Question

The solution of the inequality $|x^2-4x|<5$ is

• A. $(-1,5)$
• B. $(-4,5)$
• C. $(-5,4)$
• D. $(-1,4)$

Answer 1

Answer: (A)

We have,
$|x^2-4x|<5$
$\therefore -5<x^2-4x<5$
Case I $x^2-4x<5$
$\Rightarrow x^2-4x-5,0$
$\Rightarrow (x-5)(x+1)<0$
$\Rightarrow x\in (-1,5)$
Case II $x^2-4x>-\mathrm{5...}(i)$
which is true $\forall x\in R$
$\Rightarrow x\in (-\infty ,\infty )....$ (ii)
From Eqn. (i) and (ii),
$x\in (-1,5)$