The area (in sq. units) of the region {(x, y) | vedclass

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(B) The region is defined by $0 \leq y \leq \min(2|x|+1, x^2+1)$ for $x \in [-3, 3]$.Due to symmetry about the $y$-axis,the total area is $2 	imes \i

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$\frac{64}{3}$

Vedclass · Answer

(B) The region is defined by $0 \leq y \leq \min(2|x|+1, x^2+1)$ for $x \in [-3, 3]$.Due to symmetry about the $y$-axis,the total area is $2 	imes \int_0^3 \min(2x+1, x^2+1) dx$.First,find the intersection of $y = 2x+1$ and $y = x^2+1$ for $x > 0$:$x^2+1 = 2x+1 \Rightarrow x^2 - 2x = 0 \Rightarrow x(x-2) = 0$.So,the curves intersect at $x = 0$ and $x = 2$.For $x \in [0, 2]$,$x^2+1 \leq 2x+1$,so $\min(2x+1, x^2+1) = x^2+1$.For $x \in [2, 3]$,$2x+1 \leq x^2+1$,so $\min(2x+1, x^2+1) = 2x+1$.Thus,the area is $2 \left[ \int_0^2 (x^2+1) dx + \int_2^3 (2x+1) dx ight]$.$= 2 \left[ \left( \frac{x^3}{3} + x ight)_0^2 + \left( x^2 + x ight)_2^3 ight]$$= 2 \left[ \left( \frac{8}{3} + 2 ight) + ((9+3) - (4+2)) ight]$$= 2 \left[ \frac{14}{3} + (12 - 6) ight] = 2 \left[ \frac{14}{3} + 6 ight] = 2 \left[ \frac{14+18}{3} ight] = 2 \left( \frac{32}{3} ight) = \frac{64}{3}$.

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 (in sq. units) of the region $\{(x, y): 0 \leq y \leq 2|x|+1, 0 \leq y \leq x^2+1, |x| \leq 3\}$ is

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