Question

Find the set of all possible real values of a such that the inequality x - ( a - 1 )) ( x - ( a² + 2 ) ) < 0 holds for all x ∈ ( -1 , 3 )

Answer 1

$\textbf{Given:}$

$\textsf{Inequality}$

$\mathsf{(x-(a-1))\;(x-(a^2+2))\;<\;0\;holds\;for\;all\;x\;\in\;(-1,3)}$

$\textbf{To find:}$

$\textsf{The set of all possible real values of a}$

$\textbf{Solution:}$

$\underline{\textsf{Concept used:}}$

$\boxed{\mathsf{(x-a)(x-b)\;<\;0\;is\;true\;when\;a\;<\;x\;<\;b}}$

$\mathsf{Consider,}$

$\mathsf{(x-(a-1))\;(x-(a^2+2))\;<\;0}$

$\implies\mathsf{a-1\,<\,x\,<\,a^2+2}$

$\mathsf{But\,-1\,<\,x\,<\,3}$

$\implies\mathsf{a-1=-1}$

$\implies\mathsf{a=0}$

$\mathsf{and}$

$\mathsf{a^2+2=3}$

$\mathsf{a^2=1}$

$\implies\mathsf{a=\pm\,1}$

$\textsf{a=-1 does not satisfy the given inequality}$

$\therefore\textbf{The possible values of a are all real no. from 0 to 1}$