Question

If complete set of values of a for which $f(x)=x^2-(a+2)x+4$ is not injective in $[-1,1]$, satisfies the inequality $a^2-2ab+b<0$ then greatest possible integral value of $b$ will be

• A. -2
• B. -1
• C. 0
• D. 1

Answer 1

Answer: (A)

$\Theta f(x)$ is not injective.
$\therefore -1<\frac{a+2}{2}<1\Rightarrow -2<a+2<2\Rightarrow -4<a<0$
$\Theta a\in (-4,0)$ satisfies $a^2-2ab+b<0$
$\therefore (-4{)}^2+8b+b\le 0\Rightarrow b\le -\frac{16}{9}$
\& $b\le 0$
$\therefore b\in \left(-\infty ,-\frac{16}{9}\right]$
$\therefore$ greatest possible integral value of $b=-2$