Question

The complete set of values of $x$ satisfying the inequation $(x-3)<\sqrt{x^2+4x-5}$ is

• A. $(-\infty ,-5]\cup [1,\infty )$
• B. $(-5,3]$
• C. $[3,5)$
• D. $(-5,3)$

Answer 1

Answer: (A)

$(-\infty ,-5]\cup [1,\infty )$

The domain of the given function:

$x^2+4x-5\ge 0$
$\Rightarrow x\in (-\infty ,-5]\cup [1,\infty )$
To solve this inequality, let us consider the sign of $(x-3)$.

Case 1: When $(x-3)$ is negative i.e. $x\in (-\infty ,-5]\cup [1,3)$
$L.H.S<R.H.S$
Hence, this is true for all the values of $x$. Hence, $x\in (-\infty ,-5]\cup [1,3)$ ..........(1)
Case 2: When, $x$ is non-negative, i.e. $x\in [3,\infty )$
$x-3<\sqrt{x^2+4x-5}$

Since both the sides of the inequality are non-negative, we can square both the sides. On squaring we get,

$\Rightarrow x^2-6x+9<x^2+4x-5$
$\Rightarrow x>\frac{7}{5}$

The common set of $x$ is $[3,\infty )$ ........(2)
The complete set is union of the two sets in (1) and (2)
$(-\infty ,-5]\cup [1,\infty )$.