JEE Challenger
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MathHardAdv 2020 • Paper 2

Evaluation of Initial Value of Function using Definite Integral

Let f:R→Rf: \mathbb{R} \rightarrow \mathbb{R} be a differentiable function with a continuous derivative f′f' such that f(π)=−6f(\pi) = -6. Define F:[0,π]→RF: [0, \pi] \rightarrow \mathbb{R} by F(x)=∫0xf(t) dtF(x) = \int_{0}^{x} f(t) \, dt. If the integral condition ∫0π(f′(x)+F(x))cos⁡x dx=2\int_{0}^{\pi} \left( f'(x) + F(x) \right) \cos x \, dx = 2 holds true, evaluate f(0)f(0).

Official Numerical Answer4

Topics & Concepts

IntegralsDefinite IntegrationProperties of Definite Integrals

Step-by-Step Solution

To find the value of f(0)f(0), we start by splitting the given definite integral into two separate integrals:

∫0π(f′(x)+F(x))cos⁡x dx=∫0πf′(x)cos⁡x dx+∫0πF(x)cos⁡x dx=2\int_{0}^{\pi} \left( f'(x) + F(x) \right) \cos x \, dx = \int_{0}^{\pi} f'(x) \cos x \, dx + \int_{0}^{\pi} F(x) \cos x \, dx = 2

Let: I1=∫0πf′(x)cos⁡x dxI_1 = \int_{0}^{\pi} f'(x) \cos x \, dx I2=∫0πF(x)cos⁡x dxI_2 = \int_{0}^{\pi} F(x) \cos x \, dx

Evaluating I1I_1:

Using integration by parts on I1I_1 with u=cos⁡xu = \cos x and dv=f′(x) dxdv = f'(x) \, dx:

  • du=−sin⁡x dxdu = -\sin x \, dx
  • v=f(x)v = f(x)

Applying the formula ∫u dv=uv−∫v du\int u \, dv = uv - \int v \, du: I1=[f(x)cos⁡x]0π−∫0πf(x)(−sin⁡x) dxI_1 = \left[ f(x) \cos x \right]_{0}^{\pi} - \int_{0}^{\pi} f(x) (-\sin x) \, dx I1=(f(π)cos⁡π−f(0)cos⁡0)+∫0πf(x)sin⁡x dxI_1 = \left( f(\pi) \cos \pi - f(0) \cos 0 \right) + \int_{0}^{\pi} f(x) \sin x \, dx

Given that f(π)=−6f(\pi) = -6, cos⁡π=−1\cos \pi = -1, and cos⁡0=1\cos 0 = 1: I1=((−6)(−1)−f(0)(1))+∫0πf(x)sin⁡x dxI_1 = \left( (-6)(-1) - f(0)(1) \right) + \int_{0}^{\pi} f(x) \sin x \, dx I1=6−f(0)+∫0πf(x)sin⁡x dxI_1 = 6 - f(0) + \int_{0}^{\pi} f(x) \sin x \, dx

Evaluating I2I_2:

Given F(x)=∫0xf(t) dtF(x) = \int_{0}^{x} f(t) \, dt, by the Fundamental Theorem of Calculus, we have F′(x)=f(x)F'(x) = f(x) and F(0)=0F(0) = 0.

Using integration by parts on I2I_2 with u=F(x)u = F(x) and dv=cos⁡x dxdv = \cos x \, dx:

  • du=F′(x) dx=f(x) dxdu = F'(x) \, dx = f(x) \, dx
  • v=sin⁡xv = \sin x

Applying the integration by parts formula: I2=[F(x)sin⁡x]0π−∫0πf(x)sin⁡x dxI_2 = \left[ F(x) \sin x \right]_{0}^{\pi} - \int_{0}^{\pi} f(x) \sin x \, dx I2=(F(π)sin⁡π−F(0)sin⁡0)−∫0πf(x)sin⁡x dxI_2 = \left( F(\pi) \sin \pi - F(0) \sin 0 \right) - \int_{0}^{\pi} f(x) \sin x \, dx

Since sin⁡π=0\sin \pi = 0 and sin⁡0=0\sin 0 = 0: I2=0−∫0πf(x)sin⁡x dx=−∫0πf(x)sin⁡x dxI_2 = 0 - \int_{0}^{\pi} f(x) \sin x \, dx = -\int_{0}^{\pi} f(x) \sin x \, dx

Combining I1I_1 and I2I_2:

I1+I2=(6−f(0)+∫0πf(x)sin⁡x dx)+(−∫0πf(x)sin⁡x dx)I_1 + I_2 = \left( 6 - f(0) + \int_{0}^{\pi} f(x) \sin x \, dx \right) + \left( -\int_{0}^{\pi} f(x) \sin x \, dx \right) I1+I2=6−f(0)I_1 + I_2 = 6 - f(0)

We are given that I1+I2=2I_1 + I_2 = 2: 6−f(0)=26 - f(0) = 2 f(0)=4f(0) = 4

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