Question ID: #974
Let $f$ be a differentiable function satisfying $f(x)=1-2x+\int_{0}^{x}e^{(x-t)}f(t)dt,x\in \mathbb{R}$ and let $g(x)=\int_{0}^{x}(f(t)+2)^{15}(t-4)^{6}(t+12)^{17}dt,x\in \mathbb{R}.$
If $p$ and $q$ are respectively the points of local minima and local maxima of $g$, then the value of $|p+q|$ is equal to ________.
Solution:
Rewrite the given integral equation for $f(x)$ by isolating the $x$ and $t$ terms:
$$f(x) = 1 – 2x + e^x \int_{0}^{x} e^{-t}f(t)dt$$
Multiply the entire equation by $e^{-x}$:
$$e^{-x}f(x) = (1 – 2x)e^{-x} + \int_{0}^{x} e^{-t}f(t)dt$$
Differentiate both sides with respect to $x$ using the product rule and Leibniz rule:
$$-e^{-x}f(x) + e^{-x}f'(x) = -2e^{-x} – (1 – 2x)e^{-x} + e^{-x}f(x)$$
Divide the entire equation by $e^{-x}$ (since $e^{-x} \neq 0$):
$$-f(x) + f'(x) = -2 – 1 + 2x + f(x)$$
$$f'(x) – 2f(x) = 2x – 3$$
This is a first-order linear differential equation of the form $\frac{dy}{dx} + Py = Q$.
Here, $P = -2$ and $Q = 2x – 3$.
Integrating Factor (IF):
$$IF = e^{\int P dx} = e^{\int -2 dx} = e^{-2x}$$
The solution is given by $y \cdot IF = \int (Q \cdot IF) dx$:
$$f(x) e^{-2x} = \int (2x – 3) e^{-2x} dx$$
Apply integration by parts $\int u v dx = u \int v dx – \int (u’ \int v dx) dx$:
$$f(x) e^{-2x} = (2x – 3) \left( \frac{e^{-2x}}{-2} \right) – \int (2) \left( \frac{e^{-2x}}{-2} \right) dx$$
$$f(x) e^{-2x} = \frac{3 – 2x}{2} e^{-2x} + \int e^{-2x} dx$$
$$f(x) e^{-2x} = \frac{3 – 2x}{2} e^{-2x} – \frac{1}{2} e^{-2x} + C$$
Divide by $e^{-2x}$:
$$f(x) = \frac{3 – 2x}{2} – \frac{1}{2} + C e^{2x}$$
$$f(x) = 1 – x + C e^{2x}$$
To find $C$, use the initial condition from the original integral equation at $x = 0$:
$$f(0) = 1 – 2(0) + \int_{0}^{0} e^{(0-t)}f(t)dt = 1$$
$$f(0) = 1 – 0 + C e^0 = 1 \Rightarrow C = 0$$
Therefore, $f(x) = 1 – x$.
Now substitute $f(t) = 1 – t$ into $g(x)$:
$$g(x) = \int_{0}^{x} (1 – t + 2)^{15} (t – 4)^6 (t + 12)^{17} dt$$
$$g(x) = \int_{0}^{x} (3 – t)^{15} (t – 4)^6 (t + 12)^{17} dt$$
Differentiate $g(x)$ using Leibniz rule to find critical points:
$$g'(x) = (3 – x)^{15} (x – 4)^6 (x + 12)^{17}$$
$$g'(x) = -(x – 3)^{15} (x – 4)^6 (x + 12)^{17}$$
Find the critical points by setting $g'(x) = 0$:
$x = 3, x = 4, x = -12$.
Analyze the sign changes of $g'(x)$ around the critical points using the wavy curve method:
For $x \in (-\infty, -12)$, $g'(x) = – (-) (+) (-) = -$ (Negative)
For $x \in (-12, 3)$, $g'(x) = – (-) (+) (+) = +$ (Positive)
For $x \in (3, 4)$, $g'(x) = – (+) (+) (+) = -$ (Negative)
For $x \in (4, \infty)$, $g'(x) = – (+) (+) (+) = -$ (Negative)
Sign changes from Negative to Positive at $x = -12 \Rightarrow$ Local Minima ($p = -12$).
Sign changes from Positive to Negative at $x = 3 \Rightarrow$ Local Maxima ($q = 3$).
No sign change at $x = 4 \Rightarrow$ Point of Inflection.
Calculate $|p + q|$:
$$|p + q| = |-12 + 3| = |-9| = 9$$
Ans. 9
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