TS EAMCET · Maths · Functions
For \(x \in R\), the least value of \(\frac{x^2-6 x+5}{x^2+2 x+1}\) is
- A -1
- B \(-\frac{1}{2}\)
- C \(-\frac{1}{4}\)
- D \(-\frac{1}{3}\)
Answer & Solution
Correct Answer
(D) \(-\frac{1}{3}\)
Step-by-step Solution
Detailed explanation
Let \(f(x)=\frac{x^2-6 x+5}{x^2+2 x+1}\) for \(x \in R\) Let \(\quad y=\frac{x^2-6 x+5}{x^2+2 x+1}\) \(y x^2+2 y x+y=x^2-6 x+5\) \((y-1) x^2+(2 y+6) x+(y-5)=0\) \(x=\frac{-2(y+3)+\sqrt{4(y+3)^2-4(y-1)(y-5)}}{2 \times(y-1)}\) Since, \(x\) is real. Then, its discriminant should be…
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