Question
If $$x$$ is real, the maximum value of $$\frac{{3{x^2} + 9x + 17}}{{3{x^2} + 9x + 7}}\,\,{\text{is}}$$
A.
$$\frac{1}{4}$$
B.
$$41$$
C.
$$1$$
D.
$$\frac{{17}}{7}$$
Answer :
$$41$$
Solution :
$$\eqalign{
& y = \frac{{3{x^2} + 9x + 17}}{{3{x^2} + 9x + 7}} \cr
& 3{x^2}\left( {y - 1} \right) + 9x\left( {y - 1} \right) + 7y - 17 = 0 \cr
& D \geqslant 0\,\,\,\,\,\,\,\,\,\because \,\,x\,\,{\text{is real}} \cr
& {\text{81}}{\left( {y - 1} \right)^2} - 4 \times 3\left( {y - 1} \right)\left( {7y - 17} \right) \geqslant 0 \cr
& \Rightarrow \,\,\left( {y - 1} \right)\left( {y - 41} \right) \leqslant 0 \cr
& \Rightarrow \,\,1 \leqslant y \leqslant 41 \cr
& \therefore \,\,{\text{Max value of }}y{\text{ is 41}}{\text{.}} \cr} $$