JEE Main 2022 — Mathematics Calculus

We first find the points of intersection of the curves $y=|{x}^{2}-9|$y=∣x2−9∣ and $y=3$y=3

which gives, $|{x}^{2}-9|=3$∣x2−9∣=3 $\Rightarrow {x}^{2}-9=\pm 3$⇒x2−9=±3

$\Rightarrow {x}^{2}=12$⇒x2=12 (or) ${x}^{2}=6$x2=6 $\Rightarrow x=\pm \sqrt{12}$⇒x=±12​ (or) $x=\pm \sqrt{6}$x=±6​

The graphs of the two curves are shown in figure

We have to find the area of the shaded region.

Required area $=2{A}_{1}+2{A}_{2}$=2A1​+2A2​ ($\because$∵ The areas ${A}_{1}$A1​ & ${A}_{2}$A2​ are symmetrical about $y-$y− axis )

where $\displaystyle {A}_{1}=\int _{0}^{\sqrt{6}}[(9-{x}^{2})-(3)]dx$A1​=∫06​​[(9−x2)−(3)]dx

$\displaystyle {A}_{1}=\int _{0}^{\sqrt{6}}(6-{x}^{2})dx$A1​=∫06​​(6−x2)dx$={[6x-\frac{{x}^{3}}{3}]}_{0}^{\sqrt{6}}$=[6x−3x3​]06​​ $=6\sqrt{6}-\frac{6\sqrt{6}}{3}=\frac{12\sqrt{6}}{3}=4\sqrt{6}$=66​−366​​=3126​​=46​

$${A}_{2}=\int _{\sqrt{6}}^{3}[3-(9-{x}^{2})]dx+\int _{3}^{\sqrt{12}}[3-({x}^{2}-9)]dx$$A2​=∫6​3​[3−(9−x2)]dx+∫312​​[3−(x2−9)]dx

$$={\int }_{\sqrt{6}}^{3}(-6+{x}^{2})dx+{\int }_{3}^{\sqrt{12}}(12-{x}^{2})dx$$=∫6​3​(−6+x2)dx+∫312​​(12−x2)dx

$$={[-6x+\frac{{x}^{3}}{3}]}_{\sqrt{6}}^{3}+{[12x-\frac{{x}^{3}}{3}]}_{3}^{\sqrt{12}}$$=[−6x+3x3​]6​3​+[12x−3x3​]312​​

$$=(-18+9)-(-6\sqrt{6}+2\sqrt{6}))+(12\sqrt{12}-4\sqrt{12})-(36-9)$$=(−18+9)−(−66​+26​))+(1212​−412​)−(36−9)

$$=-36+4\sqrt{6}+8\sqrt{12}$$=−36+46​+812​

Hence, required area $=2{A}_{1}+2{A}_{2}$=2A1​+2A2​

$$=2(4\sqrt{6})+2(-36+4\sqrt{6}+8\sqrt{12})$$=2(46​)+2(−36+46​+812​)

$$=8\sqrt{6}-72+8\sqrt{6}+16\sqrt{12}$$=86​−72+86​+1612​

$$=16\sqrt{6}+16\sqrt{12}-72$$=166​+1612​−72