Correct Answer - Option 4 : -a
Concept:
Real Functions:
If f(x) = a0x0 + a1x1 + a2x2 + ... anxn, where a0, a1, ... an are constants, then, f(y) = a0y0 + a1y1 + a2y2 + ... anyn.
Calculation:
Given that f(x) = \(\rm\frac{x}{x-1}\).
∴ f(a) = \(\rm\frac{a}{a-1}\) and \(\rm f\left(\frac1a\right)=\frac{\frac1a}{\frac1a-1}=\frac{1}{1-a}\).
Now, \(\rm \frac{f(a)}{f\left(\frac{1}{a}\right)}=\frac{\frac{a}{a-1}}{\frac{1}{1-a}}\) = -a.