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Let f : R → R be a differentiable function satisfying f (x) = f (x – y) f (y) " x, y Î R and f ¢ (0) = a, f ¢ (2) = b then f ¢ (-2) is

Maths-General

  1. fraction numerator b over denominator a end fraction    
  2. fraction numerator a to the power of 2 end exponent over denominator b end fraction    
  3. fraction numerator a over denominator b end fraction    
  4. fraction numerator b to the power of 2 end exponent over denominator a end fraction    

    Answer:The correct answer is: fraction numerator a to the power of 2 end exponent over denominator b end fraction

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    P open parentheses fraction numerator 15 over denominator 8 end fraction cos invisible function application theta comma 0 close parentheses, Q open parentheses 0 comma fraction numerator negative 15 over denominator 7 end fraction sin invisible function application theta close parentheses

    3h = fraction numerator 15 over denominator 8 end fraction cos invisible function application theta, 3k = negative fraction numerator 30 over denominator 7 end fraction sin invisible function application theta
    costheta = fraction numerator 8 h over denominator 5 end fraction, sintheta = fraction numerator negative 7 k over denominator 10 end fraction
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