Question

If log2(ax2+x+a)>=1 then exhaustive set of values of a is

Answer 1

$log_2(ax^2+x+a)\geq 1$

for log to be defined,

(ax² + x + a) > 0

it is possible only when coefficient of x² > 0 and discriminant < 0

i.e., a > 0

and 1² - 4a² < 0

or, (1 - 2a)(1 + 2a) < 0

or, -1/2 < a < 1/2

also $log_2(ax^2+x+a)\geq log_22$

here base is greater than 1 so, doesn't change sign of inequality if we remove log.

i.e., (ax² + x + a) ≥ 2

or, ax² + x + a - 2 ≥ 0

it is possible when a > 0 and D ≤ 0

D = 1² - 4a(a - 2) ≤ 0

or, 1 - 4a² + 8a ≤ 0

or, 4a² - 8a - 1 ≥ 0

or, {a - (8 +√80)/8}{a - (8 - √80)/8} ≥ 0

or, a ≥ (8 + √80)/8 , a ≤ (8 - √80)/8

or, a ≥ 1 + √5/2 , a ≤ 1 - √5/2

here we get,

a > 0

-1/2 < a < 1/2

a ≥ 1 + √5/2 or, a ≤ 1 - √5/2

putting all values in number line to find exhaustive set of a.

then we get

$a\in\left(-\frac{1}{2},1-\frac{\sqrt{5}}{2}\right]\cup\left(0,\frac{1}{2}\right)\cup\left[1+\frac{\sqrt{5}}{2},\infty\right)$