A) \[(-1,1)\]
B) \[\left( -\sqrt{\frac{2}{3},}0 \right)\cup \left( \sqrt{\frac{2}{3}},\infty \right)\]
C) \[\left( -\sqrt{\frac{2}{3}},\sqrt{\frac{2}{3}} \right)\]
D) None of these
Correct Answer: B
Solution :
[b] \[g'(x)=xf'(2{{x}^{2}}-1)-xf'(1-{{x}^{2}})\] \[=x(f'(2{{x}^{2}}-1)-f'(1-{{x}^{2}}))\] \[g'(x)>0\] If \[x>0,\,\,\,2{{x}^{2}}-1>1-{{x}^{2}}\] (as f? is an increasing function) or \[3{{x}^{2}}>2\] or \[x\in \left( -\infty ,-\sqrt{\frac{2}{3}} \right)\cup \left( \sqrt{\frac{2}{3}},\infty \right)\] or \[x\in \left( \sqrt{\frac{2}{3}},\infty \right)\] If \[x<0,2{{x}^{2}}-1<1-{{x}^{2}}\] or \[3{{x}^{2}}<2\] or \[x\in \left( -\sqrt{\frac{2}{3}},\sqrt{\frac{2}{3}} \right)or\,\,x\in \,\,\left( -\sqrt{\frac{2}{3}},0 \right)\]
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