If x is not equal to 0, 1 or −1, is x^2 − 1/x^2 + 1 x^2 + 1/x^2 − 1

Question

If x is not equal to 0, 1 or −1, is  \frac{x^2 − 1}{ x^2 + 1}>\frac{x^2 + 1}{x^2 − 1} ?

(1)  x^2 < x

(2)  x > 0

chetan2u · Answer

Logical LHS: \frac{x^2 − 1}{ x^2 + 1} Numerator > Denominator So |LHS|<1 RHS: \frac{x^2 +1}{ x^2 - 1} Numerator < Denominator So, |RHS|>1 Thus, how can the LHS > RHS? => When RHS is negative , that is \frac{x^2 +1}{ x^2 - 1}<0 x^2+1>0 , so x^2-1<0……x^2<1…….|x|<1 Let us see the statements (1) x^2 < x……..x^2-x<0…….x(x-1)<0 x is positive as x>x^2, so x-1<0 or x<1. So 0 0 if 01, then the answer no. Insufficient A Algebraic \frac{x^2 − 1}{ x^2 + 1}>\frac{x^2 + 1}{x^2 − 1} Subtract 1 from both sides \frac{x^2 − 1}{ x^2 + 1}-1>\frac{x^2 + 1}{x^2 − 1}-1 \frac{x^2 − 1-(x^2+1)}{ x^2 + 1}>\frac{x^2 + 1-(x^2-1)}{x^2 − 1} \frac{− 2}{ x^2 + 1}>\frac{2}{x^2 − 1} \frac{− 1}{ x^2 + 1}>\frac{1}{x^2 − 1} -1>\frac{x^2 + 1}{x^2 − 1} 0>\frac{x^2 + 1}{x^2 − 1}+1 0>\frac{x^2 + 1-(x^2-1)}{x^2 − 1} 0>\frac{2}{x^2 − 1} x^2-1<0 x^2<1 Rest of the solution is same as shown above

If x is not equal to 0, 1 or −1, is x^2 − 1/x^2 + 1 > x^2 + 1/x^2 − 1

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