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(A) Given $f(x)=1-2x+\int_{0}^{x}e^{(x-t)}f(t)dt$. Multiplying by $e^{-x}$, we get $e^{-x}f(x) = (1-2x)e^{-x} + \int_{0}^{x}e^{-t}f(t)dt$. Differentia

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$9$

Vedclass · Answer

(A) Given $f(x)=1-2x+\int_{0}^{x}e^{(x-t)}f(t)dt$. Multiplying by $e^{-x}$, we get $e^{-x}f(x) = (1-2x)e^{-x} + \int_{0}^{x}e^{-t}f(t)dt$. Differentiating with respect to $x$ using Leibniz rule: $e^{-x}f'(x) - e^{-x}f(x) = -2e^{-x} - (1-2x)e^{-x} + e^{-x}f(x)$. $f'(x) - f(x) = -2 - 1 + 2x + f(x) \Rightarrow f'(x) - 2f(x) = 2x - 3$. This is a linear differential equation with integrating factor $I.F. = e^{\int -2 dx} = e^{-2x}$. $f(x)e^{-2x} = \int (2x-3)e^{-2x} dx = (2x-3)\frac{e^{-2x}}{-2} - \int 2 \cdot \frac{e^{-2x}}{-2} dx = -\frac{2x-3}{2}e^{-2x} - \frac{1}{2}e^{-2x} + C$. $f(x) = -x + \frac{3}{2} - \frac{1}{2} + Ce^{2x} = 1-x + Ce^{2x}$. Since $f(0) = 1-2(0) + 0 = 1$, we have $1 = 1-0 + C \Rightarrow C=0$. So, $f(x) = 1-x$. Now, $g(x) = \int_{0}^{x} (1-t+2)^{15}(t-4)^6(t+12)^{17} dt = \int_{0}^{x} (3-t)^{15}(t-4)^6(t+12)^{17} dt$. $g'(x) = (3-x)^{15}(x-4)^6(x+12)^{17} = -(x-3)^{15}(x-4)^6(x+12)^{17}$. Analyzing the sign of $g'(x)$ around critical points $-12, 3, 4$: For $x For $-12  0$. For $3 For $x > 4$, $g'(x) Local minimum at $p = -12$ and local maximum at $q = 3$. Thus, $|p+q| = |-12+3| = |-9| = 9$.

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