many-one. 

(iii) ∵ f (x) = [x] = n for all n ≤ x < n + 1, so, 1 ≤ x < 2 ⇒ f (x) = 1 

∴ The elements 1.1, 1.25, 1.4, 1.9 .... ∈ domain are mapped onto to the same value 1 in the range. 

Hence the function f is many-one.



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