Q.Consider a function $f: \left[0, \frac{\pi}{2}\right] \to \mathbb{R}$ given by $f(x) = \sin x$ and $g: \left[0, \frac{\pi}{2}\right] \to \mathbb{R}$ given by $g(x) = \cos x$. Show that $f$ and $g$ are one-one, but $f + g$ is not one-one.

One-one function. A function is one-one (injective) when different inputs always give different outputs. A strictly increasing or strictly decreasing function is automatically one-one, so we check monotonicity for $f$ and $g$, then look for a repeated output for $f+g$.

Reasoning:

1. $f(x) = \sin x$ is strictly increasing on $\left[0, \frac{\pi}{2}\right]$, so distinct inputs give distinct outputs $\Rightarrow$ $f$ is one-one.
2. $g(x) = \cos x$ is strictly decreasing on $\left[0, \frac{\pi}{2}\right]$, so it too is one-one (decreasing still means no repeats).
3. For $h(x) = f(x) + g(x) = \sin x + \cos x$, test the endpoints: $h(0) = \sin 0 + \cos 0 = 1$ and $h\!\left(\frac{\pi}{2}\right) = \sin\frac{\pi}{2} + \cos\frac{\pi}{2} = 1$. …

The core idea here is monotonicity and symmetry. A function is one‑one (injective) on an interval if it is strictly increasing or strictly decreasing — no two distinct inputs give the same output. Both $\sin x$ and $\cos x$ behave beautifully on $[0,\pi/2]$: one climbs, the other falls, each strictly. But their sum? That’s a different story.

Let’s walk through it.

1. Check $f(x) = \sin x$ on $[0,\pi/2]$.

   The derivative is $f'(x) = \cos x$. On $(0,\pi/2)$, $\cos x > 0$, so $f$ is strictly increasing. A strictly monotonic function is always one‑one.

   Tip

   You don’t even need calculus: $\sin x$ is known to increase from $0$ to $1$ on this interval — no repeats possible.

2. Check $g(x) = \cos x$ on $[0,\pi/2]$.

   Here $g'(x) = -\sin x$, which is negative on $(0,\pi/2)$. So $g$ is strictly decreasing, hence also one‑one.

   Watch out

   A common mistake is to think “decreasing” means not one‑one — but strictly decreasing is perfectly injective.

3. Now examine $h(x) = f(x) + g(x) = \sin x + \cos x$.

   To test injectivity, we look for two distinct $x_1, x_2$ in $[0,\pi/2]$ with $h(x_1) = h(x_2)$.

   A useful identity:

This shifts the sine wave left by $\pi/4$. On $[0,\pi/2]$, the argument $x + \pi/4$ runs from $\pi/4$ to $3\pi/4$.

Over $[\pi/4, 3\pi/4]$, $\sin$ is not one‑one — it rises to a maximum at $\pi/2$ then falls. So $h$ is symmetric:

$h(x) = h\!\left(\frac{\pi}{2} - x\right)$ …

Method: Testing One-One (Injectivity), Including for a Sum of Functions

To prove a function is one-one, show strict monotonicity or use the algebraic test $f(x_1)=f(x_2)\Rightarrow x_1=x_2$. To prove it is not one-one, find two distinct inputs giving the same output. This problem needs both directions.

Steps

Step 1: For each individual function, argue strict monotonicity.

A strictly increasing or strictly decreasing function is automatically one-one. On $[0,\tfrac{\pi}{2}]$, $\sin x$ strictly increases (derivative $\cos x > 0$) and $\cos x$ strictly decreases (derivative $-\sin x < 0$), so each is one-one. Note that decreasing still counts as injective.

Step 2: For the sum, do not assume the property carries over.

Injectivity is not preserved under addition, so treat $f+g$ as a brand-new function and test it directly rather than inferring anything from $f$ and $g$.

Step 3: Look for hidden symmetry using an identity. …

Common Mistakes

Mistake 1: Assuming the sum of one-one functions is one-one.

Why it's wrong: $\sin x$ and $\cos x$ are each one-one on $[0,\tfrac{\pi}{2}]$, but $\sin x + \cos x$ is not — injectivity is not preserved under addition. Correct approach: test $f+g$ separately, e.g. via the identity $\sin x+\cos x=\sqrt{2}\sin(x+\tfrac{\pi}{4})$.

Mistake 2: Believing a decreasing function cannot be one-one.

Why it's wrong: $\cos x$ is strictly decreasing on the interval, and strictly decreasing functions are perfectly injective. Correct approach: one-one requires strict monotonicity in either direction, so decreasing counts. …