1. **Statement of the problem:** We study the function $g(x) = 2x - 1 - \ln x$ defined on $]0; +\infty[$ and analyze its behavior, zeros, and relation to $f(x) = x^2 - x\ln x$. Then, we study $f$'s monotonicity, tangents at a point $\alpha$, and related limits. 2. **Study of $g(x)$:** - Domain: $]0; +\infty[$ since $\ln x$ is defined for $x>0$. - Derivative: $$g'(x) = 2 - \frac{1}{x}$$ - Sign of $g'(x)$: - $g'(x) = 0 \Rightarrow 2 - \frac{1}{x} = 0 \Rightarrow x = \frac{1}{2}$ - For $x < \frac{1}{2}$, $g'(x) < 0$ (decreasing) - For $x > \frac{1}{2}$, $g'(x) > 0$ (increasing) 3. **Variation table of $g$:** - $g$ decreases on $(0, \frac{1}{2})$ and increases on $(\frac{1}{2}, +\infty)$. - Limits: - $\lim_{x \to 0^+} g(x) = -\infty$ (since $\ln x \to -\infty$ dominates) - $\lim_{x \to +\infty} g(x) = +\infty$ 4. **Check $g(1)$:** $$g(1) = 2(1) - 1 - \ln 1 = 2 - 1 - 0 = 1$$ 5. **Existence of another root:** - Since $g(1) = 1 > 0$ and $\lim_{x \to 0^+} g(x) = -\infty < 0$, by Intermediate Value Theorem, $g$ has a root $\alpha$ in $(0,1)$. 6. **Function $f(x) = x^2 - x \ln x$:** - Domain: $]0; +\infty[$ - Relation to $g$: Derivative of $f$ is $$f'(x) = 2x - \ln x - 1 = g(x)$$ - So zeros of $g$ correspond to critical points of $f$. 7. **Check values of $f$ at $0$ and $9$:** - $f(0) = \lim_{x \to 0^+} x^2 - x \ln x = 0$ (since $x \ln x \to 0$) - $f(9) = 81 - 9 \ln 9$ 8. **Monotonicity of $f$:** - Since $f'(x) = g(x)$, $f$ decreases on $(0, \alpha)$ (where $g<0$) and increases on $(\alpha, +\infty)$ (where $g>0$). 9. **Limits for $f$ and $g$ at $0$ and $+\infty$:** - $\lim_{x \to 0^+} f(x) = 0$ - $\lim_{x \to +\infty} f(x) = +\infty$ 10. **Tangents at $\alpha$:** - $f(\alpha) = 1$, $g(\alpha) = 0$ - Limits of difference quotients: $$\lim_{x \to \alpha^-} \frac{f(x) - f(\alpha)}{x - \alpha} = m_1$$ $$\lim_{x \to \alpha^+} \frac{f(x) - f(\alpha)}{x - \alpha} = m_2$$ - Since $f'(\alpha) = g(\alpha) = 0$, the tangent slope at $\alpha$ is 0. 11. **Equations of tangents $T$ and $T'$:** - Given: $$f(x) = x[g(x) - x + 1] + f(\alpha)$$ - Tangent $T$ at $\alpha$ has equation: $$y = f(\alpha) + m(x - \alpha)$$ - Since $m = 0$, tangent line $T$ is horizontal: $$y = 1$$ 12. **Calculate $\alpha$:** - Solve $g(\alpha) = 0$: $$2\alpha - 1 - \ln \alpha = 0$$ - Numerically, $0.1 < \alpha < 0.3$ (given) 13. **Slopes of $T$ and $T'$:** - Both tangents have slope 0 at $\alpha$. 14. **Conclusion:** - $g$ is decreasing then increasing with a root $\alpha$ in $(0,1)$. - $f$ has a minimum at $\alpha$ with $f(\alpha) = 1$. - Tangent at $\alpha$ is horizontal. - $f$ is defined and continuous on $]0; +\infty[$.