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(C) To find the area bounded by the curves $y = x + \sin x$ and $y = x$,we first find their points of intersection by setting $x + \sin x = x$.This si

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$4$

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(C) To find the area bounded by the curves $y = x + \sin x$ and $y = x$,we first find their points of intersection by setting $x + \sin x = x$.This simplifies to $\sin x = 0$,which gives $x = 0, \pi, 2\pi$ in the interval $[0, 2\pi]$.The area $A$ is given by the integral of the absolute difference between the curves:$A = \int_{0}^{2\pi} |(x + \sin x) - x| \, dx = \int_{0}^{2\pi} |\sin x| \, dx$.We split the integral at the point where $\sin x$ changes sign:$A = \int_{0}^{\pi} \sin x \, dx + \int_{\pi}^{2\pi} (-\sin x) \, dx$.Evaluating the integrals:$\int_{0}^{\pi} \sin x \, dx = [-\cos x]_{0}^{\pi} = -(-1 - 1) = 2$.$\int_{\pi}^{2\pi} -\sin x \, dx = [\cos x]_{\pi}^{2\pi} = (1 - (-1)) = 2$.Total Area $A = 2 + 2 = 4$.

Column-$I$	Column-$II$
$(A)$ In a triangle $\triangle XYZ$,let $a, b$ and $c$ be the lengths of the sides opposite to the angles $X, Y$ and $Z$,respectively. If $2(a^2-b^2)=c^2$ and $\lambda=\frac{\sin(X-Y)}{\sin Z}$,then possible values of $n$ for which $\cos(n\pi\lambda)=0$ is (are)	$(P)$ $1$
$(B)$ In a triangle $\triangle XYZ$,let $a, b$ and $c$ be the lengths of the sides opposite to the angles $X, Y$ and $Z$,respectively. If $1+\cos 2X-2\cos 2Y=2\sin X\sin Y$,then possible value$(s)$ of $\frac{a}{b}$ is (are)	$(Q)$ $2$
$(C)$ In $\mathbb{R}^2$,let $\sqrt{3}\hat{i}+\hat{j}$,$\hat{i}+\sqrt{3}\hat{j}$ and $\beta\hat{i}+(1-\beta)\hat{j}$ be the position vectors of $X, Y$ and $Z$ with respect to the origin $O$,respectively. If the distance of $Z$ from the bisector of the acute angle of $\overline{OX}$ with $\overline{OY}$ is $\frac{3}{\sqrt{2}}$,then possible value$(s)$ of $\|\beta\|$ is (are)	$(R)$ $3$
$(D)$ Suppose that $F(\alpha)$ denotes the area of the region bounded by $x=0, x=2, y^2=4x$ and $y=\|\alpha x-1\|+\|\alpha x-2\|+\alpha x$,where $\alpha \in \{0, 1\}$. Then the value$(s)$ of $F(\alpha)+\frac{8}{3}\sqrt{2}$,when $\alpha=0$ and $\alpha=1$,is (are)	$(S)$ $5$
	$(T)$ $6$

For $x \in [0, 2\pi]$,the area of the region bounded by the curves $y = x + \sin x$ and $y = x$ is:

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