Step 1: For \( f(x) \) to be continuous at \( x = 2 \), the left-hand limit and the right-hand limit must be equal.
Hence, we equate the two functions at \( x = 2 \): \[ \lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x). \] Step 2: The left-hand limit for \( x<2 \) is \( f(x) = ax + 1 \), so: \[ \lim_{x \to 2^-} f(x) = 2a + 1. \] The right-hand limit for \( x \geq 2 \) is \( f(x) = x^2 + 7 \), so: \[ \lim_{x \to 2^+} f(x) = 2^2 + 7 = 4 + 7 = 11. \] Step 3: Equating both limits for continuity: \[ 2a + 1 = 11. \] Step 4: Solving for \( a \): \[ 2a = 10 \quad \Rightarrow \quad a = 5. \]
Let \[ f(t)=\int \left(\frac{1-\sin(\log_e t)}{1-\cos(\log_e t)}\right)dt,\; t>1. \] If $f(e^{\pi/2})=-e^{\pi/2}$ and $f(e^{\pi/4})=\alpha e^{\pi/4}$, then $\alpha$ equals