Mathematics - Functions Question with Solution | TestHub

Since the function $f$ is bijective, therefore $f$ is onto. Therefore range of $f=B$. Let $y=-x^2+6x-8$ $\Rightarrow {\mathrm{x}}^2-6\mathrm{x}+(8+\mathrm{y})=0$ $\Rightarrow x=\frac{6\pm \sqrt{36-4(8+y)}}{2}=\frac{6\pm \sqrt{4-4y}}{2}$ For $x$ to be real, $4-4y\ge 0\Rightarrow y\le 1$ $\therefore \mathrm{B}=$ range of $\mathrm{F}=(-\infty ,1]$