For any two distinct elements x1 and x2 in [0, π/2]
We know that
sin x1 ≠ sin x2 and cos x1 ≠ cos x2
So we get
f(x1) ≠ f(x2) and g(x1) ≠ g(x2)
Here, both f and g are one-one
We know that
(f + g) (x) = f( x) + g(x) = sin x + cos x
By substituting the values
(f + g) (0) = sin 0 + cos 0 = 1
(f + g) (π/2) = sin π/2 + cos π/2 = 1
We know that 0 ≠ π/2
Here, (f + g) (0) = (f + g) (π/2)
So, f + g is not one-one.