A) 0
B) 2
C) 10
D) ? 5
Correct Answer: D
Solution :
\[{}^{\left( x\frac{dy}{dx}-y \right)}/{}_{{{x}^{2}}=\frac{1}{x}-\frac{1}{a+bx}b=\frac{a}{x(a+bx)}}\] Integrate w.r.t. x, \[{f}'(x)-{g}'(x)=c\] At \[x=1\], \[{f}'(1)-{g}'(1)=c\] Þ \[2-4=c\] Þ \[c=-2\] Hence, \[{f}'(x)-{g}'(x)=-2\]. Again integrate w.r.t. x, \[f(x)-g(x)=-2x+{{c}_{1}}\]. At\[x=2\],\[f(2)-g(2)=-2\times 2+{{c}_{1}}\]Þ \[3-9+4={{c}_{1}}\] Þ \[{{c}_{1}}=-2\] Then \[f(x)-g(x)=-\,2x-2=-\,(2x+2)\] \ \[f(3/2)-g(3/2)=-\,\left( 2\times \frac{3}{2}+2 \right)=-\,5\].
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