Question

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

A $[3, \infty)$
B $[-3,0] \cup [3, \infty)$
C $[-3,-1] \cup [3, \infty)$
D $[-3,1] \cup [3, \infty)$

✨ Free AI Tools for You

Powered by Chatterbot AI

Answer

The correct option is C: [$-3, -1] \cup [3, \infty)$.

We begin with the given equation: $$ \left| x^2 - 2x - 3 \right| + |x + 3| = \left| x^2 - x \right| $$

Define $a$ and $b$ as follows: $$ a = x^2 - 2x - 3 \quad \text{and} \quad b = x + 3 $$

It can be observed that: $$ a + b = x^2 - x $$

From the properties of absolute values, the equality $|a| + |b| = |a + b|$ holds only when: $$ ab \geq 0 $$

Substituting back the values of $a$ and $b$ we get: $$ (x^2 - 2x - 3)(x + 3) \geq 0 $$

Upon factoring, this inequality becomes: $$ (x - 3)(x + 1)(x + 3) \geq 0 $$

To find the solution to this inequality, we must determine the intervals where this product is non-negative. The critical points of the inequality are $-3$, $-1$, and $3$. Testing the intervals defined by these points, we find that:

  1. $x \in (-\infty, -3)$ gives a negative product.
  2. $x \in [-3, -1]$ gives a non-negative product.
  3. $x \in (-1, 3)$ gives a negative product.
  4. $x \in [3, \infty)$ gives a non-negative product.

Combining these intervals where the product is non-negative, we conclude that: $$ x \in [-3, -1] \cup [3, \infty) $$

Thus, the correct answer is C: $[-3, -1] \cup [3, \infty)$.


Was this helpful?

India's 1st AI Doubt Solver for CBSE, JEE, and NEET

Ask a Question for Free

and then it's just ₹212 a month