A) \[{{x}^{2}}\]
B) \[1-{{x}^{2}}\]
C) \[1+{{x}^{2}}\]
D) \[{{x}^{2}}+x+1\]
Correct Answer: B
Solution :
| \[{{x}^{2}}F(x)+f(1-x)=2x-{{x}^{4}}\] |
| Replacing \[x\] by \[1-x,\] gives \[{{(1-x)}^{2}}F(1-x)+F(x)\] |
| \[=2(1-x)-{{(1-x)}^{4}}\] ?(2) |
| Multiplying (1) by \[{{\left( 1-x \right)}^{2}}\]and subtracting (2) from it gives \[[{{x}^{2}}{{(1-x)}^{2}}-1]F(x)\] |
| \[=2x{{(1-x)}^{2}}-{{x}^{4}}{{(1-x)}^{2}}\]\[-2(1-x)+{{(1-x)}^{4}}\] |
| \[\Rightarrow [x(1-x)-1][x(1-x)+1]F(x)\] |
| \[=2(1-x)[x(1-x)-1]\]\[-{{(1-x)}^{2}}[{{x}^{4}}-{{(1-x)}^{2}}]\] |
| \[\Rightarrow \]\[[x-1-{{x}^{2}}][x+1-{{x}^{2}}]F(x)\] |
| \[=2(1-x)[x-1-{{x}^{2}}]\]\[-{{(1-x)}^{2}}({{x}^{2}}-1+x)({{x}^{2}}+1-x)\] |
| \[\Rightarrow \]\[(x+1-{{x}^{2}})F(x)\] |
| \[=(1-x)[2+(1-x)({{x}^{2}}+x-1)]\] |
| \[=(1-x)[2+2x-{{x}^{3}}-1]\] |
| \[=(1-x)[2(1+x)-(x+1)({{x}^{2}}-x+1)]\] |
| \[=(1-{{x}^{2}})[1+x-{{x}^{2}}]\] |
| \[\therefore F\left( x \right)=1-{{x}^{2}}\] |
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