A) \[\frac{18}{25}\]
B) \[\frac{6}{25}\]
C) \[\frac{24}{25}\]
D) \[\frac{4}{5}\]
Correct Answer: C
Solution :
\[f(x)\,=\,2x-{{x}^{2}}f(x)\] \[f(x)\,=\,\frac{2x}{1+{{x}^{2}}},\,\,f\left( \frac{1}{2} \right)\,\,=\,\,\frac{1}{1+(1/4)}\,\,=\,\,\frac{4}{5}\,\,\] \[f'(x)\,=\,\frac{{{(1+{{x}^{2}})}^{2'}}-2x(2x)}{{{(1+{{x}^{2}})}^{2}}}\,\,=\,\,2\left[ \frac{1-{{x}^{2}}}{{{(1-{{x}^{2}})}^{2}}} \right]\] \[f'\left( \frac{1}{2} \right)\,=\,\frac{2\left( 1-\frac{1}{4} \right)}{{{\left( 1+\frac{1}{4} \right)}^{2}}}\,\,=\,\,\frac{2(3/4)}{25/16}\,\,=\,\,\frac{24}{25}\]
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