Question:

Step 1: Let $g(x) = f(x)e^{-mx}$. Then the derivative $g'(x)$ is given by: $$g'(x) = f'(x)e^{-mx} - mf(x)e^{-mx} = e^{-mx}(f'(x)-mf(x))$$ Given that $f'(x) \leq mf(x)$, it follows that $f'(x) - mf(x) \leq 0$. Since $e^{-mx} > 0$ for all real $x$, we conclude that $g'(x) \leq 0$. Therefore, $g(x)$ is a non-increasing function for $x \geq 1$. Step 2: The function $f(x)$ is non-negative, so $f(x) \geq 0$. Consequently, $g(x) = f(x)e^{-mx} \geq 0$. We are given $f(1)=0$. Evaluating $g(x)$ at $x=1$: $$g(1) = f(1)e^{-m(1)} = 0 \cdot e^{-m} = 0$$ Since $g(x)$ is non-increasing for $x \geq 1$, $g(1)=0$, and $g(x) \geq 0$, it must be that $g(x) = 0$ for all $x \geq 1$. From $g(x) = f(x)e^{-mx} = 0$, and $e^{-mx} \neq 0$, we conclude that $f(x) = 0$ for all $x \geq 1$. Step 3: The function $f(x)$ is defined only for $x \geq 1$. For a function to be classified as odd or even, its domain must be symmetric about the origin. Since the domain $x \geq 1$ is not symmetric, $f(x)$ cannot be classified as odd or even. Step 4: Evaluate the limit: $$L = \lim_{x \to 2}(5 + f(x) + x^2 - 4x)^{\frac{e^2}{e^x - e^2(x-1)}}$$ Since $f(x)=0$ for $x \geq 1$, we have $f(2)=0$. Substituting $f(x)=0$: $$L = \lim_{x \to 2}(5 + 0 + x^2 - 4x)^{\frac{e^2}{e^x - e^2(x-1)}} = \lim_{x \to 2}(x^2 - 4x + 5)^{\frac{e^2}{e^x - e^2(x-1)}}$$ As $x \to 2$, the base approaches $2^2 - 4(2) + 5 = 4 - 8 + 5 = 1$. The exponent approaches $\frac{e^2}{e^2 - e^2(2-1)} = \frac{e^2}{e^2 - e^2} = \frac{e^2}{0}$. To evaluate this $1^\infty$ indeterminate form, we use the form $e^{\lim_{x \to a} g(x) \ln(h(x))}$. Let $x = 2+h$ as $h \to 0$. The base becomes $(2+h)^2 - 4(2+h) + 5 = 4+4h+h^2 - 8-4h+5 = 1+h^2$. The exponent becomes $\frac{e^2}{e^{2+h} - e^2(2+h-1)} = \frac{e^2}{e^2 e^h - e^2(1+h)} = \frac{1}{e^h - (1+h)}$. So the limit is: $$L = \lim_{h \to 0}(1+h^2)^{\frac{1}{e^h - (1+h)}}$$ This can be written as $e^{\lim_{h \to 0} \frac{\ln(1+h^2)}{e^h - (1+h)}}$. Using Taylor series expansions for small $h$: $\ln(1+h^2) = h^2 - \frac{(h^2)^2}{2} + \dots \approx h^2$. $e^h = 1 + h + \frac{h^2}{2!} + \frac{h^3}{3!} + \dots$. So, $e^h - (1+h) = (1 + h + \frac{h^2}{2} + \dots) - (1+h) = \frac{h^2}{2} + \dots$. Therefore, the exponent of $e$ is: $$\lim_{h \to 0} \frac{h^2}{\frac{h^2}{2}} = 2$$ Thus, the limit $L = e^2$. Step 5: Consider the equation $f(x) = e^x - x^2$. Since $f(x)=0$ for $x \geq 1$, the equation becomes $0 = e^x - x^2$. Let $h(x) = e^x - x^2$. We seek the number of solutions for $h(x)=0$ in the domain $x \geq 1$. First, evaluate $h(x)$ at $x=1$: $$h(1) = e^1 - 1^2 = e - 1$$ Since $e \approx 2.718$, $h(1) = e-1 > 0$. Next, consider the first derivative of $h(x)$: $$h'(x) = e^x - 2x$$ Evaluate $h'(x)$ at $x=1$: $$h'(1) = e - 2$$ Since $e \approx 2.718$, $h'(1) = e-2 > 0$. Now, consider the second derivative of $h(x)$: $$h''(x) = e^x - 2$$ For $x \geq 1$, $e^x \geq e \approx 2.718$. Thus, $e^x - 2 > 0$ for all $x \geq 1$. Since $h''(x) > 0$ for $x \geq 1$, $h'(x)$ is strictly increasing for $x \geq 1$. As $h'(1) > 0$ and $h'(x)$ is strictly increasing for $x \geq 1$, it follows that $h'(x) > 0$ for all $x \geq 1$. Since $h(1) > 0$ and $h(x)$ is strictly increasing for $x \geq 1$, it follows that $h(x) > 0$ for all $x \geq 1$. Therefore, the equation $e^x - x^2 = 0$ has no solutions for $x \geq 1$. The number of solutions for $f(x) = e^x - x^2$ is 0. Step 6: Evaluate the definite integral $\displaystyle\int_{f(e)-1}^{f(e^2)+1} \frac{dx}{2^x+1}$. Since $f(x)=0$ for $x \geq 1$, we have $f(e)=0$ and $f(e^2)=0$. The limits of integration become $0-1 = -1$ and $0+1 = 1$. The integral is: $$I = \displaystyle\int_{-1}^{1} \frac{dx}{2^x+1}$$ Using the property $\displaystyle\int_a^b g(x) dx = \displaystyle\int_a^b g(a+b-x) dx$, with $a=-1$ and $b=1$: $$I = \displaystyle\int_{-1}^{1} \frac{dx}{2^{(-1+1-x)}+1} = \displaystyle\int_{-1}^{1} \frac{dx}{2^{-x}+1}$$ Multiply the numerator and denominator of the integrand by $2^x$: $$I = \displaystyle\int_{-1}^{1} \frac{2^x dx}{1+2^x}$$ Adding the two expressions for $I$: $$2I = \displaystyle\int_{-1}^{1} \left(\frac{1}{2^x+1} + \frac{2^x}{2^x+1}\right) dx = \displaystyle\int_{-1}^{1} \frac{1+2^x}{2^x+1} dx = \displaystyle\int_{-1}^{1} 1 dx$$ $$2I = [x]_{-1}^{1} = 1 - (-1) = 2$$ Therefore, $I = 1$.