MHT CET · Maths · Application of Derivatives
For all real \(x\), the minimum value of \(\frac{1-x+x^{2}}{1+x+x^{2}}\) is
- A 0
- B \(1 / 3\)
- C 1
- D 3
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
\(1-\frac{2}{1+1+1}=1-\frac{2}{3}\)
\(=\frac{1}{3}\)
\(=1-\frac{2}{\frac{1}{x}+1+x}\)
\(\Rightarrow y=1-\frac{2}{t}\)
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
\(1-\frac{2}{1+1+1}=1-\frac{2}{3}\)
\(=\frac{1}{3}\)
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