Question

If \( |x-3| < 2x+9 \), then \(x\) lies in the interval

• A. \( (-\infty,-2) \)
• B. \( (-2,0) \)
• C. \( (-2,\infty) \)
• D. \( (2,\infty) \)
• E. \( (-12,-2) \)

Hint:

For inequalities of the form \( |A| < B \), always: 1. Ensure \(B > 0\), 2. Convert into double inequality, 3. Solve both parts, 4. Take intersection of all conditions.

Answer 1

The Correct Option is C

Concept: For any real expression \(A\), the inequality \[ |A| < B \] is equivalent to the compound inequality \[ -B < A < B, \] provided that \(B > 0\). This condition is very important because absolute value is always non-negative, so if the right-hand side is not positive, the inequality may not hold for any real \(x\).

Step 1: Identify the expressions
Given: \[ |x - 3| < 2x + 9 \] Here, \[ A = x - 3, \quad B = 2x + 9 \] Before directly applying the standard inequality, we must ensure: \[ 2x + 9 > 0 \]

Step 2: Solve the condition \(2x+9>0\)
\[ 2x + 9 > 0 \Rightarrow 2x > -9 \Rightarrow x > -\frac{9}{2} \] This gives a necessary condition for the inequality to even exist.

Step 3: Apply modulus inequality
Now apply: \[ -(2x+9) < x-3 < 2x+9 \] This splits into two separate inequalities:

Step 4: Solve the left inequality
\[ -(2x+9) < x-3 \] Expanding: \[ -2x - 9 < x - 3 \] Bring like terms together: \[ -9 + 3 < x + 2x \] \[ -6 < 3x \] \[ x > -2 \]

Step 5: Solve the right inequality
\[ x - 3 < 2x + 9 \] Rearrange: \[ -3 - 9 < 2x - x \] \[ -12 < x \] \[ x > -12 \]

Step 6: Combine all conditions carefully
We now have three conditions: \[ x > -\frac{9}{2}, \quad x > -2, \quad x > -12 \] Among these, the strongest restriction is: \[ x > -2 \]

Step 7: Final interval
Thus the solution set is: \[ (-2, \infty) \] \[ \boxed{(-2,\infty)} \]

Step 8: Verification (very important for clarity)
Take a value \(x = 0\): \[ |0-3| = 3,\quad 2(0)+9 = 9 \Rightarrow 3 < 9 \quad \checkmark \] Take \(x = -3\): \[ |-6| = 6,\quad 2(-3)+9 = 3 \Rightarrow 6 < 3 \quad \times \] So values less than \(-2\) fail, confirming correctness.