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BITSAT 2009
Exam
BITSAT
Year
2009
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities

BITSAT 2009 · Mathematics · Trigonometric Ratios & Identities

If the function (f:(-∞, ∞) arrow B) defined by f(x)=-x²+6 x-8 is...

If the function (f:(-\infty, \infty) \rightarrow B) defined by f(x)=−x2+6x−8f(x)=-x²+6 x-8 is bijective, then B=B=

Options

  1. A([1, \infty))
  2. B((-\infty, 1])
  3. C((-\infty, \infty))
  4. DNone of these

Correct answer

B. ((-\infty, 1])

Step-by-step solution

Since the function ff is bijective, therefore ff is onto. Therefore range of f=Bf=B. Let y=−x2+6x−8y=-x²+6 x-8 (\Rightarrow \mathrm{x}^2-6 x+(\mathrm{x}+(8+y)=\mathrm{y})=0) (\Rightarrow x=6±36−4(8+y)2=6±4−4y2\frac{6 \pm \sqrt{36-4(8+y)}}{2}=\frac{6 \pm \sqrt{4-4 y}}{2}) For xx to be real, (4-4 y \geq 0 \Rightarrow y \leq 1) (\therefore B=\mathrm{B}=) range of (F=(−∞,\mathrm{F}=(-\infty,1])
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