Question

If $$f(x) = \begin{cases} 2 \sin x & \text{for} \ -\pi \leq x \leq -\frac{\pi}{2}, a \sin x + b & \text{for} \ -\frac{\pi}{2}<x<\frac{\pi}{2}, \cos x & \text{for} \ \frac{\pi}{2} \leq x \leq \pi,\end{cases}$$and it is continuous on $[- \pi, \pi]$, then the values of $ a $ and $ b $ are:

• A. \( a = 1 \) and \( b = 1 \)
• B. \( a = -1 \) and \( b = -1 \)
• C. \( a = -1 \) and \( b = 1 \)
• D. \( a = 1 \) and \( b = -1 \)

Hint:

When a piecewise function is continuous at the points where the function changes form, equate the corresponding parts of the function at those points and solve for the constants.

Answer 1

The Correct Option is D

To ensure the function is continuous on \([- \pi, \pi]\), the values at the points where the function changes must be the same.
Step 1: Continuity at \( x = -\frac{\pi}{2} \)
At \( x = -\frac{\pi}{2} \), the first and second parts of the function must be equal: \[ 2 \sin \left( -\frac{\pi}{2} \right) = a \sin \left( -\frac{\pi}{2} \right) + b. \] This simplifies to: \[ -2 = -a + b. \] Thus, the first equation is: \[ a - b = 2. \quad \text{(Equation 1)} \]
Step 2: Continuity at \( x = \frac{\pi}{2} \)
At \( x = \frac{\pi}{2} \), the second and third parts of the function must be equal: \[ a \sin \left( \frac{\pi}{2} \right) + b = \cos \left( \frac{\pi}{2} \right). \] This simplifies to: \[ a + b = 0. \quad \text{(Equation 2)} \]
Step 3: Solve the system of equations
We now solve the system of equations: \[ a - b = 2 \quad \text{and} \quad a + b = 0. \] From Equation 2, \( b = -a \). Substituting into Equation 1: \[ a - (-a) = 2, \] \[ 2a = 2 \quad \Rightarrow \quad a = 1. \] Substituting \( a = 1 \) into \( a + b = 0 \): \[ 1 + b = 0 \quad \Rightarrow \quad b = -1. \] Thus, the values of \( a \) and \( b \) are \( a = 1 \) and \( b = -1 \).

Answer 2

To determine the values of \( a \) and \( b \) that make the function \( f(x) \) continuous at \( x = \pm\frac{\pi}{2} \), we analyze the continuity conditions at these points.

1. Continuity at \( x = \frac{\pi}{2} \):
The function changes definition at \( \frac{\pi}{2} \). We examine both sides:

Left-hand limit (LHL):
\[ \lim_{x \to \pi/2^-} (a \sin x + b) = a \sin\left(\frac{\pi}{2}\right) + b = a + b \]

Right-hand limit (RHL):
\[ \lim_{x \to \pi/2^+} \cos x = \cos\left(\frac{\pi}{2}\right) = 0 \]

For continuity, LHL = RHL:
\[ a + b = 0 \quad \text{(Equation 1)} \]

2. Continuity at \( x = -\frac{\pi}{2} \):
The function also changes definition at \( -\frac{\pi}{2} \):

Left-hand limit (LHL):
\[ \lim_{x \to -\pi/2^-} 2 \sin x = 2 \sin\left(-\frac{\pi}{2}\right) = -2 \]

Right-hand limit (RHL):
\[ \lim_{x \to -\pi/2^+} (a \sin x + b) = a \sin\left(-\frac{\pi}{2}\right) + b = -a + b \]

For continuity, LHL = RHL:
\[ -a + b = -2 \quad \text{(Equation 2)} \]

3. Solving the System of Equations:
From Equation 1: \( b = -a \)
Substitute into Equation 2:
\[ -a + (-a) = -2 \Rightarrow -2a = -2 \Rightarrow a = 1 \]
Then \( b = -1 \)

Final Answer:
The values that make \( f(x) \) continuous are \( a = 1 \) and \( b = -1 \).