Question
If a function $\text{f}:[2,\infty)\rightarrow\ \text{B}$ defined by f(x) = x2 - 4x + 5 is a bijection, then B =
- $\text{R}$
- $[1,\infty)$
- $[4,\infty)$
- $[5,\infty)$
Solution:
Since f is a bijection, co-domain of f = range of f
⇒ B = range of f
Given: f(x) = x2 - 4x + 5
Let f(x) = y
⇒ y = x2 - 4x + 5
⇒ x2 - 4x + (5 - y) = 0
$\because$ Discrimant, $\text{D}=\text{b}^2-4\text{ac}\geq0,$
$(-4)^2-4\times1\times(5-\text{y})\geq0$
$\Rightarrow\ 16-20+4\text{y}\geq0$
$\Rightarrow\ 4\text{y}\geq4$
$\Rightarrow\ \text{y}\geq1$
$\Rightarrow\ \text{y}\in[1,\infty)$
⇒ Range of
$\text{f}=[1,\infty)$$\Rightarrow\ \text{B}=[1,\infty)$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Let $\mathrm{R}$ be a relation on $\mathrm{A} \times \mathrm{B}$ define by $(\mathrm{a}, \mathrm{b}) \mathrm{R}(\mathrm{c}, \mathrm{d})$ if and only if $3 \mathrm{ad}-7 \mathrm{bc}$ is an even integer. Then the relation $\mathrm{R}$ is