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JEE Advanced 2024 · Paper 2, Mathematics question 16

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PARAGRAPH "II"

Let f:[0,π2]→[0,1]f : \left[0, \dfrac{\pi}{2}\right] \to [0, 1] be the function defined by f(x)=sin⁡2xf(x) = \sin^2 x and let g:[0,π2]→[0,∞)g : \left[0, \dfrac{\pi}{2}\right] \to [0, \infty) be the function defined by g(x)=πx2−x2g(x) = \sqrt{\dfrac{\pi x}{2} - x^2}.

The value of 2∫0π/2f(x)g(x) dx−∫0π/2g(x) dx\displaystyle 2\int_0^{\pi/2} f(x)g(x)\,dx - \int_0^{\pi/2} g(x)\,dx is _____.

Answer and worked solution

Official answer: 0

g(x)=x(π2−x)g(x)=\sqrt{x\left(\frac\pi2-x\right)} satisfies g(π2−x)=g(x)g\left(\frac\pi2-x\right)=g(x).

Let I=∫0π/2sin⁡2x g(x) dxI=\int_0^{\pi/2}\sin^2x\,g(x)\,dx. Substituting x→π2−xx\to\frac\pi2-x gives I=∫0π/2cos⁡2x g(x) dxI=\int_0^{\pi/2}\cos^2x\,g(x)\,dx. Adding the two forms:
2I=∫0π/2g(x) dx.2I=\int_0^{\pi/2}g(x)\,dx.

Hence 2∫0π/2fg dx−∫0π/2g dx=02\int_0^{\pi/2}fg\,dx-\int_0^{\pi/2}g\,dx=0.

Topics:Properties of definite integrals
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