Question

If the function \(f: [0, \frac{\pi}{2}] \rightarrow \mathbb{R}\) is given by \(f(x) = \sin x\) and function \(g: [0, \frac{\pi}{2}] \rightarrow \mathbb{R}\) is given by \(g(x) = \cos x\), then prove that f and g are one-one but \(f + g\) is not one-one.

Hint:

To prove a function is one-one, showing it's strictly monotonic (using its derivative) is a robust method. To prove a function is *not* one-one, you only need to find a single counterexample: two different inputs that give the same output. Testing boundary points is often a good first step.

Answer 1

Step 1: Understanding the Concept:
A function is one-one (injective) if distinct elements in the domain map to distinct elements in the codomain. That is, if \(f(x_1) = f(x_2)\), then \(x_1 = x_2\). A common way to prove a differentiable function is one-one on an interval is to show that it is strictly monotonic (either strictly increasing or strictly decreasing) on that interval. This can be checked using its first derivative.
Step 2: Detailed Explanation:
Part 1: Proving f(x) = sin(x) is one-one on \([0, \frac{\pi}{2}]\).
To check for monotonicity, we find the first derivative of f(x).
\[ f'(x) = \frac{d}{dx}(\sin x) = \cos x \] In the interval \((0, \frac{\pi}{2})\), \(f'(x) = \cos x>0\).
Since the derivative is positive, the function f(x) is strictly increasing on \([0, \frac{\pi}{2}]\).
A strictly monotonic function is always one-one. Therefore, f(x) is one-one.
Part 2: Proving g(x) = cos(x) is one-one on \([0, \frac{\pi}{2}]\).
We find the first derivative of g(x).
\[ g'(x) = \frac{d}{dx}(\cos x) = -\sin x \] In the interval \((0, \frac{\pi}{2})\), \(\sin x>0\), which means \(g'(x) = -\sin x<0\).
Since the derivative is negative, the function g(x) is strictly decreasing on \([0, \frac{\pi}{2}]\).
A strictly monotonic function is always one-one. Therefore, g(x) is one-one.
Part 3: Proving (f + g)(x) is not one-one on \([0, \frac{\pi}{2}]\).
Let \(h(x) = (f + g)(x) = \sin x + \cos x\).
To prove that h(x) is not one-one, we need to find two distinct values \(x_1\) and \(x_2\) in the domain \([0, \frac{\pi}{2}]\) such that \(h(x_1) = h(x_2)\).
Let's evaluate h(x) at the endpoints of the interval: \[ h(0) = \sin(0) + \cos(0) = 0 + 1 = 1 \] \[ h(\frac{\pi}{2}) = \sin(\frac{\pi}{2}) + \cos(\frac{\pi}{2}) = 1 + 0 = 1 \] We have \(h(0) = h(\frac{\pi}{2}) = 1\), but \(0 \neq \frac{\pi}{2}\).
Since two different inputs (0 and \(\frac{\pi}{2}\)) produce the same output (1), the function \(f + g\) is not one-one.
Step 3: Final Answer:
We have shown that f(x) and g(x) are strictly monotonic and therefore one-one on the given interval, while \(f+g\) is not one-one as it has the same value at the endpoints. Hence proved.