CUET UG 2025 — Mathematics Algebra

The function $f: [0, \infty) \rightarrow \mathbb{R}$f:[0,∞)→R is defined by $f(x) = 2x^2 + 3$f(x)=2x2+3.

Domain: $[0, \infty)$[0,∞) (all non-negative real numbers)

Codomain: $\mathbb{R}$R (all real numbers)

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To check if the function is one-one, assume $f(x_1) = f(x_2)$f(x1​)=f(x2​):

$$2x_1^2 + 3 = 2x_2^2 + 3$$2x12​+3=2x22​+3

$$2x_1^2 = 2x_2^2$$2x12​=2x22​

$$x_1^2 = x_2^2$$x12​=x22​

Since both $x_1, x_2 \in [0, \infty)$x1​,x2​∈[0,∞), both are non-negative:

$$x_1 = x_2$$x1​=x2​

If outputs are equal, inputs must be equal. The function is one-one.

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To check if the function is onto, find the range.

When $x = 0$x=0:

$$f(0) = 2(0)^2 + 3 = 3$$f(0)=2(0)2+3=3

This is the minimum value.

As $x$x increases, $f(x)$f(x) increases since $x^2$x2 increases.

As $x \to \infty$x→∞, $f(x) \to \infty$f(x)→∞

Range of $f = [3, \infty)$f=[3,∞)

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Codomain: $\mathbb{R}$R (all real numbers)

Range: $[3, \infty)$[3,∞) (only numbers greater than or equal to 3)

Since Range $\neq$= Codomain, the function is not onto.

Values like $f(x) = 0$f(x)=0, $f(x) = 1$f(x)=1, or $f(x) = -5$f(x)=−5 cannot be achieved as they are less than 3.

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The function is one-one but not onto.