Question

Consider the following statements : \(\mathrm{S}_{1}\) : For an odd function \(\mathrm{f}(\mathrm{x})\), graph of \(y=f(x)\) always passes through origin. \(\mathbf{S}_{2}\) : If \(f\) and \(g\) are two bijective function then \(\mathrm{f}(\mathrm{g}(\mathrm{x}))\) is also bijective. \(\mathbf{s}_{3}:\) All points of intersection of \(y=f(x)\) and \(y=f^{-1}(x)\) lies on \(y=x\) only. State, in order, whether \(\mathrm{S}_{1}, \mathrm{~S}_{2}, \mathrm{~S}_{3}, \mathrm{~S}_{4}\) are true or false (1) T F T (2) T T F (3) \(\mathrm{FTT}\) (4) F FF

Short answer

T T T

Step-by-step solution

Evaluate Statement \ \(\mathbf{S_1}\)

An odd function \(f(x)\) satisfies the property \(f(-x) = -f(x)\). For any point \((a, f(a))\) on the graph, the point \((-a, -f(a))\) will also be on the graph. If the graph passes through the origin, then \(f(0) = 0\). Therefore, an odd function's graph must pass through the origin. Hence, \(\ \mathbf{S_1}\) is true.

Evaluate Statement \ \(\mathbf{S_2}\)

If \(f\) and \(g\) are bijective functions, they are both one-to-one and onto. The composition function \(f(g(x))\) must also be bijective. The reason is that the composition of two one-to-one functions is one-to-one, and the composition of two onto functions is onto. Hence, \(\ \mathbf{S_2}\) is true.

Evaluate Statement \ \(\mathbf{S_3}\)

For any function \(f(x)\), if \(y = f(x)\) intersects with \(y = f^{-1}(x)\), then at the point of intersection, we have \(f(a) = b\) and \(f^{-1}(b) = a\). To satisfy both equations, it must be that \(a = b\). Therefore, at any intersection point, \(y = x\). Thus, \(\ \mathbf{S_3} \) is true.

Key concepts

Odd Functions

Odd functions are a fascinating type of mathematical function with a unique property. For an odd function \(f(x)\), the property \(f(-x) = -f(x)\) must hold. This implies a kind of symmetry about the origin.
Imagine you have any point \((a, f(a))\) on the graph; the corresponding point \((-a, -f(a))\) will also be present. This symmetric behavior around the origin tells us that the graph of any odd function must pass through the point \((0,0)\).
This happens because for \(x = 0\), the equation becomes \(f(0) = -f(0)\), which simplifies to \(f(0) = 0\). Let's consider a practical example: the function \(f(x) = x^3\). This is an odd function because \((-x)^3 = -(x^3)\). We can see its graph clearly passes through the origin and showcases the odd function property.

Bijective Functions

Bijective functions are special kinds of functions crucial for many applications. A function is bijective if it is both injective (one-to-one) and surjective (onto). An injective function ensures that every element of the function's domain maps to a unique element in its codomain. In simple words, different inputs always produce different outputs.
A surjective function guarantees that every element in the codomain has at least one corresponding element in the domain. When a function is both injective and surjective, it is bijective, creating a perfect 'pairing' between elements of the domain and codomain.
For example, consider the function \(f(x) = x + 3\). This function is bijective because each input yields a unique output (injective), and every real number is an output for some input (surjective). What about the composition of bijective functions? The magic here is that if you compose two bijective functions, their composition \(f(g(x))\) also remains bijective. This is because the composition of two one-to-one functions remains one-to-one, and the composition of two onto functions remains onto.

Function Composition

Function composition is like stacking two functions to create a new function. Given two functions, \(f\) and \(g\), the composition \(f(g(x))\) means you first apply \(g\) to \(x\), and then apply \(f\) to the result of \(g(x)\). This is generally written as \((f \circ g)(x)\).
Suppose you have \(g(x) = 2x\) and \(f(x) = x + 3\). The composition \((f \circ g)(x)\) is \(f(g(x)) = f(2x) = 2x + 3\). This new function incorporates both transformations: first doubling \(x\), then adding 3.
Function composition is significant because it allows us to build complex operations out of simpler ones. When dealing with bijective functions, the beauty of composition lies in retaining the bijectivity. If both \(f\) and \(g\) are bijective, then \(f(g(x))\) will also be bijective. This is a powerful mathematical property used in various fields like cryptography, coding theory, and more.

Inverse Functions

Inverse functions are like the 'undo' button in mathematics. If you have a function \(f(x)\) that maps \(x\) to \(y\), the inverse function \(f^{-1}(x)\) maps \(y\) back to \(x\). This means \(f(f^{-1}(x)) = x\) and \(f^{-1}(f(x)) = x\).
Not all functions have inverses, though. A function must be bijective to ensure every output comes from a unique input and covers the entire range of outputs.
Think of the function \(f(x) = 2x + 3\). Its inverse would be \(f^{-1}(x) = \frac{x-3}{2}\). You can verify that by composing them: \(f(f^{-1}(x)) = 2(\frac{x-3}{2}) + 3 = x\).
Moreover, if a function intersects its inverse's graph, all points of intersection lie on the line \(y = x\). This happens because at the intersection points, \(f(a) = b\) and \(f^{-1}(b) = a\), which simplifies to \(a = b\). Thus, the points lie on the line where \(y = x\). Understanding inverse functions is critical for solving equations and modeling real-world situations where one needs to reverse processes.