CUET UG 2025 — Mathematics Algebra

A function is one-one (injective) if each output comes from exactly one input. A function is onto (surjective) if every value in the target set is achieved by some input.

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To check if $f(x) = \cos x$f(x)=cosx is one-one, consider:

$$f(0) = \cos(0) = 1$$f(0)=cos(0)=1

$$f(2\pi) = \cos(2\pi) = 1$$f(2π)=cos(2π)=1

Since $0 \neq 2\pi$0=2π but both give the same output, different inputs produce the same output.

Similarly:

$$f(\pi/3) = \cos(\pi/3) = 0.5$$f(π/3)=cos(π/3)=0.5

$$f(5\pi/3) = \cos(5\pi/3) = 0.5$$f(5π/3)=cos(5π/3)=0.5

The function is not one-one.

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To check if $f(x) = \cos x$f(x)=cosx is onto, verify that every value in $[-1, 1]$[−1,1] can be achieved.

The range of $\cos x$cosx is exactly $[-1, 1]$[−1,1]:

$$\cos(0) = 1$$cos(0)=1

$$\cos(\pi/3) = 0.5$$cos(π/3)=0.5

$$\cos(\pi/2) = 0$$cos(π/2)=0

$$\cos(2\pi/3) = -0.5$$cos(2π/3)=−0.5

$$\cos(\pi) = -1$$cos(π)=−1

For any value between $-1$−1 and $1$1, there exists an $x$x value such that $\cos x$cosx equals that value.

The function is onto.

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The function is onto but not one-one.