Question

Solve the following inequality :
$\frac{x-1}{x^2-4x+3}<1$

• A. (1,3) ∪ (5, ∞)
• B. (1,3) ∪ (6, ∞)
• C. (1,3) ∪ (4, ∞)
• D. (1,2) ∪ (4, ∞)

Answer 1

The correct option is C (1,3) ∪ (4, ∞)

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

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

$\Rightarrow \frac{x-1-(x^2-4x+3)}{x^2-4x+3}<0$

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

$\Rightarrow \frac{-x^2+5x-4}{x(x-3)-1(x-3)}<0$

$\Rightarrow \frac{-x^2+4x+x-4}{(x-3)(x-1)}<0$

$\Rightarrow \frac{-x(x-4)+1(x-4)}{(x-3)(x-1)}<0$

$\Rightarrow \frac{(x-4)(-x+1)}{(x-3)(x-1)}<0$

$\Rightarrow \frac{(x-4)(-x+1)(x-3)(x-1)}{(x-3{)}^2(x-1{)}^2}<0$


Critical points are 1,3 and 4
also $x\ne 1$ and $x\ne 3$ as denominator cannot be equal to zero.

Thus the solution set is (1,3) ∪ (4, ∞).
So, option c is correct.