CUET UG 2025 — Mathematics Calculus

To determine if $f(x) = 100x + 1$f(x)=100x+1 is one-one (injective), assume $f(x_1) = f(x_2)$f(x1​)=f(x2​):

$$100x_1 + 1 = 100x_2 + 1$$100x1​+1=100x2​+1

$$100x_1 = 100x_2$$100x1​=100x2​

$$x_1 = x_2$$x1​=x2​

Since $f(x_1) = f(x_2)$f(x1​)=f(x2​) implies $x_1 = x_2$x1​=x2​, different inputs produce different outputs.

Therefore, $f$f is one-one.

---

To determine if $f(x) = 100x + 1$f(x)=100x+1 is onto (surjective), let $y$y be any real number.

Find $x$x such that $f(x) = y$f(x)=y:

$$100x + 1 = y$$100x+1=y

$$100x = y - 1$$100x=y−1

$$x = \frac{y - 1}{100}$$x=100y−1​

For any real number $y$y, the value $x = \frac{y - 1}{100}$x=100y−1​ is a real number and satisfies $f(x) = y$f(x)=y.

Therefore, $f$f is onto.

---

Since $f$f is both one-one and onto, $f$f is a bijection.

The function $f(x) = 100x + 1$f(x)=100x+1 is both one-one and onto.