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 $a^2-2 a b+b<0$ 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 $a ^2-2 ab + b <0$
$\therefore (-4)^2+8 b+b \leq 0 \Rightarrow b \leq-\frac{16}{9}$
\& $ b \leq 0$
$\therefore b \in\left(-\infty,-\frac{16}{9}\right]$
$\therefore $ greatest possible integral value of $b=-2$