TS EAMCET · Maths · Quadratic Equation
If the range of the real valued function \(f(x)=\frac{x^2+x+k}{x^2-x+k}\) is \(\left[\frac{1}{3}, 3\right]\), then \(k=\)
- A 2
- B -2
- C 1
- D 2
Answer & Solution
Correct Answer
(C) 1
Step-by-step Solution
Detailed explanation
\(y = \frac{x^2+x+k}{x^2-x+k}\) \((y-1)x^2 - (y+1)x + k(y-1) = 0\) \(\Delta = (y+1)^2 - 4(y-1)k(y-1) \ge 0\) \((y+1)^2 \ge 4k(y-1)^2\) At the range boundaries, \((y+1)^2 = 4k(y-1)^2\) Using \(y=3\): \((3+1)^2 = 4k(3-1)^2 \Rightarrow 4^2 = 4k(2^2) \Rightarrow 16 = 16k\) \(k = 1\)
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