Question:

Step 1: Determine the function $f(x)$. Given the integral equation $$ \int e^{x^2} \cdot e^x (2x^2 + x + 1)\, dx = e^{x^2} \cdot f(x) + C $$ Differentiating both sides with respect to $x$: $$ e^{x^2} \cdot e^x (2x^2 + x + 1) = \frac{d}{dx} \left( e^{x^2} f(x) \right) $$ $$ e^{x^2} \cdot e^x (2x^2 + x + 1) = e^{x^2} (2x) f(x) + e^{x^2} f'(x) $$ Dividing by $e^{x^2}$: $$ e^x (2x^2 + x + 1) = 2x f(x) + f'(x) $$ This is a first-order linear differential equation for $f(x)$. We can attempt a solution of the form $f(x) = e^x g(x)$. Then $f'(x) = e^x g(x) + e^x g'(x)$. Substituting these into the equation: $$ e^x (2x^2 + x + 1) = 2x e^x g(x) + e^x g(x) + e^x g'(x) $$ Dividing by $e^x$: $$ 2x^2 + x + 1 = (2x+1)g(x) + g'(x) $$ By inspection, let $g(x) = x$. Then $g'(x) = 1$. Substituting $g(x)=x$ and $g'(x)=1$: $$ 2x^2 + x + 1 = (2x+1)x + 1 $$ $$ 2x^2 + x + 1 = 2x^2 + x + 1 $$ This identity confirms that $g(x) = x$ is a solution. Therefore, $f(x) = xe^x$. Step 2: Find the local minimum value $m$ of $f(x)$. The function is $f(x) = xe^x$. To find local extrema, we compute the first derivative and set it to zero: $$ f'(x) = \frac{d}{dx}(xe^x) = 1 \cdot e^x + x \cdot e^x = e^x(1+x) $$ Setting $f'(x) = 0$: $$ e^x(1+x) = 0 $$ Since $e^x > 0$ for all $x$, we must have $1+x=0$, which implies $x=-1$. To determine if this is a local minimum, we use the second derivative test: $$ f''(x) = \frac{d}{dx}(e^x(1+x)) = e^x(1+x) + e^x(1) = e^x(2+x) $$ Evaluate $f''(-1)$: $$ f''(-1) = e^{-1}(2+(-1)) = e^{-1}(1) = e^{-1} $$ Since $f''(-1) = e^{-1} > 0$, there is a local minimum at $x=-1$. The local minimum value $m$ is $f(-1)$: $$ m = f(-1) = (-1)e^{-1} = -\frac{1}{e} $$ Step 3: Evaluate the given statements. Statement 1: "$f(x)$ is increasing in $(0, \infty)$" From Step 2, $f'(x) = e^x(1+x)$. For $x \in (0, \infty)$, we have $x > 0$, which implies $1+x > 1$. Also, $e^x > 0$. Thus, $f'(x) = e^x(1+x) > 0$ for all $x \in (0, \infty)$. Therefore, $f(x)$ is increasing in $(0, \infty)$. This statement is TRUE. Statement 2: "the value of $\lim_{x \to 0}(1 + f(x))^{1/x}$ is equal to 1." We need to evaluate $\lim_{x \to 0}(1 + xe^x)^{1/x}$. This is an indeterminate form of type $1^\infty$. We can evaluate this limit using the property $\lim_{x \to a} (1+h(x))^{g(x)} = e^{\lim_{x \to a} g(x)h(x)}$ when $\lim_{x \to a} h(x) = 0$. Here, $h(x) = xe^x$ and $g(x) = 1/x$. $$ \lim_{x \to 0} \frac{1}{x} (xe^x) = \lim_{x \to 0} e^x = e^0 = 1 $$ So, the limit is $e^1 = e$. Since $e \neq 1$, the statement that the limit is equal to 1 is FALSE. Statement 3: "the value of $\int_0^1 (f(x) + e^x)\, dx$ is equal to $2e$." We need to evaluate $\int_0^1 (xe^x + e^x)\, dx$. $$ \int_0^1 (xe^x + e^x)\, dx = \int_0^1 (x+1)e^x\, dx $$ We can use integration by parts with $u = x+1$ and $dv = e^x\, dx$, so $du = dx$ and $v = e^x$. $$ \int_0^1 (x+1)e^x\, dx = \left[ (x+1)e^x \right]_0^1 - \int_0^1 e^x\, dx $$ $$ = ( (1+1)e^1 - (0+1)e^0 ) - \left[ e^x \right]_0^1 $$ $$ = (2e - 1) - (e^1 - e^0) $$ $$ = (2e - 1) - (e - 1) $$ $$ = 2e - 1 - e + 1 = e $$ Since the value of the integral is $e$, not $2e$, this statement is FALSE. Statement 4: "the value of $\left[\dfrac{-1}{m}\right]$ is equal to 2." From Step 2, $m = -1/e$. $$ \frac{-1}{m} = \frac{-1}{-1/e} = e $$ The floor function $\lfloor x \rfloor$ gives the greatest integer less than or equal to $x$. $$ \left\lfloor \frac{-1}{m} \right\rfloor = \lfloor e \rfloor $$ Since $e \approx 2.718$, we have $\lfloor e \rfloor = 2$. Therefore, the value of $\left[\dfrac{-1}{m}\right]$ is equal to 2. This statement is TRUE.