Question

If both the roots of $x^2+2ax+a=0$ are less than $2$, then the set values of ${}^{\prime }{a}^{\prime }$ is

• A. $(-2,0]\cup [1,\infty )$
• B. $(-\frac{4}{5},0]\cup [1,\infty )$
• C. $(-\infty ,0]\cup [1,\infty )$
• D. $[-\frac{4}{5},0)\cup (1,\infty )$

Answer 1

The correct option is B $(-\frac{4}{5},0]\cup [1,\infty )$

Let $f\left(x\right)=x^2+2ax+a$, whose roots are $\alpha ,\beta$
Given: $\alpha ,\beta <2$


Conditions:
$\left(i\right)\Delta \ge 0\Rightarrow 4a^2-4a\ge 0$
$\Rightarrow a(a-1)\ge 0\Rightarrow a\le 0\text{or}a\ge 1\dots \left(1\right)$

$\left(ii\right)-\frac{B}{2A}<2\Rightarrow -\frac{2a}{2}<2\Rightarrow a>-2\dots \left(2\right)$

$\left(iii\right)f\left(2\right)>0\Rightarrow 5a+4>0\Rightarrow a>-\frac{4}{5}\dots \left(3\right)$

From $\left(1\right),\left(2\right)\text{and}\left(3\right)$, we get
$a\in (-\frac{4}{5},0]\cup [1,\infty )$