A) \[p=0\]
B) \[p=1\]
C) \[p=-1\]
D) For all \[p>1\]
Correct Answer: C
Solution :
[c] The given system of equations are: \[{{p}^{3}}x+{{(p+1)}^{3}}y={{(p+2)}^{3}}\] .... (1) \[px+(p+1)y=(p+2)\] .... (2) \[x+y=1\] .... (3) This system is consistent, if values of x and y from first two equation satisfy the third equation. which \[\Rightarrow \,\,\left| \begin{matrix} {{p}^{3}} & {{(p+1)}^{3}} & {{(p+2)}^{3}} \\ p & (p+1) & (p+2) \\ 1 & 1 & 1 \\ \end{matrix} \right|=0\] \[\Rightarrow \,\,\left| \begin{matrix} {{p}^{3}} & {{(p+1)}^{3}}-{{p}^{3}} & {{(p+2)}^{3}}-{{p}^{3}} \\ p & 1 & 2 \\ 1 & 0 & 0 \\ \end{matrix} \right|=0\] \[\Rightarrow \,\,2{{(p+1)}^{3}}-2{{p}^{3}}-{{(p+2)}^{3}}+{{p}^{3}}=0\] \[\Rightarrow \,\,2({{p}^{3}}+1+3{{p}^{2}}+3p)-2{{p}^{3}}-({{p}^{3}}+8+12p+6{{p}^{2}})+{{p}^{3}}=0\]\[\Rightarrow \,\,2{{p}^{3}}+2+6{{p}^{2}}+6p-2{{p}^{3}}-{{p}^{3}}-8-12p\] \[-6{{p}^{2}}+{{p}^{3}}=0\] \[\Rightarrow \,\,-6-6p=0\,\,\,\,\,\,\,\,\Rightarrow \,\,\,\,\,\,\,\,\,\,p=-1\]
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