Suppose g(x) is a real valued differentiable functn satisfying g'(x) + 2g(x) >1
then show tat e2x ( g(x) - 1/2 ) is an increasin functn
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let 2g(x) + g'(x) = x (x>1) by given differentiate e^2x g(x) - 1/2 e^2x you'll get e^2x ( 2g(x) + g'(x)) - e^2x now 2g(x) + g'(x) >1 so e^2x (2g(x) + g'(x)) - e^2x >0 thus function is incrasing because f'(x)>0
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