A) \[A=4,\,\,B=16\]
B) \[A=6,\,\,B=24\]
C) \[A=2,\,\,B=12\]
D) \[A=8,\,B=16\]
Correct Answer: C
Solution :
Since, \[(x-1)\] is a factor of \[A{{x}^{3}}+B{{x}^{2}}-36x+22\]. \[\therefore \] \[A{{(1)}^{3}}+{{(1)}^{2}}-36(1)+22=0\] \[\Rightarrow \] \[A+B=14\] ?(i) and \[{{2}^{B}}={{({{2}^{6}})}^{A}}\Rightarrow B=6A\] ?(ii) On solving Eqs. (i) and (ii), we get \[A=2,\,\,B=12\]You need to login to perform this action.
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