The solution set of (x2 - 3x + 4/ x + 1) 1 , x ∈ R, is

Question

The solution set of (x2 - 3x + 4/ x + 1) > 1 , x ∈ R, is (A)  (3 , + ∞ ) (B)  ( -1, 1) ∪ (3, + ∞ )  (C)  [ -1, 1 ] ∪ [3, + ∞ )  (D) none of theses

Tardigrade Admin · Accepted Answer

(x2-3 x+4/x+1)>1 ⇒ (x2-3 x+4/x+1)-1>0 ⇒ (x2-4 x+3/x+1)>0 ⇒ ((x+1)(x-1)(x-3)/(x+1)2)>0 ⇒(x+1)(x-1)(x-3)>0 text and x ≠-1 Using method of interval, we get, x ∈(-1,1) ∪(3, ∞)

Q. The solution set of $\frac{x ^{2} - 3 x + 4}{x + 1} > 1, x \in R$ , is

$(3, + \infty)$

$(- 1, 1) \cup (3, + \infty)$

$[- 1, 1] \cup [3, + \infty)$

none of theses

Q. The solution set of x+1x2−3x+4​>1,x∈R, is

(3,+∞)

(−1,1)∪(3,+∞)

[−1,1]∪[3,+∞)

none of theses

x+1x2−3x+4​>1⇒x+1x2−3x+4​−1>0 ⇒x+1x2−4x+3​>0 ⇒(x+1)2(x+1)(x−1)(x−3)​>0 ⇒(x+1)(x−1)(x−3)>0 and x=−1 Using method of interval, we get, x∈(−1,1)∪(3,∞)

Q. The solution set of $\frac{x ^{2} - 3 x + 4}{x + 1} > 1, x \in R$ , is

$(3, + \infty)$

$(- 1, 1) \cup (3, + \infty)$

$[- 1, 1] \cup [3, + \infty)$