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- Thread starter jacks
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- Feb 5, 2012

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Hi jacks,If $f(x+y) = f(xy)$ and $\displaystyle f\left(-\frac{1}{2}\right) = -\frac{1}{2}$. Then $f(2012) = $

Since, \(f(x+y) = f(xy)\) we have,

\[f\left(-\frac{1}{2}\right)=f\left(-\frac{1}{2}+0\right)=f\left(-\frac{1}{2}\times 0\right)=f(0)\]

Since, \(f\left(-\dfrac{1}{2}\right) = -\dfrac{1}{2}\)

\[f(0)=-\frac{1}{2}\]

Now,

\[f(2012)=f(2012+0)=f(2012\times 0)=f(0)=-\frac{1}{2}\]

Kind Regards,

Sudharaka.

- Jan 26, 2012

- 890

Observe:If $f(x+y) = f(xy)$ and $\displaystyle f\left(-\frac{1}{2}\right) = -\frac{1}{2}$. Then $f(2012) = $

\[f(x+y)=f(xy) \Rightarrow f(x)=f(0)\]

Hence \( f(x) \) is a constant function, so \(f(-1/2)=-1/2 \Rightarrow f(2012)=-1/2\) \)

CB