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JEE Advanced 2024 Paper 2 · Mathematics · Area Under Curves · easy

JEE Advanced 2024 Paper 2 Mathematics: Paragraph 1

Consider the curves C1: y=x2C_1:\ y=x^2 and C2: y=2−∣x∣C_2:\ y=2-|x| in the xyxy-plane. The two curves enclose a bounded region RR lying above C1C_1 and below C2C_2, symmetric about the yy-axis. Use π=3.1416\pi=3.1416 wherever it is needed and give each answer correct to two decimal places.

Question 1

Numerical answer

The area of the region RR is
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Answer: 2.33

The curves meet where x2=2−∣x∣x^2=2-|x|, i.e. ∣x∣=1|x|=1, so x=±1x=\pm1. By symmetry the area is 2∫01(2−x−x2) dx=2(2−12−13)=2⋅76=73≈2.332\int_0^1(2-x-x^2)\,dx=2\left(2-\tfrac12-\tfrac13\right)=2\cdot\dfrac76=\dfrac73\approx2.33.

Question 2

Numerical answer

The volume of the solid obtained by revolving RR about the xx-axis is
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Answer: 13.40

For 0≤x≤10\le x\le1 the region lies between y=x2y=x^2 and y=2−xy=2-x, both non-negative, so the washer method applies and by symmetry V=2π∫01[(2−x)2−x4]dxV=2\pi\int_0^1\left[(2-x)^2-x^4\right]dx. Now ∫01(4−4x+x2) dx=4−2+13=73\int_0^1(4-4x+x^2)\,dx=4-2+\tfrac13=\tfrac73 and ∫01x4dx=15\int_0^1x^4dx=\tfrac15, so V=2π(73−15)=2π⋅3215=64π15V=2\pi\left(\tfrac73-\tfrac15\right)=2\pi\cdot\dfrac{32}{15}=\dfrac{64\pi}{15}. With π=3.1416\pi=3.1416 this is 13.4013.40.
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