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Question 41 of 58

Q.x is real. Using differential calculus find the maximum and minimum values of (x² - x + 1) / (x² + x + 1).

West Bengal WbchseWest Bengal HS (WBCHSE) Board 2022Subjective· 5mImportance★★★★★
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Differentiate the rational function, find critical points from the numerator, and classify each using the sign change of the derivative.

Let y=x2−x+1x2+x+1y = \dfrac{x^2-x+1}{x^2+x+1}. Note the denominator x2+x+1x^2+x+1 has discriminant 1−4=−3<01-4=-3<0, so it is always positive — yy is defined for all real xx.

Differentiate using the quotient rule:

dydx=(2x−1)(x2+x+1)−(x2−x+1)(2x+1)(x2+x+1)2\dfrac{dy}{dx} = \dfrac{(2x-1)(x^2+x+1) - (x^2-x+1)(2x+1)}{(x^2+x+1)^2}

Expand the numerator.

(2x−1)(x2+x+1)=2x3+2x2+2x−x2−x−1=2x3+x2+x−1(2x-1)(x^2+x+1) = 2x^3+2x^2+2x-x^2-x-1 = 2x^3+x^2+x-1

(x2−x+1)(2x+1)=2x3+x2−2x2−x+2x+1=2x3−x2+x+1(x^2-x+1)(2x+1) = 2x^3+x^2-2x^2-x+2x+1 = 2x^3-x^2+x+1

Numerator =(2x3+x2+x−1)−(2x3−x2+x+1)=2x2−2=2(x2−1)= (2x^3+x^2+x-1)-(2x^3-x^2+x+1) = 2x^2-2 = 2(x^2-1)

So dydx=2(x2−1)(x2+x+1)2\dfrac{dy}{dx} = \dfrac{2(x^2-1)}{(x^2+x+1)^2}.

Find critical points: set dydx=0⇒x2−1=0⇒x=±1\dfrac{dy}{dx}=0 \Rightarrow x^2-1=0 \Rightarrow x=\pm1.

Classify using the sign of dydx\dfrac{dy}{dx} (denominator is always positive, so the sign follows x2−1x^2-1):

  • x<−1x<-1: x2−1>0⇒yx^2-1>0 \Rightarrow y increasing
  • −1<x<1-1<x<1: x2−1<0⇒yx^2-1<0 \Rightarrow y decreasing
  • x>1x>1: x2−1>0⇒yx^2-1>0 \Rightarrow y increasing …
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