Aspire Faculty ID #18000 · Topic: JEE Main 2026 (23 January Morning Shift) · Just now
JEE Main 2026 (23 January Morning Shift)

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

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Next 10 Questions — JEE Main 2026 (23 January Morning Shift)

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