Question

Let \( f(x) = \begin{cases} \frac{ax^2 + 2ax + 3}{4x^2 + 4x - 3}, & x \neq -\frac{3}{2}, \frac{1}{2} \\ b, & x = -\frac{3}{2}, \frac{1}{2} \end{cases} \) be continuous at \( x = -\frac{3}{2} \). If \( f(x) = \frac{7}{5} \), then \( x \) is equal to :

• A. 0
• B. 2
• C. 1
• D. 1.4

Hint:

If a rational function is continuous at a point where the denominator is zero, that point must be a "removable discontinuity," meaning the factor causing zero in the denominator must also exist in the numerator.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
For \( f(x) \) to be continuous at \( x = -3/2 \), the limit as \( x \to -3/2 \) must exist. This means the numerator must vanish at \( x = -3/2 \) since the denominator is zero there.
Step 2: Key Formula or Approach:
Factor the denominator: \( 4x^2 + 4x - 3 = (2x + 3)(2x - 1) \).
Step 3: Detailed Explanation:
Numerator must be zero at \( x = -3/2 \): \[ a(-3/2)^2 + 2a(-3/2) + 3 = 0 \implies \frac{9a}{4} - 3a + 3 = 0 \implies a = 4. \] The function becomes: \[ f(x) = \frac{4x^2 + 8x + 3}{(2x+3)(2x-1)} = \frac{(2x+3)(2x+1)}{(2x+3)(2x-1)} = \frac{2x+1}{2x-1} \] To find \( x \) when \( f(x) = 7/5 \): \[ \frac{2x+1}{2x-1} = \frac{7}{5} \implies 10x + 5 = 14x - 7 \implies 4x = 12 \implies x = 3. \] Step 4: Final Answer:
\( x = 1\).