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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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For all real values of \(x\), the minimum value of \( \large\frac{1-x+x^2}{1+x+x^2}\) is

\[(A)\; 0 \quad (B)\; 1 \quad (C)\; 3 \quad (D)\; \frac{1}{3}\]

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1 Answer

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Toolbox:
  • $\large\frac{d}{dx}$$(x^n)=nx^{n-1}$
Step 1:
Let $y=\large\frac{1-x+x^2}{1+x+x^2}$
$\large\frac{dy}{dx}=\frac{(-1+2x)(1+x+x^2)-(1-x+2x)(1+2x)}{(1+x+x^2)^2}$
Numerator of $\large\frac{dy}{dx}=$$(-1+2x)(1+x+x^2)-(1+2x)(1-x+x^2)$
$\Rightarrow (-1-x-x^2)+(2x+2x^2+2x^3)-(1-x+x^2)-(2x-2x^2+2x^3)$
$\Rightarrow -2+2x^2=2(x^2-1)$
$\Rightarrow 2(x-1)(x+1)$
Step 2:
$\large\frac{dy}{dx}=\frac{2(x-1)(x+1)}{(x^2+x+1)^2}$
$\large\frac{dy}{dx}$$=0$ at $x=1,-1$
At $x=1$
$\large\frac{dy}{dx}$ changes sign from -ve to +ve.
[-+=-ve,++=+ve]
$\therefore y$ is minimum at $x=1$
Minimum value of $\large\frac{1-x+x^2}{1+x+x^2}=\frac{1-1+1}{1+1+1}=\frac{1}{3}$
Part (D) is the correct answer.
answered Aug 9, 2013 by sreemathi.v
 

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