f(x) is a real valued function where f(xy) = [f(x)] + [f(y)] + f(x)f(y) for all real x,y. If f(2) = 1, then what is the value of $f (\frac{1}{2})$ ?

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Solution

The correct option is C -1/2
f(2) = f(2 $\times$ 1) = f(2) + f(1) + f(2).f(1) = 1

f(1) = 0

$f (1) = f (2 \times \frac{1}{2}) = f (2) + f (\frac{1}{2}) + f (2) . f (\frac{1}{2}) = 0$

$2 \times f (\frac{1}{2}) = - 1 \Rightarrow f (\frac{1}{2}) = - \frac{1}{2}$

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f(x) is a real valued function where f(xy) = [f(x)] + [f(y)] + f(x)f(y) for all real x,y. If f(2) = 1, then what is the value of f(12) ?

The correct option is C -1/2f(2) = f(2 × 1) = f(2) + f(1) + f(2).f(1) = 1 f(1) = 0 f(1)=f(2×12)=f(2)+f(12)+f(2).f(12)=0 2×f(12)=−1⇒f(12)=−12

f(x) is a real valued function where f(xy) = [f(x)] + [f(y)] + f(x)f(y) for all real x,y. If f(2) = 1, then what is the value of $f (\frac{1}{2})$ ?

The correct option is C -1/2
f(2) = f(2 $\times$ 1) = f(2) + f(1) + f(2).f(1) = 1

f(1) = 0

$f (1) = f (2 \times \frac{1}{2}) = f (2) + f (\frac{1}{2}) + f (2) . f (\frac{1}{2}) = 0$

$2 \times f (\frac{1}{2}) = - 1 \Rightarrow f (\frac{1}{2}) = - \frac{1}{2}$