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MHT CET · Maths · Application of Derivatives

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

  1. A 0
  2. B \(1 / 3\)
  3. C 1
  4. D 3
Verified Solution

Answer & Solution

Correct Answer

(B) \(1 / 3\)

Step-by-step Solution

Detailed explanation

Let \(y=\frac{1-x+x^{2}}{1+x+x^{2}}=1-\frac{2 x}{1+x+x^{2}}\)
\(=1-\frac{2}{\frac{1}{x}+1+x}\)
\(\Rightarrow y=1-\frac{2}{t}\)
where
\(
t=\frac{1}{x}+1+x
\)
Now, \(y\) is minimum, when \(\frac{2}{t}\) is \(\max \Rightarrow t\) is \(\mathrm{min}\).
\(\therefore \frac{d t}{d x}=-\frac{1}{x^{2}}+1=0\)
\(\Rightarrow x=\pm 1\)
\(\frac{d^{2} t}{d x^{2}}=\frac{2}{x^{3}}>0, \text { for } x=1\)
\(\therefore\) Minimum value of \(y\) is
\(1-\frac{2}{1+1+1}=1-\frac{2}{3}\)
\(=\frac{1}{3}\)