A) in finite
B) 1
C) 2
D) 3
Correct Answer: B
Solution :
\[\Delta =\left| \begin{matrix} k+1 & 8 \\ k & k+3 \\ \end{matrix} \right|={{k}^{2}}+4k+3-8k\] \[={{k}^{2}}-4k+3\] \[=(k-3)(k-1)\] \[{{\Delta }_{1}}=\left| \begin{matrix} 4k & 8 \\ 3k-1 & k+3 \\ \end{matrix} \right|=4{{k}^{2}}+12k-24k+8\] \[=4{{k}^{2}}-12k+8\] \[=4({{k}^{2}}-3k+2)\] \[=4(k-2)(k-1)\] \[{{\Delta }_{2}}=\left| \begin{matrix} k+1 & 4k \\ k & 3k-1 \\ \end{matrix} \right|=3{{k}^{2}}+2k-1-4{{k}^{2}}\] \[=-{{k}^{2}}+2k-1\] \[=-{{(k-1)}^{2}}\] As given no solution\[\Rightarrow {{\Delta }_{1}}\And {{\Delta }_{2}}\ne 0\] But \[\Delta =0\] \[\Rightarrow \]\[k=3\]
You need to login to perform this action.
You will be redirected in
3 sec
