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MHT CET · Maths · Functions

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

  1. A \(0\)
  2. B \(1\)
  3. C \(\frac{1}{3}\)
  4. D \(3\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(\frac{1}{3}\)

Step-by-step Solution

Detailed explanation

\(f(x)=\frac{1-x+x^2}{1+x+x^2} \)
\( \therefore f^{\prime}(x)=\frac{\left(1+x+x^2\right)(-1+2 x)-\left(1-x+x^2\right)(1+2 x)}{\left(1+x+x^2\right)^2} \)
\( = \frac{\left(-1+2 x-x+2 x^2-x^2+2 x^3\right)}{\left(1+x+x^2\right)^2} \)
\( = \frac{-2+2 x^2}{\left(1+x+x^2\right)^2}\)
If \(\mathrm{f}^{\prime}(x)=0\), then \(\frac{-2+2 x^2}{\left(1+x+x^2\right)^2}=0 \Rightarrow x^2=1\)
\(\Rightarrow x= \pm 1\)
\(\therefore \mathrm{f}(x)\) at \(x=1\) is \(\frac{1}{3}\) and \(\mathrm{f}(x)\) at \(x=-1\) is 1 ,
\(\therefore \) Minimum value of \(\mathrm{f}(x)\) is \(\frac{1}{3}\).
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