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: Proving \(f(x)=\cos x\) is one-one:
On \(\left[0,\dfrac\pi2\right]\), \(\cos x\) is a strictly decreasing function (it steadily falls from \(\cos0=1\) to \(\cos(\pi/2)=0\)). A strictly monotonic function can never take the same value twice, so \(f\) is one-one.

Step 2: Proving \(g(x)=\sin x\) is one-one:
On the same interval, \(\sin x\) is strictly increasing (from \(\sin0=0\) to \(\sin(\pi/2)=1\)), so by the same reasoning \(g\) is one-one.

Step 3: Checking \(f+g\):
\((f+g)(x) = \cos x+\sin x\). Evaluate it at the two endpoints:
\[ (f+g)(0) = \cos0+\sin0 = 1+0 = 1 \]
\[ (f+g)\left(\frac{\pi}{2}\right) = \cos\frac{\pi}{2}+\sin\frac{\pi}{2} = 0+1 = 1 \]

Step 4: Drawing the conclusion:
Both \(x=0\) and \(x=\pi/2\) give the same output, \(1\), for \(f+g\), even though \(0\ne\pi/2\). Since two different inputs give the same output, \(f+g\) is NOT one-one.

Final Answer:
\(f\) and \(g\) are individually one-one, but \((f+g)(0)=(f+g)(\pi/2)=1\) shows \(f+g\) is not one-one. \[ \boxed{f,g \text{ one-one}; f+g \text{ not one-one, since } (f+g)(0)=(f+g)(\pi/2)=1} \]