Question

If
\( f(x) = \begin{cases} 1, & 0 < x \le \dfrac{3\pi}{4} \\ 2\sin\left(\dfrac{2x}{9}\right), & \dfrac{3\pi}{4} < x < \pi \end{cases} \)
then:

• A. \(f(x)\) is continuous at \(x=0\)
• B. \(f(x)\) is continuous at \(x=\pi\)
• C. \(f(x)\) is continuous at \(x=\frac{3\pi}{4}\)
• D. f(x) is discontinuous at x=(3π)/(4)

Hint:

Check continuity only at boundary points of piecewise functions.

Answer 1

The Correct Option is C

Step 1:
Left-hand limit at \( x = \dfrac{3\pi}{4} \) is \( 1 \).
Step 2:
Right-hand limit:
\( 2\sin\left(\dfrac{2}{9} \cdot \dfrac{3\pi}{4}\right) \)
\( = 2\sin\left(\dfrac{\pi}{6}\right) = 1 \)
Step 3:
Since LHL = RHL = \( f\left(\dfrac{3\pi}{4}\right) \), function is continuous there.