Question

The solution set of $\left(\frac{x^2-3x+4}{x+1}\right)>1,x\in R$ is

• A. $(3,+\infty )$
• B. $(-1,1)\cup (3,+\infty )$
• C. $[-1,1]\cup [3,+\infty )$
• D. none of these

Answer 1

The correct option is B $(-1,1)\cup (3,+\infty )$

Given,

$\frac{x^2-3x+4}{x+1}>1$

$\Rightarrow \frac{x^2-3x+4}{x+1}-1>0$

$\Rightarrow \frac{x^2-3x+4-x-1}{x+1}>0\Rightarrow \frac{x^2-4x+3}{x+1}>0$

$\Rightarrow \frac{x^2-3x-x+3}{x+1}>0\Rightarrow \frac{x(x-3)-1(x-3)}{x+1}>0$

$\Rightarrow \frac{(x-3)(x-1)}{x+1}>0\Rightarrow (x-3)(x-1)(x+1)>0$

$\Rightarrow x\in (-1,1)\cup (3,\infty )$

Hence, option 'B' is correct.