UPSC NDA II 2018 Mathematics Solved Paper

Concept:The range of a quadratic function on an open interval is found using its vertex as the minimum (since coefficient of $x^2$ is positive) and the boundary values, excluding endpoints if the domain is open.Explanation:The given function is $f(x) = x^2 - 4x + 5$. This is a quadratic with $a=1 > 0$, so its graph is a convex parabola opening upwards. Vertex: $x = \frac{-b}{2a} = \frac{-(-4)}{2 \times 1} = 2$, $y = f(2) = 2^2 - 4 \times 2 + 5 = 1$. Thus the minimum value is $1$, attained at $x=2$, which lies inside $(1,4)$. Now evaluate $f$ at the endpoints (but not included): $f(1) = 1 - 4 + 5 = 2$, $f(4) = 16 - 16 + 5 = 5$. Since the parabola opens upward and the vertex is the lowest point, the function increases to $5$ as $x$ approaches $4$ from the left, but $5$ is never actually reached because $4$ is excluded. Therefore, the range includes $1$ (attained) and all values from $1$ up to but not including $5$. Answer:$[1, 5)$ Thus option C is correct.