Problem 143 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

If a function \(f:[2, \infty) \rightarrow B\) defined by \(f(x)=x^{2}-4 x+5\) is a bijection, then find \(B\).

Step-by-Step Solution

Answer

The range of the function is \(B = [1, \infty)\).

Step 1: Find the vertex of the parabola

The given function is \(f(x)=x^{2}-4x+5\). Rewrite it in vertex form, which is \(y=a(x-h)^2 + k\), where (h, k) is the vertex of the parabola. The given function can be rewritten as \(f(x) = (x - 2)^2 + 1\). So the vertex of the parabola is (2,1).

Step 2: Find the direction of the parabola

Since the coefficient of \(x^2\) is positive in the original equation (\(f(x)=x^{2}-4x+5\)), the parabola opens upwards.

Step 3: Find the range B

Knowing the vertex and the direction the parabola opens, we can find the range B. As the parabola opens upwards and the vertex is at y=1, its lowest point is at y=1. Because the function is defined only for \(x \geq 2\), and those x values correspond to \(y \geq 1\) we have the range \(B = [1, \infty)\).

Key Concepts

Bijection

A function is considered a bijection if it is both injective (one-to-one) and surjective (onto). To check for injectiveness, each input must map to a unique output. This means no two different inputs should yield the same output. Surjectiveness ensures that every element in the output set is mapped to by at least one input from the domain. For our function to be bijective, we require both properties to hold with the function defined over its specific domain and codomain.

In the given exercise, the function must be a bijection between the domain \( [2, \infty) \) and a suitable set \( B \) in order to fulfill the properties of both injective and surjective mapping. The relationships are crucial especially when dealing with quadratic functions, as they inform our understanding of how this function behaves and where it takes its values.

Vertex of Parabola

The vertex of a parabola described by a quadratic function provides important information about its graph. This point, indicated as ( h, k), acts as either the maximum or minimum point of the parabola, depending on its direction. The general form \( y = a(x - h)^2 + k \) efficiently helps in rewriting quadratic equations to quickly identify this point.

For our function \( f(x) = x^2 - 4x + 5 \), rewriting it as \( (x-2)^2 + 1 \) gives us the vertex at the point (2,1). Since this vertex is the lowest point when your parabola opens upwards, it becomes vital in determining the range of the function and understanding its overall behavior.

Range of a Function

The range of a function is all of the possible output values it can produce. This is determined by analyzing the behavior of the function over its entire domain. For a quadratic function like \( f(x) = x^2 - 4x + 5 \), finding the range involves looking at the vertex and the direction in which the parabola opens.

In our exercise, the parabola opens upwards because the leading coefficient is positive. The vertex at y=1 is the lowest point, so all function values will be greater than or equal to this value. Given the domain starts at x=2, these conditions inform us that the range \( B \) for a bijection would be \([1, \infty)\). This means starting from the minimum, every value upward is included in the range.