RRB Technician Grade I 20 Dec 2024 Shift 1 Paper

Concept:For a quadratic expression $ax^2 + bx + c$ where $a > 0$, the minimum value occurs at $x = -\frac{b}{2a}$ and equals $\frac{4ac - b^2}{4a}$.Explanation:First, rewrite the expression in standard quadratic form:$8x + 14 + 3x^2 = 3x^2 + 8x + 14$Here, $a = 3$, $b = 8$, and $c = 14$.Since $a = 3 > 0$, the parabola opens upwards, so the expression has a minimum value.Using the formula for the minimum value:$\frac{4ac - b^2}{4a} = \frac{4(3)(14) - (8)^2}{4(3)}$$\Rightarrow \frac{168 - 64}{12} = \frac{104}{12} = \frac{26}{3}$Converting to a mixed fraction, $\frac{26}{3} = 8\frac{2}{3}$.Alternatively, substituting $x = -\frac{4}{3}$ into $3x^2 + 8x + 14$ also gives the same result.Answer:The minimum value of $8x + 14 + 3x^2$ is $8\frac{2}{3}$, which matches option A.