TS EAMCET · Maths · Quadratic Equation
If \(x\) is real, then the maximum and minimum values of \(\frac{x^2+14 x+9}{x^2+2 x+3}\) are respectively
- A 4,-5
- B 5,-4
- C 9,3
- D 24,6
Answer & Solution
Correct Answer
(A) 4,-5
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \text { Let } y=\frac{x^2+14 x+9}{x^2+2 x+3} \\ & \Rightarrow \quad x^2 y+2 x y+3 y=x^2+14 x+9 \\ & \Rightarrow \quad x^2(y-1)+2 x(y-7)+3 y-9=0 \end{aligned}\) Here, \(x \in \mathbf{R}\)…
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