IPMAT Indore 2019Algebra > Medium(12,1)\left( \frac{1}{2}, 1 \right)(21,1)(12,∞)\left( \frac{1}{2}, \infty \right)(21,∞)(−∞,12)\left( -\infty, \frac{1}{2} \right)(−∞,21)(−∞,12)∪(1,∞)\left( -\infty, \frac{1}{2} \right) \cup \left( 1, \infty \right)(−∞,21)∪(1,∞)✅ Correct Option: 4Related questions:Consider the following statements: (i) When (0<x<1)(0 < x < 1)(0<x<1), then (11+x<1−x+x2)(\frac{1}{1+x} < 1 - x + x^2)(1+x1<1−x+x2) (ii) When (0<x<1)(0 < x < 1)(0<x<1), then (11+x>1−x+x2)(\frac{1}{1+x} > 1 - x + x^2)(1+x1>1−x+x2) (iii) When (−1<x<0)(-1 < x < 0)(−1<x<0), then (11+x<1−x+x2)(\frac{1}{1+x} < 1 - x + x^2)(1+x1<1−x+x2) (iv) When (−1<x<0)(-1 < x < 0)(−1<x<0), then (11+x>1−x+x2)(\frac{1}{1+x} > 1 - x + x^2)(1+x1>1−x+x2) Then the correct statements are:The set of all real values of x satisfying the inequality x2(x+1)(x−1)(2x+1)3>0\dfrac{x^2(x+1)}{(x-1)(2x+1)^3} > 0(x−1)(2x+1)3x2(x+1)>0 isIf x∈(a,b)x ∈ (a, b)x∈(a,b) satisfies the inequality x−3x2+3x+2≥1\dfrac{x - 3}{x^2 + 3x + 2} \geq 1x2+3x+2x−3≥1, then the largest possible value of b−ab - ab−a is