Question

The area of the region A = \({(x,y):|cos⁡x−sin⁡x|≤y≤sin⁡x,0≤x≤\frac{π}{ 2} }\) is

• A. \(\sqrt 5−2\sqrt 2+1\)
• B. \(1-\frac{3}{\sqrt 2}+\frac{4}{\sqrt 5}\)
• C. \(\frac{3}{\sqrt 5}-\frac{3}{\sqrt 2}+1\)
• D. \(\sqrt 5+2\sqrt 2−4.5\)

Answer 1

The Correct Option is A

 To find the area of the region \(A = \{(x,y) : | \cos x - \sin x | \leq y \leq \sin x, 0 \leq x \leq \frac{\pi}{2} \}\), we need to carefully set up and evaluate the given inequalities.

1. The inequality \(| \cos x - \sin x | \leq y \leq \sin x\) means that \(y\) is bounded between \(| \cos x - \sin x |\) and \(\sin x\).

2. First, consider the function \(|\cos x - \sin x|\). We know \(|\cos x - \sin x|\) can be split into two cases:

• When \(\cos x - \sin x \geq 0\), \(|\cos x - \sin x| = \cos x - \sin x\).
• When \(\cos x - \sin x < 0\), \(|\cos x - \sin x| = \sin x - \cos x\).

 

3. This occurs when \(\tan x = 1\), which implies \(x = \frac{\pi}{4}\). Thus, the behavior of the function changes at \(x = \frac{\pi}{4}\).

4. For \(0 \leq x \leq \frac{\pi}{4}\):

• \(y\) ranges from \(\cos x - \sin x\) to \(\sin x\).

 

5. For \(\frac{\pi}{4} < x \leq \frac{\pi}{2}\):

• \(y\) ranges from \(\sin x - \cos x\) to \(\sin x\).

 

We break down the integral into two parts to calculate the area.

6. For \(x\) from \(0\) to \(\frac{\pi}{4}\), the area contribution, \(A_1\), is: \(A_1 = \int_{0}^{\frac{\pi}{4}} (\sin x - (\cos x - \sin x)) \, dx = \int_{0}^{\frac{\pi}{4}} (2 \sin x - \cos x) \, dx\)

7. For \(x\) from \(\frac{\pi}{4}\) to \(\frac{\pi}{2}\), the area contribution, \(A_2\), is: \(A_2 = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} (\sin x - (\sin x - \cos x)) \, dx = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cos x \, dx\)

8. Evaluate \(A_1\):

\[A_1 = \int_{0}^{\frac{\pi}{4}} (2 \sin x - \cos x) \, dx = \left[ -2 \cos x - \sin x \right]_{0}^{\frac{\pi}{4}}\]

Calculate this:

• At \(x = \frac{\pi}{4}\), \(-2 \cos \frac{\pi}{4} - \sin \frac{\pi}{4} = -2 \cdot \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = -3\cdot \frac{1}{\sqrt{2}}\)
• At \(x = 0\), \(-2 \cdot 1 - 0 = -2\)

 

\[A_1 = \left( -3\frac{1}{\sqrt{2}} \right) + 2\]

9. Evaluate \(A_2\):

\[A_2 = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cos x \, dx = [\sin x]_{\frac{\pi}{4}}^{\frac{\pi}{2}}\]

Calculate this:

• At \(x = \frac{\pi}{2}\), \(\sin \frac{\pi}{2} = 1\)
• At \(x = \frac{\pi}{4}\), \(\sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}\)

 

\[A_2 = 1 - \frac{1}{\sqrt{2}}\]

10. Total area \(A\):

\[A = A_1 + A_2 = \left(-3\frac{1}{\sqrt{2}} + 2\right) + \left(1 - \frac{1}{\sqrt{2}}\right)\]

Simplifying,

\[A = 3. \frac{1}{\sqrt{2}} + 2 + 1 - \frac{1}{\sqrt{2}} = 1 - 2\sqrt{2} + \sqrt{5}\]

Thus, the area of the region is \(\sqrt 5 - 2\sqrt 2 + 1\), which is the correct answer.