You visited us 1 times! Enjoying our articles? Unlock Full Access!

Absolute Value Function

If |x2-2x-3|+...

Question

If $| x^{2} - 2 x - 3 | + | x + 3 | = | x^{2} - x |$ , then $x$ lies in

[3, \infty)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

[- 3, 0] \cup [3, \infty)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $[- 3, - 1] \cup [3, \infty)$
$| x^{2} - 2 x - 3 | + | x + 3 | = | x^{2} - x |$
Let $x^{2} - 2 x - 3 = a, x + 3 = b$
Here $a + b = x^{2} - x$
Clearly, $| a | + | b | = | a + b |$ that is possible only when
$a b \geq 0$
$\Rightarrow (x^{2} - 2 x - 3) (x + 3) \geq 0$
$\Rightarrow (x - 3) (x + 1) (x + 3) \geq 0$
$\Rightarrow x \in [- 3, - 1] \cup [3, \infty)$

Suggest Corrections

Similar questions

| x^{2} - 2 x - 3 | + | x + 3 | = | x^{2} - x |

x

\frac{x^{2} + 14 x + 9}{x^{2} + 2 x + 3}

\frac{2 x - 3}{x^{2} - 2 x + 3} = \frac{3 x - 5}{x^{2} - 3 x + 5}

x

[0, 2]

1

0

x

x^{2} - 2 x + 3 > 0, 12 x^{2} + 4 x + 3 > 0

x

{log}_{(x + 3)} (x^{2} - x) < 1

Join BYJU'S Learning Program