A) \[\frac{1}{2}\]
B) \[\frac{7}{2}\]
C) \[\frac{3}{2}\]
D) None of these
Correct Answer: C
Solution :
Here, \[f(a)=4{{a}^{3}}-12{{a}^{2}}+14a-3\] and \[g(a)=2a-1\] Put \[g(a)=0\Rightarrow a=\frac{1}{2}\] Remainder = \[f\left( \frac{1}{2} \right)=4{{\left( \frac{1}{2} \right)}^{3}}-12\times {{\left( \frac{1}{2} \right)}^{2}}+14\times \left( \frac{1}{2} \right)-3\] \[=4\times \frac{1}{8}-12\times \frac{1}{4}+7-3=\frac{3}{2}\]
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