A) \[-3\]
B) \[3\]
C) \[-4\]
D) \[-9\]
Correct Answer: A
Solution :
\[f(x)=2{{x}^{3}}+a{{x}^{2}}+3x-5\] \[g(x)={{x}^{3}}+{{x}^{2}}-2x+a\] By Remainder Theorem, \[f(2)=(2\times {{2}^{3}}+a\times {{2}^{2}}+3\times 2-5)=17+4a\] Again,\[g(2)=({{2}^{3}}+{{2}^{2}}-2\times 2+a)=8+a\] \[\therefore \] \[17+4a=8+a\] \[\Rightarrow \] \[3a=-9\] \[\Rightarrow \] \[a=-3\]You need to login to perform this action.
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