Subject: General Aptitude · Chapter: Number System · Exam: · Marks: · Difficulty:
If $x$ is a real number, then $x^{2}+x+1$ is
(a)less than $\frac{3}{4}$
(b)zero for at least one value of $x$
(c)always negative
(d)greater than or equal to $\frac{3}{4}$
Answer
Answer (as printed): D
Explanation
$\left(x^{2}+x+1\right)=\left(x^{2}+x+\frac{1}{4}\right)+\frac{3}{4}$ $=\left(x+\frac{1}{2}\right)^{2}+\frac{3}{4} \geq \frac{3}{4} \quad\left[\because\left(x+\frac{1}{2}\right)^{2} \geq 0\right]$ Hence, $\left(x^{2}+x+1\right)$ is greater than or equal to $\frac{3}{4}$.
Explanation as extracted from the printed page; notation may be imperfect.