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RELATIONS AND FUNCTIONS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsRELATIONS AND FUNCTIONS1 Mark
Question
The function f : A → B defined by f(x) = -x2 + 6x- 8 is a bijection if,
  1. $\text{A}=(-\infty,3]$ and $\text{B}=(-\infty,1]$
  2. $\text{A}=[-3,\infty)$ and $\text{B}=(-\infty,1]$
  3. $\text{A}=(-\infty,3]$ and $\text{B}=[1,\infty)$
  4. $\text{A}=[3,\infty)$ and $\text{B}=[1,\infty)$
✓

Answer

  1. $\text{A}=(-\infty,3]$ and $\text{B}=(-\infty,1]$

Solution:

f(x) = -x2 + 6x - 8, is a polynomial function and the domain of polynomial function is real number.

$\therefore\ \text{x}\in\text{R}$

f(x) = -x2 + 6x - 8

= -(x2 - 6x + 8)

= -(x2 - 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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