Question

The function $f:A \to B$ defined by $f(x) = - {x^2} + 6x - 8$ is a bijection, if
${\text{A}}{\text{. A = ( - }}\infty {\text{,3];B = ( - }}\infty {\text{,1]}}$
${\text{B}}{\text{. A = [ - 3,}}\infty ){\text{;B = ( - }}\infty {\text{,1]}}$
${\text{C}}{\text{. A = ( - }}\infty {\text{,3];B = [1,}}\infty )$
${\text{D}}{\text{. A = [3,}}\infty ){\text{;B = [1,}}\infty )$

Answer 1

Hint – A bijective function is a function between the elements of two sets where each element of one set is paired with exactly one element of another set, and each element of another set is paired with exactly one element of the first set.

Complete step-by-step solution -
Use the concept that the given function $f(x) = - {x^2} + 6x - 8$ is a polynomial and domain of polynomial function is real i.e., $x \in R$.
We have been given that function $f:A \to B$defined by $f(x) = - {x^2} + 6x - 8$.

Drawing the graph of f(x), we get the sketch as-

Now, if we simplify the polynomial-
$
  f(x) = - {x^2} + 6x - 8 \\
   = - ({x^2} - 6x + 8) \\
   = - ({x^2} - 6x + 9 - 1) \\
   = - {(x - 3)^2} + 1 \\
$
Now, we can say the maximum value of $ - {(x - 3)^2}$would be 0.
Therefore, the maximum value of $ - {(x - 3)^2} + 1$ would be 1.
$\therefore f(x) \in ( - \infty ,1]$$[\because \min (f(x)) = - \infty ]$
We can also say from the graph that the function is symmetrical about x = 3 and the given function is bijective.
So, x would be either $( - \infty ,3]or[3,\infty )$.
The correct option which satisfies A and B both is:
Option ${\text{A}}{\text{. A = ( - }}\infty {\text{,3];B = ( - }}\infty {\text{,1]}}$.
Note – Whenever such types of questions appear, then first draw the graph of f(x), and then simplify it. From the simplified equation, we can find the maximum value of f(x) and from the graph we can see where it is symmetrical. Hence, in this way we can get both A and B ranges for which the function is bijection.