28) For all real values of x, the minimum value of
\frac{1- x + x^2 }{1+ x +x^2}
is
(A) 0    (B)   1   (C) 3 (D) 1/3 

Answers (1)

Given function is
f(x)= \frac{1- x + x^2 }{1+ x +x^2}
f^{'}(x)= \frac{(-1+2x)(1+x+x^2)-(1-x+x^2)(1+2x)}{(1+ x +x^2)^2}
            = \frac{-1-x-x^2+2x+2x^2+2x^3-1-2x+x+2x^2-x^2-2x^3}{(1+ x +x^2)^2} = \frac{-2+2x^2}{(1+ x +x^2)^2}
f^{'}(x)=0\\ \frac{-2+2x^2}{(1+ x +x^2)^2} = 0\\ x^2 = 1\\ x= \pm 1
Hence, x = 1 and x = -1 are the critical points
Now,
f^{''}(x)= \frac{4x(1+ x +x^2)^2-(-2+2x^2)2(1+x+x^2)(2x+1)}{(1+ x +x^2)^4} \\ f^{''}(1) = \frac{4\times(3)^2}{3^4} = \frac{4}{9} > 0
Hence, x = 1 is the point of minima and the minimum value is 
f(1)= \frac{1- 1 + 1^2 }{1+ 1 +1^2} = \frac{1}{3}

f^{''}(-1) =-4 < 0
Hence, x = -1 is the point of maxima
Hence,  the minimum value of
\frac{1- x + x^2 }{1+ x +x^2}  is   \frac{1}{3}
Hence, (D) is the correct answer

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