A) -3
B) -2
C) 2
D) None of the above
Correct Answer: D
Solution :
\[\therefore \] (x + 2) is a factor of \[2{{x}^{3}}+a{{x}^{2}}+bx-2\] \[\Rightarrow \,2{{(-2)}^{3}}+a{{\left( -2 \right)}^{2}}+b\left( -2 \right)-2=0\] \[\Rightarrow \,\,-16+4a-2b-2=0\] \[\Rightarrow \,\,4a-2b=18\,\,\,\therefore \,\,\,2a-b=9\,\,\,\,\,\,\,\,\,...(i)\] Also when\[~2{{x}^{3}}+a{{x}^{2}}+bx-2\] is divided by \[2x-3,\] it leaves remainder 7. \[\Rightarrow \]\[2{{\left( \frac{3}{2} \right)}^{3}}+a{{\left( \frac{3}{2} \right)}^{2}}+b\left( \frac{3}{2} \right)-9=0\] \[\Rightarrow \]\[\,\frac{9}{4}a+\frac{3}{2}b=9-\frac{27}{4}=\frac{9}{4}\] \[\Rightarrow \]\[\,\frac{9a+6b}{4}=\frac{9}{4}\Rightarrow 3a+2b=3\,\,\,\,\,...(ii)\] On solving (i) and (ii), we have a = 3
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