If x is real the maximum value of 3x2+9x+173x2+9x+7 is

Let 3x2+9x+173x2+9x+7=y⇒ 3x2+9x+17=3x2y+9xy+7y⇒ 3(y−1)x2+9(y−1)x+(7y−17)=0∀ x∈R ⇒ D≥0⇒ 81(y−1)2−4(3)(y−1)(7y−17)≥0⇒ 27(y2+1−2y)−4(7y2−17y−7y+17)≥0⇒ 27y2+27−54y−28y2+68y+28y−68≥0⇒ −y2+42y−41≥0⇒ y2−42y+41≤0⇒(y−41)(y−1)≤0⇒ 1≤y≤41∴ maximum value = 41

If x is real the maximum value of 3x2+9x+173x2+9x+7 is

Let 3x2+9x+173x2+9x+7=y⇒ 3x2+9x+17=3x2y+9xy+7y⇒ 3(y−1)x2+9(y−1)x+(7y−17)=0∀ x∈R ⇒ D≥0⇒ 81(y−1)2−4(3)(y−1)(7y−17)≥0⇒ 27(y2+1−2y)−4(7y2−17y−7y+17)≥0⇒ 27y2+27−54y−28y2+68y+28y−68≥0⇒ −y2+42y−41≥0⇒ y2−42y+41≤0⇒(y−41)(y−1)≤0⇒ 1≤y≤41∴ maximum value = 41

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$I f x i s r e a l t h e \max i m u m v a l u e o f \frac{3 x^{2} + 9 x + 17}{3 x^{2} + 9 x + 7} i s$

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$\frac{17}{7}$

$\frac{1}{4}$

answer is C.

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Detailed Solution

$\begin{array}{l} L e t \frac{3 x^{2} + 9 x + 17}{3 x^{2} + 9 x + 7} = y \\ \Rightarrow 3 x^{2} + 9 x + 17 = 3 x^{2} y + 9 x y + 7 y \\ \Rightarrow 3 (y - 1) x^{2} + 9 (y - 1) x + (7 y - 17) = 0 \\ \forall x \in R \Rightarrow D \geq 0 \\ \Rightarrow 81 {(y - 1)}^{2} - 4 (3) (y - 1) (7 y - 17) \geq 0 \\ \Rightarrow 27 (y^{2} + 1 - 2 y) - 4 (7 y^{2} - 17 y - 7 y + 17) \geq 0 \\ \Rightarrow 27 y^{2} + 27 - 54 y - 28 y^{2} + 68 y + 28 y - 68 \geq 0 \\ \Rightarrow - y^{2} + 42 y - 41 \geq 0 \\ \Rightarrow y^{2} - 42 y + 41 \leq 0 \\ \Rightarrow (y - 41) (y - 1) \leq 0 \\ \Rightarrow 1 \leq y \leq 41 \\ ∴ \max i m u m v a l u e = 41 \end{array}$