A) \[a=-1,\,b=-2\]
B) \[a=2,\,b=5\]
C) \[a=5,\,b=2\]
D) \[a=2,\,b=0\]
Correct Answer: C
Solution :
[c] \[(x+1)\]is a factor of \[2{{x}^{3}}+a{{x}^{2}}+2bx+1\] |
\[\therefore \,\,\,\,\,\,\,\,\,2{{(-1)}^{3}}+a{{(-1)}^{2}}-2b+1=0\] |
\[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,a-2b=1\] (1) |
Also, \[2a-3b=4\] (2) |
On solving eqs. (1) and (2) we get, |
\[a=5,\,b=2\] |
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