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Let f: [0, π/2] → R: f(x) = sin x and g: [0, π/2] → R: g(x) = cos x. Show that each one of f and g is one-one but (f + g) is not one-one.

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For any two distinct elements x1 and xin [0, π/2]

We know that

sin x≠ 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.

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