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Functions — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSFunctions1 Mark
MCQ
The function $f : A \rightarrow B$ defined by $f(x) = -x^2 + 6x- 8$ is a bijection if,
  • ✓
    $\text{A}=(-\infty,3]$ and $\text{B}=(-\infty,1]$
  • B
    $\text{A}=[-3,\infty)$ and $\text{B}=(-\infty,1]$
  • C
    $\text{A}=(-\infty,3]$ and $\text{B}=[1,\infty)$
  • D
    $\text{A}=[3,\infty)$ and $\text{B}=[1,\infty)$
✓

Answer

Correct option: A.
$\text{A}=(-\infty,3]$ and $\text{B}=(-\infty,1]$
$f(x) = -x^2 + 6x - 8,$ is a polynomial function and the domain of polynomial function is real number.
$\therefore\ \text{x}\in\text{R}$
$f(x) = -x^2 + 6x - 8$
$= -(x^2 - 6x + 8)$
$= -(x^2 - 6x + 9 - 1)$
$= -(x - 3)^2 + 1$
Maximum value of $-(x - 3)^2$ woud be $0$
$\therefore$ Maximum value of $-(x - 3)^2 + 1$ woud be $1$
$\therefore\ \text{f(x)}\in(-\infty,1]$

We can see from the given graph that function is symmetrical about $x = 3$ and the given function is bijective.
So$, x$ would be either $(-\infty,3]\text{ or }[3,\infty)$
The correct option which satisfy $A$ and $B$ both is :
$\text{A}=(-\infty,3]$ and $\text{B}=(-\infty,1]$

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