1. **Problem statement:** Consider the function $f$ defined on $]0; +\infty[$ by $f(x) = 2x(1 - \ln x)$. We need to find the limits of $f$ as $x \to 0$ and $x \to +\infty$. 2. **Limits:** - As $x \to 0^+$, $\ln x \to -\infty$, so $1 - \ln x \to +\infty$, but $x \to 0$, so consider the product: $$\lim_{x \to 0^+} 2x(1 - \ln x) = 2 \lim_{x \to 0^+} x - 2 \lim_{x \to 0^+} x \ln x.$$ We know $\lim_{x \to 0^+} x = 0$ and $\lim_{x \to 0^+} x \ln x = 0$ (since $x \ln x \to 0$). Thus, $$\lim_{x \to 0^+} f(x) = 0.$$ - As $x \to +\infty$, $\ln x \to +\infty$, so $1 - \ln x \to -\infty$, and $x \to +\infty$, so: $$\lim_{x \to +\infty} f(x) = 2 \lim_{x \to +\infty} x (1 - \ln x) = 2 \lim_{x \to +\infty} (x - x \ln x).$$ Since $x \ln x$ grows faster than $x$, $x - x \ln x \to -\infty$, so $$\lim_{x \to +\infty} f(x) = -\infty.$$ 3. **Intersection with x-axis:** - To find point $A$, solve $f(x) = 0$: $$2x(1 - \ln x) = 0 \implies x=0 \text{ or } 1 - \ln x = 0.$$ Since $x > 0$, $x=0$ is excluded. So, $$1 - \ln x = 0 \implies \ln x = 1 \implies x = e.$$ Thus, $A = (e, 0)$. 4. **Derivative and variations:** - Compute $f'(x)$: $$f(x) = 2x(1 - \ln x) = 2x - 2x \ln x,$$ so $$f'(x) = 2 - 2(\ln x + 1) = 2 - 2 \ln x - 2 = -2 \ln x.$$ - Since $\ln x$ is increasing, $f'(x) = -2 \ln x$ is zero at $x=1$, positive on $(0,1)$, negative on $(1, +\infty)$. - Therefore, $f$ is increasing on $(0,1)$, decreasing on $(1, +\infty)$. 5. **Tangent at $A$:** - The slope at $A$ is $$f'(e) = -2 \ln e = -2 \times 1 = -2.$$ - Equation of tangent $T$ at $A(e,0)$: $$y - 0 = -2(x - e) \implies y = -2x + 2e.$$ 6. **Inverse function $g$ on $(1, +\infty)$:** - Since $f$ is strictly decreasing on $(1, +\infty)$ and continuous, it has an inverse $g$ on $f((1, +\infty))$. - The domain of $g$ is the image of $f$ on $(1, +\infty)$, which is $(-\infty, f(1)]$. - Calculate $f(1) = 2 \times 1 \times (1 - 0) = 2$. - So domain of $g$ is $(-\infty, 2]$. 7. **Variations of $g$:** - Since $f$ is decreasing on $(1, +\infty)$, $g$ is also decreasing on $(-\infty, 2]$. 8. **Integration by parts for $\int x \ln x \, dx$:** - Let $u = \ln x$, $dv = x dx$. - Then $du = \frac{1}{x} dx$, $v = \frac{x^2}{2}$. - Integration by parts formula: $$\int u \, dv = uv - \int v \, du,$$ so $$\int x \ln x \, dx = \frac{x^2}{2} \ln x - \int \frac{x^2}{2} \times \frac{1}{x} dx = \frac{x^2}{2} \ln x - \frac{1}{2} \int x \, dx = \frac{x^2}{2} \ln x - \frac{1}{2} \times \frac{x^2}{2} + C = \frac{x^2}{2} \ln x - \frac{x^2}{4} + C.$$ 9. **Integral of $f(x)$ from 1 to $e$:** - Recall $f(x) = 2x(1 - \ln x) = 2x - 2x \ln x$. - So $$\int_1^e f(x) dx = \int_1^e 2x dx - \int_1^e 2x \ln x dx = 2 \int_1^e x dx - 2 \int_1^e x \ln x dx.$$ - Calculate each: $$\int_1^e x dx = \left[ \frac{x^2}{2} \right]_1^e = \frac{e^2}{2} - \frac{1}{2}.$$ $$\int_1^e x \ln x dx = \left[ \frac{x^2}{2} \ln x - \frac{x^2}{4} \right]_1^e = \left( \frac{e^2}{2} \times 1 - \frac{e^2}{4} \right) - \left( 0 - \frac{1}{4} \right) = \frac{e^2}{2} - \frac{e^2}{4} + \frac{1}{4} = \frac{e^2}{4} + \frac{1}{4}.$$ - Substitute back: $$\int_1^e f(x) dx = 2 \left( \frac{e^2}{2} - \frac{1}{2} \right) - 2 \left( \frac{e^2}{4} + \frac{1}{4} \right) = (e^2 - 1) - \left( \frac{e^2}{2} + \frac{1}{2} \right) = e^2 - 1 - \frac{e^2}{2} - \frac{1}{2} = \frac{e^2}{2} - \frac{3}{2} = \frac{e^2 - 3}{2}.$$ 10. **Area of shaded region bounded by $(C)$, $(T)$, and $(d)$:** - The shaded region is bounded by $x=1$, the curve $f$, and the tangent $T$. - The area $A$ is: $$A = \int_1^e [f(x) - T(x)] dx,$$ where $T(x) = -2x + 2e$. - Compute: $$A = \int_1^e [2x(1 - \ln x) - (-2x + 2e)] dx = \int_1^e [2x - 2x \ln x + 2x - 2e] dx = \int_1^e [4x - 2x \ln x - 2e] dx.$$ - Split the integral: $$A = \int_1^e 4x dx - \int_1^e 2x \ln x dx - \int_1^e 2e dx = 4 \int_1^e x dx - 2 \int_1^e x \ln x dx - 2e (e - 1).$$ - Using previous results: $$\int_1^e x dx = \frac{e^2 - 1}{2}, \quad \int_1^e x \ln x dx = \frac{e^2 + 1}{4}.$$ - Substitute: $$A = 4 \times \frac{e^2 - 1}{2} - 2 \times \frac{e^2 + 1}{4} - 2e(e - 1) = 2(e^2 - 1) - \frac{e^2 + 1}{2} - 2e^2 + 2e.$$ - Simplify: $$2e^2 - 2 - \frac{e^2}{2} - \frac{1}{2} - 2e^2 + 2e = (2e^2 - 2e^2) + (-2 - \frac{1}{2}) - \frac{e^2}{2} + 2e = -\frac{5}{2} - \frac{e^2}{2} + 2e = \frac{-5 - e^2 + 4e}{2}.$$ **Final answers:** - $\lim_{x \to 0^+} f(x) = 0$ - $\lim_{x \to +\infty} f(x) = -\infty$ - $A = (e, 0)$ - $f'(x) = -2 \ln x$ - Tangent at $A$: $y = -2x + 2e$ - $f$ has inverse $g$ on $(-\infty, 2]$ - $\int_1^e f(x) dx = \frac{e^2 - 3}{2}$ - Area of shaded region $= \frac{-5 - e^2 + 4e}{2}$