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Question
If a function\[f : [2, \infty )\text{ to B defined by f}\left( x \right) = x^2 - 4x + 5\] is a bijection, then B =
Options
R
[1, ∞)
[4, ∞)
[5, ∞)
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Solution
Since f is a bijection, co-domain of f = range of f
⇒ B = range of f
\[\text{Given}: f\left( x \right) = x^2 - 4x + 5\]
\[\text{Let}f\left( x \right) = y\]
\[ \Rightarrow y = x^2 - 4x + 5\]
\[ \Rightarrow x^2 - 4x + \left( 5 - y \right) = 0\]
\[ \because \text{Discrimant}, D = b^2 - 4ac \geq 0, \]
\[ \left( - 4 \right)^2 - 4 \times 1 \times \left( 5 - y \right) \geq 0\]
\[ \Rightarrow 16 - 20 + 4y \geq 0\]
\[ \Rightarrow 4y \geq 4\]
\[ \Rightarrow y \geq 1\]
\[ \Rightarrow y \in [1, \infty )\]
\[ \Rightarrow \text{Range of f} = [1, \infty )\]
\[ \Rightarrow B = [1, \infty )\]
So, the answer is (b).
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