Let y = f(x) be function such that f \left(\right. x \left.\right) = 2 x - x^{2} + \int_{0}^{x} f \left(\right. t \left.\right) \textrm{ } dt + 1, then number of real roots of the equation f(x) = 0 is/are

Engineering

Mathematics

Linear Differential Equation

Question

Let y = f(x) be function such that $f (x) = 2 x - x^{2} + \int_{0}^{x} f (t) dt + 1$ , then number of real roots of the equation f(x) = 0 is/are

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$f (x) = 2 x - x^{2} + \int_{0}^{x} f (t) dt + 1$

f '(x) = 2 – 2x + f(x)

$\frac{dy}{dx} + (- 1) y = 2 (1 - x)$

${ye}^{- x} = 2 \int (1 - x) e^{- x} dx - 2 (1 - x) e^{- x} + 2 e^{- x} + C$

f(x) = 2x + Ce^x but f(0) = 1

⇒ f(x) = 2x + e^x.

f(x) = 2x + e^x = 0 ⇒ e^x = – 2x.