Question:

Step 1: Understand the region and conditions. The region $A$ is defined by the inequalities $|\cos x-\sin x|\leq y\leq\sin x$ for $0\leq x\leq\dfrac{\pi}{2}$. This means the area is bounded above by $y_U = \sin x$ and below by $y_L = |\cos x - \sin x|$. For the region to be well-defined, we must have $y_L \leq y_U$. Step 2: Analyze the lower bound and the condition $y_L \leq y_U$. The lower bound is $y_L = |\cos x - \sin x|$. We need to consider two cases for this absolute value within the interval $0 \leq x \leq \frac{\pi}{2}$. Case 1: $0 \leq x \leq \frac{\pi}{4}$. In this interval, $\cos x \geq \sin x$, so $y_L = \cos x - \sin x$. The condition $y_L \leq y_U$ becomes $\cos x - \sin x \leq \sin x$. $$ \cos x \leq 2\sin x $$ $$ \cot x \leq 2 $$ $$ \tan x \geq \frac{1}{2} $$ Let $\alpha = \arctan\left(\frac{1}{2}\right)$. For $x \in [0, \frac{\pi}{4}]$, the condition $\tan x \geq \frac{1}{2}$ holds for $x \in [\alpha, \frac{\pi}{4}]$. For $x \in [0, \alpha)$, $\tan x < \frac{1}{2}$, which implies $y_L > y_U$, so the region is not defined in $[0, \alpha)$. Thus, the integration must start from $x=\alpha$. Case 2: $\frac{\pi}{4} \leq x \leq \frac{\pi}{2}$. In this interval, $\sin x \geq \cos x$, so $y_L = \sin x - \cos x$. The condition $y_L \leq y_U$ becomes $\sin x - \cos x \leq \sin x$. $$ -\cos x \leq 0 $$ $$ \cos x \geq 0 $$ This condition is true for all $x \in [\frac{\pi}{4}, \frac{\pi}{2}]$. Combining both cases, the interval over which the region is defined and we need to integrate is $\left[\alpha, \frac{\pi}{2}\right]$, where $\alpha = \arctan\left(\frac{1}{2}\right)$. Step 3: Set up the definite integral for the area. The area of the region is given by the integral of the difference between the upper and lower bounds: $$ \text{Area} = \int_{\alpha}^{\pi/2} (y_U - y_L) \, dx = \int_{\alpha}^{\pi/2} (\sin x - |\cos x - \sin x|) \, dx $$ We need to split the integral at $x=\frac{\pi}{4}$ because of the absolute value function: $$ \text{Area} = \int_{\alpha}^{\pi/4} (\sin x - (\cos x - \sin x)) \, dx + \int_{\pi/4}^{\pi/2} (\sin x - (\sin x - \cos x)) \, dx $$ Simplifying the integrands: $$ \text{Area} = \int_{\alpha}^{\pi/4} (2\sin x - \cos x) \, dx + \int_{\pi/4}^{\pi/2} \cos x \, dx $$ Step 4: Evaluate the first integral. The antiderivative of $(2\sin x - \cos x)$ is $(-2\cos x - \sin x)$. Evaluating from $\alpha$ to $\frac{\pi}{4}$: $$ \left[-2\cos x - \sin x\right]_{\alpha}^{\pi/4} = \left(-2\cos\left(\frac{\pi}{4}\right) - \sin\left(\frac{\pi}{4}\right)\right) - (-2\cos\alpha - \sin\alpha) $$ $$ = \left(-2\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}\right) - (-2\cos\alpha - \sin\alpha) $$ $$ = -\frac{3}{\sqrt{2}} + 2\cos\alpha + \sin\alpha $$ Since $\alpha = \arctan\left(\frac{1}{2}\right)$, we can construct a right triangle with opposite side 1 and adjacent side 2. The hypotenuse is $\sqrt{1^2+2^2} = \sqrt{5}$. Therefore, $\sin\alpha = \frac{1}{\sqrt{5}}$ and $\cos\alpha = \frac{2}{\sqrt{5}}$. Substitute these values: $$ -\frac{3}{\sqrt{2}} + 2\left(\frac{2}{\sqrt{5}}\right) + \frac{1}{\sqrt{5}} = -\frac{3}{\sqrt{2}} + \frac{4}{\sqrt{5}} + \frac{1}{\sqrt{5}} $$ $$ = -\frac{3}{\sqrt{2}} + \frac{5}{\sqrt{5}} = -\frac{3}{\sqrt{2}} + \sqrt{5} $$ Step 5: Evaluate the second integral. The antiderivative of $\cos x$ is $\sin x$. Evaluating from $\frac{\pi}{4}$ to $\frac{\pi}{2}$: $$ \left[\sin x\right]_{\pi/4}^{\pi/2} = \sin\left(\frac{\pi}{2}\right) - \sin\left(\frac{\pi}{4}\right) $$ $$ = 1 - \frac{1}{\sqrt{2}} $$ Step 6: Calculate the total area. Add the results from Step 4 and Step 5: $$ \text{Area} = \left(-\frac{3}{\sqrt{2}} + \sqrt{5}\right) + \left(1 - \frac{1}{\sqrt{2}}\right) $$ $$ \text{Area} = \sqrt{5} + 1 - \frac{3}{\sqrt{2}} - \frac{1}{\sqrt{2}} $$ $$ \text{Area} = \sqrt{5} + 1 - \frac{4}{\sqrt{2}} $$ To rationalize the denominator, multiply $\frac{4}{\sqrt{2}}$ by $\frac{\sqrt{2}}{\sqrt{2}}$: $$ \frac{4}{\sqrt{2}} = \frac{4\sqrt{2}}{2} = 2\sqrt{2} $$ So, the total area is: $$ \text{Area} = \sqrt{5} + 1 - 2\sqrt{2} $$ The final answer is $\sqrt{5} - 2\sqrt{2} + 1$. The correct option is Option 4.