Question

Consider the equation $x^2+x+a-9<0$.

The values of the real parameter $a$ so that the given inequation is true $\forall x\in (-1,3)$

• A. $(-\infty ,-3)$
• B. $(-3,\infty )$
• C. $[9,\infty )$
• D. $(-\infty ,\frac{37}{4})$

Answer 1

The correct option is A $(-\infty ,-3)$

If the given inequality is correct $\forall xϵ(-1,3)$

Then,

$f(-1)<0$

$\Rightarrow 1-1+a-9<0$

$a<9$

Also, $f\left(3\right)<0$

$12+a-9<0$

$\Rightarrow a+3<0$

$a<-3$

Taking intersection of these two conditions .

Clearly the answer will be

$\left(A\right)aϵ(-\infty ,-3)$