Question

If \( f(x) = \begin{cases} \sin\left(\frac{\pi x}{2}\right), & x<1 \\ 3 - 2x, & x \ge 1 \end{cases} \), then \( f(x) \) has:

• A. local minimum at \(x=1\)
• B. local maximum at \(x=1\)
• C. Both local maximum and local minimum at \(x=1\)
• D. None of the above

Hint:

Local maximum if function increases before and decreases after the point.

Answer 1

The Correct Option is B

Concept: Check left-hand limit, right-hand limit, and value at \(x=1\).

Step 1: Left-hand limit as \(x \to 1^-\). \[ \lim_{x \to 1^-} f(x) = \sin\left(\frac{\pi \cdot 1}{2}\right) = \sin\frac{\pi}{2} = 1 \]

Step 2: Value at \(x=1\). \[ f(1) = 3 - 2(1) = 1 \]

Step 3: Right-hand limit as \(x \to 1^+\). \[ \lim_{x \to 1^+} f(x) = 3 - 2(1) = 1 \] Function is continuous at \(x=1\).

Step 4: Check derivative sign. For \(x<1\), \(f'(x) = \frac{\pi}{2} \cos\left(\frac{\pi x}{2}\right)>0\) near \(x=1^-\). For \(x>1\), \(f'(x) = -2<0\). Thus, \(f\) increases up to \(x=1\) and then decreases → local maximum at \(x=1\).