Let A = {x \in R: x is not a positive integer}. Define a function f: A → R as f (x) =\frac{2 x}{x

Engineering

Mathematics

Introduction to Functions

Types of Functions

domain of function

Question

Let A = {x $\in$ R: x is not a positive integer}. Define a function f: A → R as f (x) = $\frac{2 x}{x - 1}$ , then f is

Injective but not surjective

Not injective

Neither injective nor surjective

Surjective but not injective

$f (x) = 2 (1 + \frac{1}{x - 1})$

$f' (x) = \frac{2}{{(x - 1)}^{2}}$

⇒ of is one – one but not onto

Let's analyze the function f: A → R defined by f(x) = $\frac{2 x}{x - 1}$ , where A = {x ∈ R: x is not a positive integer}. We need to determine if f is injective (one-to-one), surjective (onto), both, or neither.

Step 1: Check Injectivity (One-to-One)

A function is injective if f(a) = f(b) implies a = b. Let's assume f(a) = f(b):

$\frac{2 a}{a - 1} = \frac{2 b}{b - 1}$

Cross-multiplying:

$2 a (b - 1) = 2 b (a - 1)$

$2 a b - 2 a = 2 a b - 2 b$

Subtracting 2ab from both sides:

$- 2 a = - 2 b$

Dividing by -2:

$a = b$

So, f(a) = f(b) implies a = b. Therefore, f is injective.

Step 2: Check Surjectivity (Onto)

A function is surjective if for every y in the codomain R, there exists an x in A such that f(x) = y. Let y be any real number. We solve for x:

$y = \frac{2 x}{x - 1}$

Multiply both sides by (x - 1):

$y (x - 1) = 2 x$

$y x - y = 2 x$

Bring terms with x to one side:

$y x - 2 x = y$

$x (y - 2) = y$

So, $x = \frac{y}{y - 2}$ , provided y ≠ 2.

Now, we must ensure that this x is in the domain A, meaning x is not a positive integer. Also, note that x = 1 is not allowed because it makes the denominator zero, but x=1 is excluded from A since 1 is a positive integer. However, we must check if for some y, the computed x becomes a positive integer.

Suppose x = n, where n is a positive integer. Then from x = y/(y-2), we get n = y/(y-2). Solving for y: n(y-2) = y ⇒ ny - 2n = y ⇒ ny - y = 2n ⇒ y(n-1) = 2n ⇒ y = 2n/(n-1). For n>1, this gives a real number y. For example, if n=2, y=4; if n=3, y=3; etc. So for y=4, x=2 which is a positive integer and hence not in A. Similarly, for y=3, x=3 which is also not in A. Therefore, for these y values (like 3,4, etc.), there is no x in A such that f(x)=y. Also, note that y=2 is not attained because if we set y=2 in the equation, we get 2 = 2x/(x-1) ⇒ 2(x-1)=2x ⇒ 2x-2=2x ⇒ -2=0, which is impossible. So y=2 is also not in the range.

Thus, the range of f is R \ {2} and also excluding those y for which x becomes a positive integer (like y=3,4, etc.). Therefore, f is not surjective onto R.

Conclusion

f is injective but not surjective.

Detailed Solution

Step 1: Check Injectivity (One-to-One)

Step 2: Check Surjectivity (Onto)

Conclusion

Related Topics and Formulae

Detailed Solution

Step 1: Check Injectivity (One-to-One)

Step 2: Check Surjectivity (Onto)

Conclusion

Related Topics and Formulae

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