A) \[-f(2)\]
B) \[-f(-\,2)\]
C) \[f\left( \frac{1}{2} \right)\]
D) \[-f\left( \frac{1}{2} \right)\]
Correct Answer: B
Solution :
| We have, \[{{x}^{2}}f(x)-2f\left( \frac{1}{x} \right)=g\,(x),\] |
| Where \[g(x)\] is odd function. |
| \[\therefore \]\[g(x)+g(-x)=0\] |
| \[{{x}^{2}}[f(x)+f(-x)]-2\left[ f\left( \frac{1}{x} \right)+f\left( -\frac{1}{x} \right) \right]=0\] |
| \[x\to \frac{1}{x},\]we get \[\frac{1}{{{x}^{2}}}\left[ f\left( \frac{1}{x} \right)+f\left( -\frac{1}{x} \right)-2f(x)+f(-x) \right]=0\] |
| From Eqs. (i) and (ii), we get |
| \[f(x)=-f(-x)\] |
| \[f(2)=-f(-\,2)\] |
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