CBSE 2024, Set 2

Question

Show that a function $f: R 	o R$ defined by $f(x) = \frac{2x}{1+x^2}$ is neither one-one nor onto. Further, find set A so that the given function $f: R 	o A$ becomes an onto function.

Edvaya · Accepted Answer

The correct answer is Subjective. **Step 1: Check for One-One (Injectivity)**

Let's test the function with two different real values, say $x = 2$ and $x = \frac{1}{2}$.

$$f(2) = \frac{2(2)}{1 + 2^2} = \frac{4}{1 + 4} = \frac{4}{5}$$

$$f\left(\frac{1}{2}ight) = \frac{2(1/2)}{1 + (1/2)^2} = \frac{1}{1 + 1/4} = \frac{1}{5/4} = \frac{4}{5}$$

Since $f(2) = f\left(\frac{1}{2}ight)$ but $2 
e \frac{1}{2}$, distinct elements in the domain have the same image.
Therefore, the function $f(x)$ is **not one-one**.

**Step 2: Check for Onto (Surjectivity)**

Let $y \in \mathbb{R}$ (codomain) such that $y = f(x)$.
$$y = \frac{2x}{1+x^2}$$
$$y(1 + x^2) = 2x$$
$$yx^2 - 2x + y = 0$$

For $x$ to be a real number (since domain is $\mathbb{R}$), the discriminant of this quadratic equation in $x$ must be greater than or equal to zero ($D \ge 0$).
$$b^2 - 4ac \ge 0$$
$$(-2)^2 - 4(y)(y) \ge 0$$
$$4 - 4y^2 \ge 0$$
$$4(1 - y^2) \ge 0$$
$$y^2 \le 1 \implies -1 \le y \le 1$$

The range of the function is $[-1, 1]$.
Since the range $[-1, 1]$ is a proper subset of the codomain $\mathbb{R}$ (Range $
e$ Codomain), the function is **not onto**.

**Step 3: Make the function onto**

For a function to be onto, its codomain must exactly equal its range.
We found the range of $f(x)$ to be $[-1, 1]$.

Therefore, for the function $f: R 	o A$ to become an onto function, the set $A$ must be defined as:
**$A = [-1, 1]$**

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