Q59 / 125 · BITSAT 2009 · Mathematics

If the function \(f:(-\infty, \infty) \rightarrow B\) defined by \(f(x)=-x^2+6 x-8\) is bijective, then \(B=\)

  • A
    \([1, \infty)\)
  • Correct answer
    B
    \((-\infty, 1]\)
  • C
    \((-\infty, \infty)\)
  • D
    None of these

Solution

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 \(\mathrm{F}=(-\infty, 1]\)

Hence, the correct answer is option: 'B'.

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