A) \[\frac{3}{14}\]
B) \[{{10}^{o}}\]
C) \[{{20}^{o}}\]
D) \[{{30}^{o}}\]
Correct Answer: D
Solution :
Given \[\frac{7}{2}\] \[\frac{{{x}^{2}}+{{x}^{2}}-2x+1}{{{x}^{2}}-x}=\frac{5}{2}\]\[\frac{7}{2}\] \[\frac{2{{x}^{2}}-2x+1}{{{x}^{2}}-x}=\frac{5}{2}\]\[2(2{{x}^{2}}-2x+1)=5({{x}^{2}}-x)\]\[\frac{7}{2}\] \[6x+3{{x}^{2}}=8x+8\] Cross-multiplying, we get \[\frac{7}{2}\] \[\Rightarrow \] \[\frac{7}{2}\] \[\Rightarrow \] \[\frac{7}{2}\] \[\Rightarrow \] \[{{\left( \frac{{{x}^{a}}}{{{x}^{b}}} \right)}^{a+b}}\times {{\left( \frac{{{x}^{b}}}{{{x}^{c}}} \right)}^{b+c}}\times {{\left( \frac{{{x}^{c}}}{{{x}^{a}}} \right)}^{c+a}}\] \[\Rightarrow \] \[{{\left( \frac{a+b}{a-b} \right)}^{2}}\] \[\Rightarrow \] \[(0,\frac{1}{2})\] \[\Rightarrow \] \[(\frac{1}{2},1)\] \[\Rightarrow \] \[x-2=0\] or \[x+1=0\] \[\Rightarrow \] \[x=2\] or \[x=-1\] \[4{{x}^{2}}-20x+25=0\] The solution is ?1, 2.
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