Question

If the function

\[f(x)=\left\{\begin{array}{cc} x+a \sqrt{2} \sin x & \text { if } 0 \leq x \leq \frac{\pi}{4} \\ 2 x \cot x+b & \text { if } \frac{\pi}{4} < x \leq \frac{\pi}{2} \\ a \cos 2 x-b \sin x & \text { if } \frac{\pi}{2} < x \leq \pi \end{array}\right.\]

is continuous in \([0, \pi]\) then \(a-b=\)

• A. \(\frac{\pi}{4}\)
• B. \(\frac{\pi}{12}\)
• C. \(\frac{5\pi}{12}\)
• D. \(\frac{7\pi}{12}\)

Hint:

Equate the left and right limits at x = pi/4 and x = pi/2 to get two equations in a and b.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
The function is continuous on \([0,\pi]\) if the pieces join at \(x=\dfrac{\pi}{4}\) and \(x=\dfrac{\pi}{2}\).

Step 2: At x = pi/4
Left: \(\dfrac{\pi}{4}+a\sqrt2\sin\dfrac{\pi}{4}=\dfrac{\pi}{4}+a\). Right: \(2\cdot\dfrac{\pi}{4}\cot\dfrac{\pi}{4}+b=\dfrac{\pi}{2}+b\).

Step 3: Equate
\(\dfrac{\pi}{4}+a=\dfrac{\pi}{2}+b\Rightarrow a-b=\dfrac{\pi}{4}\).

Step 4: Check at x = pi/2
Middle: \(2\cdot\dfrac{\pi}{2}\cot\dfrac{\pi}{2}+b=b\). Right: \(a\cos\pi-b\sin\dfrac{\pi}{2}=-a-b\). So \(a=-2b\), giving \(b=-\dfrac{\pi}{12}\) and \(a=\dfrac{\pi}{6}\), consistent with \(a-b=\dfrac{\pi}{4}\).

Final Answer:
The value of a - b is pi/4. \[ \boxed{\text{(A) }\dfrac{\pi}{4}} \]