A) \[3,\frac{1}{2}\]
B) 3, -1
C) \[0,\frac{1}{3}\]
D) \[1,\frac{1}{3}\]
Correct Answer: D
Solution :
If (x- 1) is a factor, then for x = 1, So, \[4{{x}^{5}}+9{{x}^{4}}-7{{x}^{3}}-5{{x}^{2}}-4kx+3{{k}^{2}}=0\] \[\Rightarrow \,\,\,\,\,\,4\times {{(1)}^{5}}+9\times {{(1)}^{4}}-7\times {{(1)}^{3}}-5\times {{(1)}^{2}}-4\times k\times (1)+3\times {{k}^{2}}=0\]\[\begin{align} & \Rightarrow \,\,\,\,\,4+9-7-5-4k+3{{k}^{3}}=0 \\ & \Rightarrow \,\,\,\,\,3{{k}^{2}}-4k+1=0 \\ & \Rightarrow \,\,\,\,\,3{{k}^{2}}-3k-k+1=0 \\ & \Rightarrow \,\,\,\,\,3k(k-1)-1(k-1)=0 \\ \end{align}\] \[\Rightarrow \,\,\,\,\,(3k-1)(k-1)=0\therefore k\frac{\mathbf{1}}{\mathbf{3}},\mathbf{1}\]
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