1.

If \( \log _{2}\left(a x^{2}+x+a\right) \geq 1 \forall x \in R \), then exhaustive set of values of ' \( a \) ' is:

Answer»

\(log_2 (ax^2 + x +a) \ge 1\)

⇒ \(ax^2 + x +a \ge 2^1\)

⇒ \(ax^2 +x +a -2 \ge 0\)

Then \(\triangle \le 0\)

\(b^2 - 4ac \le 0\)

⇒ \(1 - 4\times a \times (a -2)\le 0\)

⇒ \(4a^2 - 8a \ge 1\)

⇒ \(4a^2 - 8a -1 \ge 0\)

⇒ \(a \in \left(- \infty , \frac{2-\sqrt5}{2}\right] \cup \left[\frac{2 + \sqrt5}{2}, \infty\right)\)    ........(1)

Also to defined log function, we have 

\(ax^2 + x +a > 0\)

\(\therefore D < 0\)

⇒ \(b^2 - 4ac <0\)

⇒ \(1 - 4 \times a \times a <0\)

⇒ \(4a^2 >1\)

⇒ \(a \in \left(- \infty , \frac12\right)\cup \left(\frac12, \infty\right)\)        ......(2)

From (1) & (2), we have

\(a \in \left(\left(- \infty , \frac{2-\sqrt5}{2}\right] \cup \left[\frac{2 + \sqrt5}{2}, \infty\right)\right)\cup\left(\left(- \infty , \frac{2-\sqrt5}{2}\right] \cup \left[\frac{2 + \sqrt5}{2}, \infty\right)\right)\)

⇒ \(a \in \left(- \infty , \frac{-1}2\right)\cup \left(\frac{2+\sqrt5}{2}, \infty\right)\).


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