Question

If log2 $\left( {a{x^2} + x + a} \right) \geqslant 1$ $x \in R$ then find the exhaustive set of values of a.

Answer 1

Hint: We know log a is defined only when a > 0. Here log2 $\left( {a{x^2} + x + a} \right) \geqslant 1$ is defined when $a{x^2} + x + a \geqslant 0$ & $a{x^2} + x + a \geqslant 2$. Use both inequalities to find the exhaustive set of value of a.

Complete step by step solution: We are given,
${\log _2}\left( {a{x^2} + x + a} \right) \geqslant 1$
For log to be defined, we must have
$a{x^2} + x + a \geqslant 0$

It is only possible when coefficient of ${x^2} > 0$ & Discriminant of quadratic equation $a{x^2} + x + a < 0$

i.e.$\boxed{a > 0}$ →(1)

& Discriminant of $a{x^2} + x + a$

is equal to $1^{2}-4a^{2}$

i.e. Discriminant < 0

or, ${\text{D}} < 0$
or, $1 - 4{a^2} < 0$

$\left( {1 - 2a} \right)\left( {1 + 2a} \right) < 0$

Since ${\log _2}\left( {a{x^2} + x + a} \right) \geqslant 1$

Here, the base of log is 2 which is greater than 1.
So, it won’t change the sign of inequality when we remove the log.

Removing log from ${\log _2}\left( {a{x^2} + x + a} \right) \geqslant 1$
We get,
$a{x^2} + x + a \geqslant {2^1}$
$ \Rightarrow a{x^2} + x + a \geqslant 2$

It is possible only when the coefficient of x2 is greater than O and Discriminant of the quadratic equation should be less than or equal to 0

i.e. $a > 0$

& Discriminant $ \leqslant 0$

$ \Rightarrow D < 0$
${1^2} - 4a(a - 2) \leqslant 0$
$\underbrace {1 - 4{a^2} + 8a}_ \downarrow \leqslant 0$
Carrying this to right of inequality we get,

$4{a^2} - 8a - 1 \geqslant 0$

Now we will calculate roots of quadratic equation $4{a^2} - 8a - 1 = 0$

We are going to use the formula to find the roots of quadratic eq. which is

$\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
i.e, [for quadratic equation $a{x^2} + bx + c = 0$ roots are$\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$]

Solving roots of $4{a^2} - 8a - 1 = 0$

We get,
$a = - \dfrac{{( - 8) \pm \sqrt {{{\left( { - 8} \right)}^2} - \left( { - 1} \right)\left( 4 \right)\left( 4 \right)} }}{{2 \times 4}}$
$ = \dfrac{{8 \pm \sqrt {64 + 16} }}{8}$
$ = \dfrac{{8 \pm \sqrt {80} }}{8}$
$ = \dfrac{{1 \pm \sqrt {80} }}{8}$
$ = \dfrac{{1 \pm 4\sqrt 5 }}{8}$
$a = \dfrac{{1 \pm \sqrt 5 }}{2}$
$\therefore {\text{Roots}}\;\;{\text{are}}\;\;{\text{1 + }}\dfrac{{\sqrt 5 }}{2},\;\;\;1 - \dfrac{{\sqrt 5 }}{2}$

We had $4{a^2} - 8a - 1 \geqslant 0$
\[ \Rightarrow \left[ {a - \left( {1 + \dfrac{{\sqrt 5 }}{2}} \right)} \right]\left[ {a - \left( {1 - \dfrac{{\sqrt 5 }}{2}} \right)} \right] \geqslant 0\]
\[ \Rightarrow \boxed{a \geqslant 1 + \dfrac{{\sqrt 5 }}{2}\& a \leqslant 1 - \dfrac{{\sqrt 5 }}{2}}\]→(3)

From 1, 2 and 3

i.e $a > 0,\;\;\dfrac{{ - 1}}{2} < a < \dfrac{1}{2}$ and
\[a \geqslant 1 + \dfrac{{\sqrt 5 }}{2}\& a \leqslant 1 - \dfrac{{\sqrt 5 }}{2}\]

the exhaustive set of values of a are

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

Note: We can’t include a $ \in \left( {\dfrac{{ - 1}}{2},\;0} \right)$ and $a \leqslant 1 - \dfrac{{\sqrt 5 }}{2}$ to exhaustive set because a > 0 always. So we have to rule out $a \in \left( {\dfrac{{ - 1}}{2},0} \right)$ i.e in the interval $\left( {\dfrac{{ - 1}}{2},0} \right)$ a is less than 0. So you must really take care when you are writing the exhaustive set of values of x for some function