A) \[-1\]
B) 12
C) 15
D) 17
Correct Answer: D
Solution :
Given, \[\left| \begin{matrix} x-1 & 5x & 7 \\ {{x}^{2}}-1 & x-1 & 8 \\ 2x & 3x & 0 \\ \end{matrix} \right|=a{{x}^{2}}+b{{x}^{2}}+cx+d\] LHS\[=(x-1)(0-24x)-5x(0-16x)\] \[+\,7(3{{x}^{2}}-3x-2{{x}^{2}}+2x)\] \[=-24{{x}^{2}}+24x+80{{x}^{2}}+21{{x}^{3}}-14{{x}^{2}}-7x\] \[=21{{x}^{3}}+42{{x}^{2}}+17x\] \[\therefore \] \[21{{x}^{3}}+42{{x}^{2}}+17x\] \[=a{{x}^{3}}+b{{x}^{2}}+cx+d\] \[\Rightarrow \] \[c=17\]You need to login to perform this action.
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