The area enclosed by y=sqrt{5-x^{2}} and y=|x | vedclass

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(C) The curves are $y=\sqrt{5-x^2}$ (a semicircle of radius $\sqrt{5}$) and $y=|x-1|$.To find the intersection points, set $\sqrt{5-x^2} = |x-1|$.Squa

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$\left(\frac{5 \pi}{4}-\frac{1}{2}ight) 	ext{ sq units}$

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(C) The curves are $y=\sqrt{5-x^2}$ (a semicircle of radius $\sqrt{5}$) and $y=|x-1|$.To find the intersection points, set $\sqrt{5-x^2} = |x-1|$.Squaring both sides: $5-x^2 = x^2-2x+1 \implies 2x^2-2x-4=0 \implies x^2-x-2=0 \implies (x-2)(x+1)=0$.So, the intersection points are $x=-1$ and $x=2$.The area $A$ is given by $\int_{-1}^{2} (\sqrt{5-x^2} - |x-1|) dx$.$A = \int_{-1}^{2} \sqrt{5-x^2} dx - \int_{-1}^{2} |x-1| dx$.For the first integral, $\int_{-1}^{2} \sqrt{5-x^2} dx = \left[ \frac{x}{2}\sqrt{5-x^2} + \frac{5}{2}\sin^{-1}\left(\frac{x}{\sqrt{5}}\right) \right]_{-1}^{2} = (1 + \frac{5}{2}\sin^{-1}\frac{2}{\sqrt{5}}) - (-1 + \frac{5}{2}\sin^{-1}\frac{-1}{\sqrt{5}}) = 2 + \frac{5}{2}(\sin^{-1}\frac{2}{\sqrt{5}} + \sin^{-1}\frac{1}{\sqrt{5}})$.Since $\sin^{-1}\frac{2}{\sqrt{5}} + \sin^{-1}\frac{1}{\sqrt{5}} = \sin^{-1}(\frac{2}{\sqrt{5}}\sqrt{1-\frac{1}{5}} + \frac{1}{\sqrt{5}}\sqrt{1-\frac{4}{5}}) = \sin^{-1}(1) = \frac{\pi}{2}$, the first integral is $2 + \frac{5\pi}{4}$.For the second integral, $\int_{-1}^{2} |x-1| dx = \int_{-1}^{1} (1-x) dx + \int_{1}^{2} (x-1) dx = [x - \frac{x^2}{2}]_{-1}^{1} + [\frac{x^2}{2} - x]_{1}^{2} = (1 - \frac{1}{2} - (-1 - \frac{1}{2})) + (2 - 2 - (\frac{1}{2} - 1)) = (1.5) + (0.5) = 2$.Thus, $A = 2 + \frac{5\pi}{4} - 2 = \frac{5\pi}{4}$ sq units. Wait, re-evaluating the integral $\int_{-1}^{2} |x-1| dx$: $\int_{-1}^{1} (1-x) dx = [x - x^2/2]_{-1}^{1} = (1-0.5) - (-1-0.5) = 0.5 + 1.5 = 2$. $\int_{1}^{2} (x-1) dx = [x^2/2 - x]_1^2 = (2-2) - (0.5-1) = 0.5$. Total is $2.5 = 5/2$.$A = 2 + \frac{5\pi}{4} - 2.5 = \frac{5\pi}{4} - 0.5$ sq units.

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)	$(S)$ $5$
	$(T)$ $6$

The area enclosed by $y=\sqrt{5-x^{2}}$ and $y=|x-1|$ is

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