Question:
Consider a function f :(0, pie/2)...R given by f (x) = sinx and g :(0, pie/2)...R given by g (x)=cos x show that f and g are one- one , but f+g is not one- one.
Answer:
We know that for all x1 , x2 ∈ [0, π/2], x1 ≠ x2
=> sin x1 ≠ sin x2 and cos x1 ≠ cos x2
So, f and g both are one-one
Again, we know that f + g: [0, π/2] -> R is defined by
=> (f + g)(x) = f(x) + g(x) = sin x + cos x
So, (f + g)(π/2) = sin π/2 + cos π/2 = 1 + 0 = 1
Since 0 ≠ π/2 but (f + g)(0) = (f + g)(π/2)
Therefore, f + g is not one-one.
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