Question

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

Hint:

cos x is strictly decreasing and sin x strictly increasing on [0,π/2], so both are one-one; but f+g=cosx+sinx repeats a value at the two endpoints.

Answer 1

Step 1: Use the derivative sign to prove $f$ and $g$ are one-one, instead of just describing the trend:
$f'(x)=-\sin x$, which is $\le 0$ on $[0,\pi/2]$ and strictly negative on the open interval, so $f$ is strictly decreasing, hence one-one. $g'(x)=\cos x \ge 0$ on $[0,\pi/2]$ and strictly positive on the open interval, so $g$ is strictly increasing, hence one-one.

Step 2: Combine $f$ and $g$ and simplify using the auxiliary-angle identity:
$(f+g)(x)=\cos x+\sin x=\sqrt2\sin\left(x+\dfrac{\pi}{4}\right)$.

Step 3: Study this combined function on $[0,\pi/2]$:
As $x$ runs from $0$ to $\pi/2$, the angle $x+\pi/4$ runs from $\pi/4$ to $3\pi/4$. Since $\sin$ increases up to $\pi/2$ then decreases afterward, and $\pi/2$ lies inside $(\pi/4,3\pi/4)$, the function $\sqrt2\sin(x+\pi/4)$ rises then falls over this interval, so it is not monotonic.

Step 4: Exhibit the repeated value explicitly:
At $x=0$: $\sqrt2\sin(\pi/4)=\sqrt2\cdot\dfrac{1}{\sqrt2}=1$. At $x=\pi/2$: $\sqrt2\sin(3\pi/4)=\sqrt2\cdot\dfrac{1}{\sqrt2}=1$. Same value at two different points confirms $f+g$ is not one-one.

Final Answer:
$f,g$ are one-one individually, but $f+g$ repeats the value 1 at $x=0$ and $x=\pi/2$, so it is not one-one. \[ \boxed{f,g \text{ one-one}; f+g \text{ not one-one}} \]