Question

Let $$f(x) = \begin{array}{cc} x + a\sqrt{2}\sin x & , 0 \le x \lt \frac{\pi}{4} \\ 2x\cot x + b & , \frac{\pi}{4} \le x \lt \frac{\pi}{2} \\ a\cos 2x - b\sin x & , \frac{\pi}{2} \le x \le \pi \end{array}$$ If $f(x)$ is continuous for $0 \le x \le \pi$, then

• A. $a = \frac{\pi}{6}, b = \frac{\pi}{12}$
• B. $a = -\frac{\pi}{6}, b = -\frac{\pi}{12}$
• C. $a = -\frac{\pi}{6}, b = \frac{\pi}{12}$
• D. $a = \frac{\pi}{6}, b = -\frac{\pi}{12}$

Hint:

When checking continuity for piecewise trigonometric functions, check the point where terms easily vanish first. At $x = \frac{\pi}{2}$, $\cot x$ becomes 0, which immediately yields a simple relation between $a$ and $b$ ($a = -2b$), instantly eliminating options (A) and (B).

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The piecewise function f(x) must be continuous on [0, π], so we enforce continuity conditions at the boundary points x = π/4 and x = π/2 to determine a and b.

Step 2: Key Formula or Approach:
Continuity at a point x = c requires Left-Hand Limit = Right-Hand Limit = f(c). We equate the adjoining piecewise expressions at each junction.

Step 3: Detailed Explanation:
At x = π/4: (π/4) + a = (π/2) + b → a – b = π/4. At x = π/2: b = –a – b → a = –2b. Solving the system yields b = –π/12 and a = π/6.

Step 4: Final Answer:
The values are a = π/6, b = –π/12, corresponding to option (D).