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Question Number 27524 by julli deswal last updated on 08/Jan/18

If x is real, the maximum value of  ((3x^2 +9x+17)/(3x^2 +9x+7))  is

$$\mathrm{If}\:{x}\:\mathrm{is}\:\mathrm{real},\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{value}\:\mathrm{of} \\ $$$$\frac{\mathrm{3}{x}^{\mathrm{2}} +\mathrm{9}{x}+\mathrm{17}}{\mathrm{3}{x}^{\mathrm{2}} +\mathrm{9}{x}+\mathrm{7}}\:\:\mathrm{is} \\ $$

Answered by prakash jain last updated on 08/Jan/18

y=((3x^2 +9x+17)/(3x^2 +9x+7))  3x^2 y+9xy+7y=3x^2 +9x+17  3x^2 (y−1)+9x(y−1)+(7y−17)=0  x∈R⇒Δ≥0  81(y−1)^2 −4×3(y−1)(7y−17)≥0  81(y^2 −2y+1)−12(7y^2 −24y+17)≥0  −3y^2 +126y−123≥0  −y^2 +42y−41≥0  −(y−41)(y−1)≥0  1≤y≤41  min. value of y=1  max. value of y=41

$${y}=\frac{\mathrm{3}{x}^{\mathrm{2}} +\mathrm{9}{x}+\mathrm{17}}{\mathrm{3}{x}^{\mathrm{2}} +\mathrm{9}{x}+\mathrm{7}} \\ $$$$\mathrm{3}{x}^{\mathrm{2}} {y}+\mathrm{9}{xy}+\mathrm{7}{y}=\mathrm{3}{x}^{\mathrm{2}} +\mathrm{9}{x}+\mathrm{17} \\ $$$$\mathrm{3}{x}^{\mathrm{2}} \left({y}−\mathrm{1}\right)+\mathrm{9}{x}\left({y}−\mathrm{1}\right)+\left(\mathrm{7}{y}−\mathrm{17}\right)=\mathrm{0} \\ $$$${x}\in\mathbb{R}\Rightarrow\Delta\geqslant\mathrm{0} \\ $$$$\mathrm{81}\left({y}−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{4}×\mathrm{3}\left({y}−\mathrm{1}\right)\left(\mathrm{7}{y}−\mathrm{17}\right)\geqslant\mathrm{0} \\ $$$$\mathrm{81}\left({y}^{\mathrm{2}} −\mathrm{2}{y}+\mathrm{1}\right)−\mathrm{12}\left(\mathrm{7}{y}^{\mathrm{2}} −\mathrm{24}{y}+\mathrm{17}\right)\geqslant\mathrm{0} \\ $$$$−\mathrm{3}{y}^{\mathrm{2}} +\mathrm{126}{y}−\mathrm{123}\geqslant\mathrm{0} \\ $$$$−{y}^{\mathrm{2}} +\mathrm{42}{y}−\mathrm{41}\geqslant\mathrm{0} \\ $$$$−\left({y}−\mathrm{41}\right)\left({y}−\mathrm{1}\right)\geqslant\mathrm{0} \\ $$$$\mathrm{1}\leqslant{y}\leqslant\mathrm{41} \\ $$$$\mathrm{min}.\:\mathrm{value}\:\mathrm{of}\:{y}=\mathrm{1} \\ $$$$\mathrm{max}.\:\mathrm{value}\:\mathrm{of}\:{y}=\mathrm{41} \\ $$

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