Question

If \(=\mid \begin{array}{ll}\mathrm{x}+\mathrm{a} \sqrt{2} \operatorname{Sin} \mathrm{x}, & 0 \leq \mathrm{x} \leq(\pi / 4) \\ 2 \mathrm{x} \operatorname{Cot} \mathrm{x}+\mathrm{b}, & {[(\pi / 4)<\mathrm{x} \leq(\pi / 2)]} \\ \mathrm{a} \operatorname{Cos} 2 \mathrm{x}+\mathrm{b} \operatorname{Sin} \mathrm{x}, & (\pi / 2)<\mathrm{x} \leq \pi\end{array}\) is continuous on \([0, \pi]\), then \(\mathrm{a}=\ldots \ldots\) and \(\mathrm{b}=\ldots \ldots .\) (a) \(\mathrm{a}=(5 \pi / 2), \mathrm{b}=(5 \pi / 4)\) (b) \(\mathrm{a}=-(5 \pi / 2), \mathrm{b}=-(5 \pi / 4)\) (c) \(\mathrm{a}=(\pi / 6), \mathrm{b}=[(-\pi) / 12]\) (d) \(\mathrm{a}=-(5 \pi / 4), \mathrm{b}=(5 \pi / 2)\)

Short answer

The short answer is: \(\mathrm{a}=(\frac{5 \pi}{2}), \mathrm{b}=(\frac{5 \pi}{4})\).

Step-by-step solution

Analyze the first two parts of the function

At the point \(x = \pi/4\), the first and second parts of the function meet. To ensure continuity, we'll need to make sure that the output of the first and second parts are equal at \(x = \pi/4\): \((x+a\sqrt{2}\sin{x})|_{x=\pi/4} = (2xCot{x}+b)|_{x=\pi/4}\)

Calculate the outputs at \(x = \pi/4\)

Now, we'll calculate the outputs of both functions at \(x = \pi/4\): \(\pi/4 + a\sqrt{2}\sin{(\frac{\pi}{4})} = \frac{2(\pi/4)\cot{(\pi/4)}}{b}\)

Simplify and solve for a

We can simplify the equation further: \(\pi/4 + a\sqrt{2}\cdot\frac{\sqrt{2}}{2} = \pi/2 + b\) Now, solve for 'a': \(a = \frac{5\pi}{2} - 2b\)

Analyze the second and third parts of the function

At the point \(x = \pi/2\), the second and third parts of the function meet. To ensure continuity, we'll need to make sure that the output of the second and third parts are equal at \(x = \pi/2\): \((2xCot{x}+b)|_{x=\pi/2} = (a\cos{2x} + b\sin{x})|_{x=\pi/2}\)

Calculate the outputs at \(x = \pi/2\)

Now, we'll calculate the outputs of both functions at \(x = \pi/2\): \((a\cos{\pi} + b\sin{\frac{\pi}{2}}) = (2(\pi/2)\operatorname{Cot}{(\pi/2)}+b)\)

Simplify and solve for a

We can simplify the equation further: \(b = -a + b\) Substituting the expression of 'a' from step 3, we get: \(b=(-\frac{5\pi}{2}+2b)\) Now we can solve for 'b': \(b = \frac{5\pi}{4}\) Finally, substituting the value of b found above into the equation from step 1, we obtain the value of a: \(a = \frac{5\pi}{2} - 2 \cdot \frac{5\pi}{4} = \frac{5\pi}{2}\) After finding the values of 'a' and 'b', we can choose the correct answer. The answer is (a) \(\mathrm{a}=(5 \pi / 2), \mathrm{b}=(5 \pi / 4)\).

Key concepts

Trigonometric Identities

Understanding trigonometric identities can greatly simplify solving equations involving trigonometric functions. One key identity is the Pythagorean identity:

• \( \sin^2{x} + \cos^2{x} = 1 \)

This identity can help in transforming expressions, making it easier to solve equations. In the context of piecewise functions, familiarizing yourself with basic trigonometric values at specific angles is vital. For example, at \( x = \pi/4 \), both sine and cosine have specific values:

• \( \sin{(\pi/4)} = \cos{(\pi/4)} = \frac{\sqrt{2}}{2} \)
• \( \cot{(\pi/4)} = 1 \)

Having these values in mind simplifies the calculations, as it allows quick substitution and simplification, especially when ensuring continuity of functions across different segments.

Solving Algebraic Equations

Solving algebraic equations in piecewise functions involves equating expressions at certain points to ensure continuity. This means you'll need to solve for unknown variables by setting the outputs of different function parts equal at their boundaries.
In our example, solving the equation at \( x = \pi/4 \) involved setting \( x + a\sqrt{2}\sin{x} \) equal to \( 2x\cot{x} + b \). By substituting known values for trigonometric functions, we derive the relationship:

• \( \pi/4 + a\sqrt{2}\cdot\frac{\sqrt{2}}{2} = \pi/2 + b \)

This simplifies algebraically to find 'a' and 'b'. It's like solving two puzzles: aligning pieces such that each section flows into the next, seamlessly.

Calculating Limits

Calculating limits in piecewise functions focuses on understanding the behavior of functions at boundary points. For continuity, the left hand and right hand limits at these boundaries must be equal.
For instance, at the boundary \( x = \pi/2 \), we equate the left hand limit \( 2x\cot{x} + b \) with the right hand limit \( a\cos{2x} + b\sin{x} \).
Here:

• As \( x \to \pi/2^- \), \( \cot{(\pi/2)} \) diminishes, so careful handling of undefined expressions is necessary.

The main idea is checking the function behavior as it approaches certain values. Finding limits like these helps ensure the function is smooth—no unexpected jumps or breaks at those points. This concept of limits smoothly ties into broader calculus themes, as it ensures a natural progression from one segment of the function to the next.