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TS EAMCET · Maths · Application of Derivatives

If \(x\) is real, then the minimum value of \(\frac{x^2-x+1}{x^2+x+1}\), is

  1. A \(\frac{1}{3}\)
  2. B 3
  3. C \(\frac{1}{2}\)
  4. D 2
Verified Solution

Answer & Solution

Correct Answer

(D) 2

Step-by-step Solution

Detailed explanation

Let \(f(x)=\frac{x^2-x+1}{x^2+x+1}\) ...(i) On differentiating w.r.t. \(x\), we get \(f^{\prime}(x)=\frac{\left(x^2+x+1\right)(2 x-1)-\left(x^2-x+1\right)(2 x+1)}{\left(x^2+x+1\right)^2}\) for maximum or minimum, put \(f^{\prime}(x)=0\)…
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