JEE MainMathematicsModerateInequalities and absolute value

The number of negative integers satisfying the equation
$|2x - 4| - 2\left| x^{2} + x - 3 \right| + 2|(x - 1)(x + 1)| = 0$ is

Show answer and solution

|2x - 4| - 2\left| x^{2} + x - 3 \right| + 2\left| (x - 1)(x + 1) \right| < 0

Critical points:

x = 2,\ \frac{- 1 \pm \sqrt{13}}{2},\ 1,\ - 1

Case 1;

x \leq \frac{- 1 - \sqrt{13}}{2}

- (2x - 4) - 2\left( x^{2} + x - 3 \right) + 2(x - 1)(x + 1)

- 2x + 4 - 2x^{2} - 2x + 6 + 2x^{2} - 2 = 0

- 4x + 8 = 0

\Rightarrow n = 2

; Not possible
Case 2;

\frac{- 1 - \sqrt{13}}{2} < x \leq - 1

- (2x - 4) + 2\left( x^{2} + x + 3 \right) + 2x^{2} - 2 = 0

\Rightarrow x = x \pm 1

\Rightarrow x = - 1

Case 3;

- 1 < x \leq \frac{- 1 + \sqrt{13}}{2}

- 1 < x \leq 0

(As we are only concerned with negative values)
No values.

Source: Grass-G / M-05 Mathematics Paper 10 October.docx

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