Question:medium
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 :
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 :
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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.
Updated On: Apr 2, 2026
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The Correct Option is C
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