1. **Problem statement:** We consider the function $f$ defined on $]0; +\infty[$ by $f(x) = 2x(1 - \ln x)$. We analyze its derivative, limits, intersections, tangents, inverse function, and integrals. 2. **Derivative and limits:** - First, compute $f'(x)$ using the product rule: $$f'(x) = \frac{d}{dx}[2x(1 - \ln x)] = 2(1 - \ln x) + 2x \cdot \frac{d}{dx}(1 - \ln x)$$ - Since $\frac{d}{dx}(1 - \ln x) = -\frac{1}{x}$, we get: $$f'(x) = 2(1 - \ln x) + 2x \left(-\frac{1}{x}\right) = 2(1 - \ln x) - 2 = -2 \ln x$$ 3. **Limits of $f'(x)$:** - As $x \to 0^+$, $\ln x \to -\infty$, so: $$\lim_{x \to 0^+} f'(x) = \lim_{x \to 0^+} -2 \ln x = +\infty$$ - As $x \to +\infty$, $\ln x \to +\infty$, so: $$\lim_{x \to +\infty} f'(x) = \lim_{x \to +\infty} -2 \ln x = -\infty$$ 4. **Intersection with x-axis (point A):** - Solve $f(x) = 0$: $$2x(1 - \ln x) = 0 \implies x=0 \text{ (not in domain) or } 1 - \ln x = 0$$ - So: $$\ln x = 1 \implies x = e$$ - Coordinates of $A$ are $(e, 0)$. 5. **Variation table of $f$:** - Since $f'(x) = -2 \ln x$, the sign of $f'(x)$ depends on $\ln x$: - For $x \in ]0,1[$, $\ln x < 0$, so $f'(x) > 0$ (increasing). - For $x=1$, $f'(1) = 0$. - For $x > 1$, $\ln x > 0$, so $f'(x) < 0$ (decreasing). - Thus, $f$ increases on $]0,1]$ and decreases on $[1,+\infty[$. 6. **Equation of tangent (T) at A:** - Slope at $A$ is: $$f'(e) = -2 \ln e = -2 \times 1 = -2$$ - Point $A$ is $(e,0)$. - Equation of tangent line: $$y - 0 = -2(x - e) \implies y = -2x + 2e$$ 7. **Inverse function $g$ on $]1,+\infty[$:** - Since $f$ is strictly decreasing on $[1,+\infty[$ and continuous, it is bijective there. - Domain of $g$ is the image of $f$ on $[1,+\infty[$. - Compute $f(1) = 2 \times 1 \times (1 - 0) = 2$ and $\lim_{x \to +\infty} f(x) = \lim_{x \to +\infty} 2x(1 - \ln x) = -\infty$ (since $\ln x$ grows faster than 1). - So $g$ is defined on $]-\infty, 2]$. 8. **Variation table of $g$:** - Since $f$ decreases on $[1,+\infty[$, $g$ decreases on $]-\infty, 2]$. 9. **Graph of $g$:** - $g$ is the inverse of $f$ on $[1,+\infty[$. - The graph of $g$ is the reflection of the part of $(C)$ for $x \geq 1$ about the line $y=x$. 10. **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: $$\int x \ln x \, dx = \frac{x^2}{2} \ln x - \int \frac{x^2}{2} \cdot \frac{1}{x} dx = \frac{x^2}{2} \ln x - \frac{1}{2} \int x \, dx$$ - Evaluate: $$= \frac{x^2}{2} \ln x - \frac{1}{2} \cdot \frac{x^2}{2} + C = \frac{x^2}{2} \ln x - \frac{x^2}{4} + C$$ 11. **Integral of $f(x)$ from 1 to $e^2$:** - Recall $f(x) = 2x(1 - \ln x) = 2x - 2x \ln x$. - So: $$\int_1^{e^2} f(x) dx = \int_1^{e^2} 2x \, dx - \int_1^{e^2} 2x \ln x \, dx$$ - Compute each: $$\int_1^{e^2} 2x \, dx = [x^2]_1^{e^2} = e^4 - 1$$ $$\int_1^{e^2} 2x \ln x \, dx = 2 \left[ \frac{x^2}{2} \ln x - \frac{x^2}{4} \right]_1^{e^2} = \left[x^2 \ln x - \frac{x^2}{2} \right]_1^{e^2}$$ - Evaluate at bounds: - At $x = e^2$: $$e^4 \times 2 - \frac{e^4}{2} = 2 e^4 - \frac{e^4}{2} = \frac{3 e^4}{2}$$ - At $x=1$: $$1 \times 0 - \frac{1}{2} = -\frac{1}{2}$$ - So: $$\int_1^{e^2} 2x \ln x \, dx = \frac{3 e^4}{2} + \frac{1}{2} = \frac{3 e^4 + 1}{2}$$ - Therefore: $$\int_1^{e^2} f(x) dx = e^4 - 1 - \frac{3 e^4 + 1}{2} = \frac{2 e^4 - 2 - 3 e^4 - 1}{2} = \frac{- e^4 - 3}{2}$$ 12. **Area of shaded region bounded by $(C)$, $(T)$, and $(d)$:** - The vertical line $(d)$ is $x=1$. - The tangent $(T)$ at $A(e,0)$ has equation $y = -2x + 2e$. - The shaded area is between $x=1$ and $x=e$ bounded by $f(x)$ and the tangent. - Area $= \int_1^e [f(x) - (-2x + 2e)] dx = \int_1^e [2x(1 - \ln x) + 2x - 2e] dx$ - Simplify integrand: $$2x(1 - \ln x) + 2x - 2e = 2x - 2x \ln x + 2x - 2e = 4x - 2x \ln x - 2e$$ - Compute: $$\int_1^e (4x - 2x \ln x - 2e) dx = \int_1^e 4x \, dx - \int_1^e 2x \ln x \, dx - \int_1^e 2e \, dx$$ - Evaluate each: $$\int_1^e 4x \, dx = [2x^2]_1^e = 2e^2 - 2$$ $$\int_1^e 2x \ln x \, dx = \left[x^2 \ln x - \frac{x^2}{2}\right]_1^e = (e^2 \times 1 - \frac{e^2}{2}) - (0 - \frac{1}{2}) = e^2 - \frac{e^2}{2} + \frac{1}{2} = \frac{e^2}{2} + \frac{1}{2}$$ $$\int_1^e 2e \, dx = 2e (e - 1) = 2e^2 - 2e$$ - Sum: $$2e^2 - 2 - \left(\frac{e^2}{2} + \frac{1}{2}\right) - (2e^2 - 2e) = 2e^2 - 2 - \frac{e^2}{2} - \frac{1}{2} - 2e^2 + 2e$$ - Simplify: $$\left(2e^2 - 2e^2 - \frac{e^2}{2}\right) + \left(-2 - \frac{1}{2}\right) + 2e = -\frac{e^2}{2} - \frac{5}{2} + 2e$$ - Final area: $$2e - \frac{e^2}{2} - \frac{5}{2}$$ **Final answers:** - $f'(x) = -2 \ln x$ - $\lim_{x \to 0^+} f'(x) = +\infty$ - $\lim_{x \to +\infty} f'(x) = -\infty$ - Point $A = (e, 0)$ - Tangent at $A$: $y = -2x + 2e$ - $f$ is decreasing on $[1, +\infty[$ and admits inverse $g$ on $]-\infty, 2]$ - $\int x \ln x \, dx = \frac{x^2}{2} \ln x - \frac{x^2}{4} + C$ - $\int_1^{e^2} f(x) dx = \frac{- e^4 - 3}{2}$ - Area bounded by $(C)$, $(T)$, and $(d)$ is $2e - \frac{e^2}{2} - \frac{5}{2}$