JEE MAIN - Mathematics (2023 - 10th April Evening Shift - No. 14)
If the area of the region $$\left\{(x, \mathrm{y}):\left|x^{2}-2\right| \leq y \leq x\right\}$$ is $$\mathrm{A}$$, then $$6 \mathrm{A}+16 \sqrt{2}$$ is equal to __________.
Answer
27
Explanation
$$
\text { We have, }\left\{(x, y):\left|x^2-2\right| \leq y \leq x\right\}
$$
_10th_April_Evening_Shift_en_14_1.png)
On solving, $\left|x^2-2\right|=y$ and $y=x$, we get $(1,1)$ and $(2,2)$
$$ \begin{aligned} & A=\int\limits_1^{\sqrt{2}}\left(x-\left(2-x^2\right)\right) d x+\int\limits_{\sqrt{2}}^2\left(x-\left(x^2-2\right)\right) d x \\\\ & A=\left[\frac{x^2}{2}-2 x+\frac{x^3}{3}\right]_1^{\sqrt{2}}+\left[\frac{x^2}{2}-\frac{x^3}{3}+2 x\right]_{\sqrt{2}}^2 \end{aligned} $$
$$ \begin{aligned} A=\left(1-2 \sqrt{2}+\frac{2 \sqrt{2}}{3}\right)-\left(\frac{1}{2}-2+\frac{1}{3}\right)+ & \left(2-\frac{8}{3}+4\right) \\ & -\left(1-\frac{2 \sqrt{2}}{3}+2 \sqrt{2}\right) \end{aligned} $$
$$ \begin{aligned} & A=-4 \sqrt{2}+\frac{4 \sqrt{2}}{3}+\frac{7}{6}+\frac{10}{3} \\\\ & A=\frac{-8 \sqrt{2}}{3}+\frac{9}{2} \Rightarrow A=\frac{-16 \sqrt{2}+27}{6} \\\\ & 6 A=-16 \sqrt{2}+27 \\\\ & 6 A+16 \sqrt{2}=27 \end{aligned} $$
On solving, $\left|x^2-2\right|=y$ and $y=x$, we get $(1,1)$ and $(2,2)$
$$ \begin{aligned} & A=\int\limits_1^{\sqrt{2}}\left(x-\left(2-x^2\right)\right) d x+\int\limits_{\sqrt{2}}^2\left(x-\left(x^2-2\right)\right) d x \\\\ & A=\left[\frac{x^2}{2}-2 x+\frac{x^3}{3}\right]_1^{\sqrt{2}}+\left[\frac{x^2}{2}-\frac{x^3}{3}+2 x\right]_{\sqrt{2}}^2 \end{aligned} $$
$$ \begin{aligned} A=\left(1-2 \sqrt{2}+\frac{2 \sqrt{2}}{3}\right)-\left(\frac{1}{2}-2+\frac{1}{3}\right)+ & \left(2-\frac{8}{3}+4\right) \\ & -\left(1-\frac{2 \sqrt{2}}{3}+2 \sqrt{2}\right) \end{aligned} $$
$$ \begin{aligned} & A=-4 \sqrt{2}+\frac{4 \sqrt{2}}{3}+\frac{7}{6}+\frac{10}{3} \\\\ & A=\frac{-8 \sqrt{2}}{3}+\frac{9}{2} \Rightarrow A=\frac{-16 \sqrt{2}+27}{6} \\\\ & 6 A=-16 \sqrt{2}+27 \\\\ & 6 A+16 \sqrt{2}=27 \end{aligned} $$
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