TS EAMCET · Maths · Functions
If \(x\) is real, then the value of \(\frac{x^2-3 x+4}{x^2+3 x+4}\) lies in the interval
- A \(\left[\frac{1}{3}, 3\right]\)
- B \(\left[\frac{1}{5}, 5\right]\)
- C \(\left[\frac{1}{6}, 6\right]\)
- D \(\left[\frac{1}{7}, 7\right]\)
Answer & Solution
Correct Answer
(D) \(\left[\frac{1}{7}, 7\right]\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \text { Let } y=\frac{x^2-3 x+4}{x^2+3 x+4} \\ & \Rightarrow x^2(y-1)+x(3 y+3)+(4 y-4)=0\end{aligned}\) Since, \(x\) is real.…
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