Question

If ƒ(x+y) = ƒ(x)ƒ(y) ∀ x, y, ƒ(1) = 2 and α_n = f(n), n∈N then the equation of the circle having (α₁ ,α₂ ) and (α₃ ,α₄ ) as the ends of its one diameter is

(a) (x–2)(x–8)+(y–4)(y–16) = 0

(b) (x–4)(x–8)+(y–2)(y–16) = 0

(c) (x–2)(x–16)+(y–4)(y–8) = 0

(d) (x–6)(x–8)+(y–5)(y–6) = 0

Answer 1

Correct option is (a) (x–2)(x–8)+(y–4)(y–16) = 0

ƒ(x+y)=ƒ(x)ƒ(y)

ƒ(2)= ƒ(1+1) = ƒ(1)ƒ(1) = 2² = α₂

α₃ =ƒ(3)=2³ , ƒ(4) = 2⁴ = α₄

(α₁ , α₂)= (2,4) & (α₃ ,α₄) = (8, 16)

Equation of circle in diameter form

(x–x₁) (x–x₂) + (y–y₁)(y–y₂) = 0

(x–2) (x–8) + (y–4)(y–16) = 0