A) \[1\]
B) \[-1\]
C) \[2\]
D) \[0\]
Correct Answer: D
Solution :
[d] Given, \[({{a}^{2}}+{{b}^{2}}+{{c}^{2}}){{x}^{2}}-2(ab+bc+cd)x+{{b}^{2}}+{{c}^{2}}\]\[+{{d}^{2}}\le 0\]. \[\Rightarrow {{(ax-b)}^{2}}+{{(bx-c)}^{2}}+{{(cx-d)}^{2}}\le 0\] \[\Rightarrow {{(ax-b)}^{2}}+{{(bx-c)}^{2}}+{{(cx-d)}^{2}}=0\] \[\Rightarrow \frac{b}{a}=\frac{c}{b}=\frac{d}{c}=x\] \[\Rightarrow {{b}^{2}}=ac\,\,or\,\,2\,\,\log \,b=\log \,\,a+\log \,c\] \[Now,\,\,\,\,\,\Delta =\left| \begin{matrix} 33 & 14 & \log a \\ 65 & 27 & \log \,b \\ 97 & 40 & \log \,c \\ \end{matrix} \right|\] Apply \[{{R}_{1}}\to {{R}_{1}}+{{R}_{3}};\] \[\Delta \left| \begin{matrix} 130 & 54 & \log a+\log c \\ 65 & 27 & \log \,b \\ 97 & 40 & \log \,c \\ \end{matrix} \right|=0\] Now, \[{{R}_{1}}\to {{R}_{1}}-2{{R}_{2}};\] \[\Delta =\left| \begin{matrix} 0 & 0 & 0 \\ 65 & 27 & \log \,\,b \\ 97 & 40 & \log \,\,c \\ \end{matrix} \right|=0\]
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