Question

lf $x^2+2ax+a<0\forall x\in [1,2]$ then

• A. $a\in (-\infty ,-\frac{4}{5})$
• B. $a\in (0,\infty )$
• C. $a\in (-\frac{4}{5},0)$
• D. $a\in (\frac{4}{5},\infty )$

Answer 1

Answer: (A)

$a\in (-\infty ,-\frac{4}{5})$

$1+2a+a<0$
$a<\frac{-1}{3}$ --- (1)
$4+4a+a<0$
$a<\frac{-4}{5}$ --- (2)
$\therefore a\in (-\infty ,-\frac{4}{5})$