Question:

Step 1: Rewrite the given integral condition. The given condition is $$ \int_{2}^{150}(x-1)\ln(x-1)(2f(x)-(x-1)\ln(x-1))\,dx = \int_{2}^{150}f^2(x)\,dx $$ Rearranging the terms, we obtain $$ \int_{2}^{150}\left[f^2(x) - 2(x-1)\ln(x-1)f(x) + (x-1)^2\ln^2(x-1)\right]\,dx = 0 $$ Step 2: Recognize the integrand as a perfect square. The integrand is a perfect square: $$ \int_{2}^{150}\left[f(x) - (x-1)\ln(x-1)\right]^2 \,dx = 0 $$ Step 3: Conclude the form of $f(x)$. Since the integrand $\left[f(x) - (x-1)\ln(x-1)\right]^2$ is non-negative and its integral over the interval $[2, 150]$ is zero, it must be that $$ f(x) = (x-1)\ln(x-1) $$ for all $x \in [2, 150]$. Since $f$ is a differentiable function on $(1, \infty)$, this equality holds for all $x \in (1, \infty)$. Step 4: Determine the derivative of $f(x)$. Differentiating $f(x) = (x-1)\ln(x-1)$ with respect to $x$: $$ f'(x) = \frac{d}{dx}((x-1)\ln(x-1)) = 1 \cdot \ln(x-1) + (x-1) \cdot \frac{1}{x-1} = \ln(x-1) + 1 $$ Step 5: Analyze the monotonicity of $f(x)$. To find critical points, set $f'(x) = 0$: $$ \ln(x-1) + 1 = 0 \implies \ln(x-1) = -1 \implies x-1 = e^{-1} \implies x = 1 + \frac{1}{e} $$ For $x \in \left(1, 1+\frac{1}{e}\right)$, we have $x-1 < \frac{1}{e}$, so $\ln(x-1) < -1$. Thus, $f'(x) = \ln(x-1) + 1 < 0$, which means $f(x)$ is strictly decreasing on this interval. For $x \in \left(1+\frac{1}{e}, \infty\right)$, we have $x-1 > \frac{1}{e}$, so $\ln(x-1) > -1$. Thus, $f'(x) = \ln(x-1) + 1 > 0$, which means $f(x)$ is strictly increasing on this interval. The minimum value of $f(x)$ occurs at $x = 1+\frac{1}{e}$, where $f\left(1+\frac{1}{e}\right) = \left(1+\frac{1}{e}-1\right)\ln\left(1+\frac{1}{e}-1\right) = \frac{1}{e}\ln\left(\frac{1}{e}\right) = \frac{1}{e}(-1) = -\frac{1}{e}$. As $x \to 1^+$, $f(x) = (x-1)\ln(x-1) \to 0^-$. As $x \to \infty$, $f(x) \to \infty$. Step 6: Evaluate the given statements. 1. **Area bounded by the curve and $x$-axis is equal to $1/2$.** The function $f(x) = (x-1)\ln(x-1)$ is zero at $x=2$. The area statement is too general and does not specify an interval. For example, the area from $x=1+\frac{1}{e}$ to $x=2$ is $\int_{1+1/e}^2 (x-1)\ln(x-1)\,dx$. This value is not necessarily $1/2$. This statement is incorrect. 2. **$f(x)$ is strictly decreasing in $\left(1, 1+\frac{1}{e}\right)$.** From Step 5, $f'(x) < 0$ for $x \in \left(1, 1+\frac{1}{e}\right)$. Therefore, $f(x)$ is strictly decreasing in this interval. This statement is correct. 3. **Number of solutions of the equation $f(x)=2$ is 2.** The function $f(x)$ decreases from $0$ (as $x \to 1^+$) to a minimum of $-\frac{1}{e}$ at $x=1+\frac{1}{e}$, and then increases from $-\frac{1}{e}$ to $\infty$ (as $x \to \infty$). Since the minimum value is $-\frac{1}{e} < 0$, and $f(x)$ approaches $0$ as $x \to 1^+$, $f(x)$ never reaches $2$ on the interval $\left(1, 1+\frac{1}{e}\right)$. On the interval $\left(1+\frac{1}{e}, \infty\right)$, $f(x)$ increases from $-\frac{1}{e}$ to $\infty$. Since $2 > -\frac{1}{e}$, by the Intermediate Value Theorem, there is exactly one solution to $f(x)=2$ in this interval. Thus, there is only one solution to $f(x)=2$ in $(1, \infty)$. This statement is incorrect. 4. **$f(x)$ is monotonic in $\left(1+\frac{1}{e},\infty\right)$.** From Step 5, $f'(x) > 0$ for all $x \in \left(1+\frac{1}{e}, \infty\right)$. This means $f(x)$ is strictly increasing on this interval, and thus it is monotonic. This statement is correct. The correct statements are 2 and 4.