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NDA 2017
Exam
NDA
Year
2017
Subject
Mathematics
Chapter
Definite Integration

NDA 2017 · Mathematics · Definite Integration

If f(x) and g(x) are continuous functions satisfying f(x)=f(a-x)...

If f(x)f(\mathrm{x}) and g(x)\mathrm{g}(\mathrm{x}) are continuous functions satisfying f(x)=f(a−x)f(\mathrm{x})=f(\mathrm{a}-\mathrm{x}) and g(x)+g(a−x)=2\mathrm{g}(\mathrm{x})+\mathrm{g}(\mathrm{a}-\mathrm{x})=2, then what is ∫0af(x)g(x)dx\int_0^{\mathrm{a}} f(\mathrm{x}) \mathrm{g}(\mathrm{x}) \mathrm{dx} equal to?

Options

  1. A∫0ag(x)dx\quad \int_0^a g(x) d x
  2. B∫0af(x)dx\int_0^{\mathrm{a}} f(\mathrm{x}) \mathrm{dx}
  3. C2∫0af(x)dx\quad 2 \int_0^{\mathrm{a}} f(\mathrm{x}) \mathrm{dx}
  4. D0

Correct answer

B. ∫0af(x)dx\int_0^{\mathrm{a}} f(\mathrm{x}) \mathrm{dx}

Step-by-step solution

I=∫0af(x)⋅g(x)dxI=∫0af(a−x)⋅g(a−x)⋅dxI=∫0af(x)⋅[2−g(x)]⋅dx{∵f(a−x)=f(x) and g(a−x)=2g(x)}I=∫0a2⋅f(x)⋅dx−∫0af(x)⋅g(x)⋅dxI=2⋅∫0af(x)⋅−I⇒2I=2∫0af(x)dx2∴I=∫0af(x)dxaaf(x)=∫0af(a−x)⋅dx(∵ Given f(x)=f(x))\begin{aligned} & I=\int_0^a f(x) \cdot g(x) d x \\ & I=\int_0^a f(a-x) \cdot g(a-x) \cdot d x \\ & I=\int_0^a f(x) \cdot[2-g(x)] \cdot d x \\ & \{\because f(a-x)=f(x) \text{ and } g(a-x)=2 g(x)\} \\ & I=\int_0^a 2 \cdot f(x) \cdot d x-\int_0^a f(x) \cdot g(x) \cdot d x \\ & I=2 \cdot \int_0^a f(x) \cdot-I \\ & \Rightarrow 2 I=2 \int_0^a f(x) d x 2 \\ & \therefore I=\int_0^a f(x) d x \\ & \frac{a}{a} f(x)=\int_0^a f(a-x) \cdot d x \\ & (\because \text{ Given } f(x)=f(x))\end{aligned}
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