Question

Let $a$ be a real number and \[ f(x) = \begin{cases} -2\sin x, & x \le -\frac{\pi}{2} \\ 1 + a\sin x, & -\frac{\pi}{2}<x \le \frac{\pi}{2} \end{cases} \] If $f(x)$ is continuous, then the value of $a$ is:

• A. 0
• B. $-1$
• C. 3
• D. 2

Hint:

For piecewise functions, check continuity at boundary points by equating LHS and RHS.

Answer 1

The Correct Option is B

Concept: For continuity at $x = -\frac{\pi}{2}$, LHS = RHS.
Step 1: Left-hand value.
\[ f\left(-\frac{\pi}{2}\right) = -2\sin\left(-\frac{\pi}{2}\right) = -2(-1) = 2 \]
Step 2: Right-hand limit.
\[ \lim_{x \to -\frac{\pi}{2}^+} (1 + a\sin x) = 1 + a(-1) = 1 - a \]
Step 3: Equate.
\[ 2 = 1 - a \Rightarrow a = -1 \]
Hence, the value of $a$ is $-1$.