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 given function $f(x)$ is piecewise-defined on the interval $[0, \pi]$. We are given that $f(x)$ is continuous everywhere on this closed interval, which means it must be continuous at the critical boundary points where the definition changes, namely at $x = \frac{\pi}{4}$ and $x = \frac{\pi}{2}$.

Step 2: Key Formula or Approach:
For a function to be continuous at a point $x = c$, the Left-Hand Limit (LHL), Right-Hand Limit (RHL), and the function value must all be equal: $$\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = f(c)$$ We will apply this condition at $x = \frac{\pi}{4}$ and $x = \frac{\pi}{2}$ to form a system of linear equations in terms of $a$ and $b$.

Step 3: Detailed Explanation:
First, let's enforce continuity at $x = \frac{\pi}{4}$: $$\lim_{x \to \frac{\pi}{4}^-} f(x) = \lim_{x \to \frac{\pi}{4}^+} f(x)$$ $$\lim_{x \to \frac{\pi}{4}^-} \left(x + a\sqrt{2}\sin x\right) = \lim_{x \to \frac{\pi}{4}^+} \left(2x\cot x + b\right)$$ Substituting $x = \frac{\pi}{4}$: $$\frac{\pi}{4} + a\sqrt{2}\sin\left(\frac{\pi}{4}\right) = 2\left(\frac{\pi}{4}\right)\cot\left(\frac{\pi}{4}\right) + b$$ Since $\sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$ and $\cot\left(\frac{\pi}{4}\right) = 1$: $$\frac{\pi}{4} + a\sqrt{2}\left(\frac{1}{\sqrt{2}}\right) = \frac{\pi}{2}(1) + b$$ $$\frac{\pi}{4} + a = \frac{\pi}{2} + b$$ $$a - b = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4} \quad \text{--- (Equation 1)}$$ Next, let's enforce continuity at $x = \frac{\pi}{2}$: $$\lim_{x \to \frac{\pi}{2}^-} f(x) = \lim_{x \to \frac{\pi}{2}^+} f(x)$$ $$\lim_{x \to \frac{\pi}{2}^-} \left(2x\cot x + b\right) = \lim_{x \to \frac{\pi}{2}^+} \left(a\cos 2x - b\sin x\right)$$ Substituting $x = \frac{\pi}{2}$: $$2\left(\frac{\pi}{2}\right)\cot\left(\frac{\pi}{2}\right) + b = a\cos\left(2 \cdot \frac{\pi}{2}\right) - b\sin\left(\frac{\pi}{2}\right)$$ Since $\cot\left(\frac{\pi}{2}\right) = 0$, $\cos(\pi) = -1$, and $\sin\left(\frac{\pi}{2}\right) = 1$: $$\pi(0) + b = a(-1) - b(1)$$ $$b = -a - b$$ $$a + 2b = 0 \implies a = -2b \quad \text{--- (Equation 2)}$$ Now, substitute Equation 2 into Equation 1: $$-2b - b = \frac{\pi}{4}$$ $$-3b = \frac{\pi}{4} \implies b = -\frac{\pi}{12}$$ Using $b = -\frac{\pi}{12}$ in Equation 2 to find $a$: $$a = -2\left(-\frac{\pi}{12}\right) = \frac{\pi}{6}$$ Looking closely at the computed values, $a = \frac{\pi}{6}$ and $b = -\frac{\pi}{12}$. This matches perfectly with option (D).

Step 4: Final Answer:
The values of the parameters are $a = \frac{\pi}{6}$ and $b = -\frac{\pi}{12}$, corresponding to option (D).