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Question
The function
f : A → B defined by
f (x) = - x2 + 6x - 8 is a bijection if
Options
A = (- ∞ , 3] and B = ( - ∞, 1 ]
A = [- 3 , ∞) and B = ( - ∞, 1 ]
A = (- ∞ , 3] and B = [ 1 ,∞)
A = [3 ,∞ ) and B = [ 1 ,∞ )
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Solution
\[A = ( - \infty , 3] \text{and }B = ( - \infty , 1]\]
\[f\left( x \right) = - x^2 + 6x - 8 , \text{is a polynomial function}\]
\[\text{And the domain of polynomial function is real number} . \]
\[ \therefore x \in R\]
\[f(x) = - x^2 + 6x - 8\]
\[ = - \left( x^2 - 6x + 8 \right)\]
\[ = - \left( x^2 - 6x + 9 - 1 \right)\]
\[ = - \left( x - 3 \right)^2 + 1\]
\[\text{Maximum value of} - \left( x - 3 \right)^2 \text{woud be } 0\]
\[ \therefore \text{Maximum value of} - \left( x - 3 \right)^2 + 1 \text{woud be} 1\]
\[ \therefore f(x) \in ( - \infty , 1]\]
\[\text{We can see from the given graph that function is symmetrical about x = 3 & the given function is bijective .} \]
\[\text{So, x would be either} ( - \infty , 3 ] or [ 3, \infty )\]
\[\text{The correct option which satisfy A and B both is}: \]
\[A = ( - \infty , 3] \text{ and }B = ( - \infty , 1]\]
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