A) \[{{x}^{4}}+\frac{1}{2}{{x}^{3}}+\frac{1}{4}{{x}^{2}}+x\]
B) \[2{{x}^{4}}+\frac{1}{4}{{x}^{3}}-\frac{3}{4}{{x}^{2}}+\frac{1}{4}x\]
C) \[{{x}^{4}}+\frac{1}{4}{{x}^{3}}-\frac{3}{2}{{x}^{2}}+\frac{3}{4}x\]
D) \[2{{x}^{4}}+\frac{3}{4}{{x}^{3}}-\frac{1}{4}{{x}^{2}}+\frac{3}{4}x\]
Correct Answer: C
Solution :
We have, \[\frac{3}{2}{{x}^{2}}({{x}^{2}}-1)+\frac{1}{4}{{x}^{2}}({{x}^{2}}+x)-\frac{3}{4}x({{x}^{3}}-1)\] \[=\frac{3}{2}{{x}^{4}}-\frac{3}{2}{{x}^{2}}+\frac{1}{4}{{x}^{4}}+\frac{1}{4}{{x}^{3}}-\frac{3}{4}{{x}^{4}}+\frac{3}{4}x\] \[=\frac{3}{2}{{x}^{4}}+\frac{1}{4}{{x}^{4}}-\frac{1}{4}{{x}^{4}}+\frac{1}{4}{{x}^{3}}-\frac{3}{4}{{x}^{2}}+\frac{3}{4}x\] \[=\frac{6{{x}^{4}}+{{x}^{4}}-3{{x}^{4}}}{4}+\frac{1}{4}{{x}^{3}}-\frac{3}{2}{{x}^{2}}+\frac{3}{4}x\] \[={{x}^{4}}+\frac{1}{4}{{x}^{3}}-\frac{3}{2}{{x}^{2}}+\frac{3}{4}x\]
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