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