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If the function \(f:(-\infty, \infty) \rightarrow B\) defined by \(f(x)=-x^2+6 x-8\) is bijective, then \(B=\)
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The correct answer is:
\((-\infty, 1]\)
Since the function \(f\) is bijective, therefore \(f\) is onto. Therefore range of \(f=B\).
Let \(y=-x^2+6 x-8\)
\(\Rightarrow \mathrm{x}^2-6 \mathrm{x}+(8+\mathrm{y})=0\)
\(\Rightarrow x=\frac{6 \pm \sqrt{36-4(8+y)}}{2}=\frac{6 \pm \sqrt{4-4 y}}{2}\)
For \(x\) to be real, \(4-4 y \geq 0 \Rightarrow y \leq 1\)
\(\therefore \mathrm{B}=\) range of \(\mathrm{F}=(-\infty, 1]\)
Let \(y=-x^2+6 x-8\)
\(\Rightarrow \mathrm{x}^2-6 \mathrm{x}+(8+\mathrm{y})=0\)
\(\Rightarrow x=\frac{6 \pm \sqrt{36-4(8+y)}}{2}=\frac{6 \pm \sqrt{4-4 y}}{2}\)
For \(x\) to be real, \(4-4 y \geq 0 \Rightarrow y \leq 1\)
\(\therefore \mathrm{B}=\) range of \(\mathrm{F}=(-\infty, 1]\)
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