Question

Let \([t]\) represent the greatest integer not exceeding \( t \) and \( C = 1 - 2e^2 \).
If the function \[ f(x) = \begin{cases} [e^x], & x < 0 \\ ae^x + [x - 2], & 0 \leq x < 2 \\ [e^x] - C, & x \geq 2 \end{cases} \] is continuous at \(x = 2\), then \(f(x)\) is discontinuous at:

• A. \(x = 1\) only
• B. \(x = 0\) and \(x = 1\)
• C. \(x = 0\) only
• D. \(x = 0, x = 1\) and \(x = \frac{1}{2}\)

Hint:

Step or floor functions often create discontinuities at integer transitions—evaluate one-sided limits to confirm.

Answer 1

The Correct Option is A

We analyze each piece: 1. For \(x < 0\): \(f(x) = [e^x]\), greatest integer function, discontinuous at \(x = 0\). But at \(x = 0\), function switches to next case. 2. \(0 \leq x < 2\): \(f(x) = ae^x + [x - 2]\). For \(0 \leq x < 1\), \([x - 2] = -2\); for \(1 \leq x < 2\), \([x - 2] = -1\).
So, discontinuity possible at \(x = 1\), where jump in step function occurs. 3. At \(x = 2\): we are told \(f\) is continuous at \(x = 2\). Hence, only point of discontinuity is at \(x = 1\).