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Manipal MET2023MathematicsFunctions

Let f(x)= 1 1+x^2 , then

Options

  1. Af(x, y)=f(x) f(y)
  2. Bf(x, y) f(x) f(y)
  3. Cf(x, y) f(x) f(y)
  4. Df(x, y)=f(x)-f(y)

Correct answer

B. f(x, y) f(x) f(y)

Step-by-step solution

Given that f(x)= 1 1+x^2 f(y)= 1 1+y^2 and f(x, y)= 1 1+x^2 y^2 Now, f(x, y)=f(x) f(y) 1 1+x^2 y^2 = 1 1+x^2 1 1+y^2 1 1+x^2 y^2 = 1 (1+x^2 ) (1+y^2 ) On squaring both sides, we get 1 1+x^2 y^2 = 1 (1+x^2 ) (1+y^2 ) 1 1+x^2 y^2 1 (1+x^2 ) (1+y^2 ) [ 1+x^2 y^2 1+x^2+y^2+x^2 y^2 ] Hence, f(x, y) f(x) f(y)

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