Question:

**Step 1: Determine the form of $f(x)$.** The given equation is $f(f(x)) = x\displaystyle\int_0^x f(t)\,dt$. Differentiating both sides with respect to $x$: $$f'(f(x)) \cdot f'(x) = \displaystyle\int_0^x f(t)\,dt + x f(x)$$ Given $f(0)=0$. Substituting $x=0$ into the differentiated equation: $$f'(f(0)) \cdot f'(0) = \displaystyle\int_0^0 f(t)\,dt + 0 \cdot f(0)$$ $$f'(0) \cdot f'(0) = 0 + 0$$ $$(f'(0))^2 = 0 \implies f'(0) = 0$$ Since $f(x)$ is a polynomial with non-negative coefficients and $f(0)=0$, we can assume $f(x) = ax^n$ for some $a > 0$ and integer $n \ge 1$. Since $f'(0)=0$, the lowest power of $x$ in $f(x)$ must be at least $x^2$. Thus, $n \ge 2$. Substitute $f(x) = ax^n$ into the original equation: The left side is $f(f(x)) = f(ax^n) = a(ax^n)^n = a \cdot a^n \cdot (x^n)^n = a^{n+1} x^{n^2}$. The right side is $x\displaystyle\int_0^x f(t)\,dt = x\displaystyle\int_0^x at^n\,dt = x \left[ a \frac{t^{n+1}}{n+1} \right]_0^x = x \left( a \frac{x^{n+1}}{n+1} \right) = \frac{a}{n+1} x^{n+2}$. Equating the left and right sides: $$a^{n+1} x^{n^2} = \frac{a}{n+1} x^{n+2}$$ Comparing the powers of $x$: $$n^2 = n+2$$ $$n^2 - n - 2 = 0$$ $$(n-2)(n+1) = 0$$ Since $n \ge 2$, we must have $n=2$. Now, comparing the coefficients for $n=2$: $$a^{2+1} = \frac{a}{2+1}$$ $$a^3 = \frac{a}{3}$$ Since $a > 0$ (due to non-negative coefficients and $f(x)$ not being identically zero), we can divide by $a$: $$a^2 = \frac{1}{3}$$ $$a = \frac{1}{\sqrt{3}}$$ Thus, the polynomial function is $f(x) = \frac{1}{\sqrt{3}} x^2$. **Step 2: Verify the function $f(x) = \frac{1}{\sqrt{3}} x^2$.** Left side: $f(f(x)) = f\left(\frac{1}{\sqrt{3}} x^2\right) = \frac{1}{\sqrt{3}} \left(\frac{1}{\sqrt{3}} x^2\right)^2 = \frac{1}{\sqrt{3}} \left(\frac{1}{3} x^4\right) = \frac{1}{3\sqrt{3}} x^4$. Right side: $x\displaystyle\int_0^x f(t)\,dt = x\displaystyle\int_0^x \frac{1}{\sqrt{3}} t^2\,dt = x \left[ \frac{1}{\sqrt{3}} \frac{t^3}{3} \right]_0^x = x \left( \frac{1}{3\sqrt{3}} x^3 \right) = \frac{1}{3\sqrt{3}} x^4$. The left and right sides are equal, confirming that $f(x) = \frac{1}{\sqrt{3}} x^2$ is the correct function. **Step 3: Evaluate the given properties.** **Property 1: Number of points where $|f(|x|)|$ is non-derivable is 0.** Given $f(x) = \frac{1}{\sqrt{3}} x^2$. Then $f(|x|) = \frac{1}{\sqrt{3}} (|x|)^2 = \frac{1}{\sqrt{3}} x^2$. And $|f(|x|)| = \left| \frac{1}{\sqrt{3}} x^2 \right| = \frac{1}{\sqrt{3}} x^2$. Let $g(x) = \frac{1}{\sqrt{3}} x^2$. This is a polynomial function, which is differentiable for all $x \in \mathbb{R}$. Therefore, the number of points where $|f(|x|)|$ is non-derivable is 0. This property is correct. **Property 2: $\text{sgn}(f(x))$ is discontinuous at $x = 1$.** For $f(x) = \frac{1}{\sqrt{3}} x^2$: For $x \ne 0$, $f(x) > 0$. For $x = 0$, $f(x) = 0$. The signum function is defined as $\text{sgn}(y) = \begin{cases} 1 & y > 0 \\ 0 & y = 0 \\ -1 & y < 0 \end{cases}$. So, $\text{sgn}(f(x)) = \begin{cases} 1 & x \ne 0 \\ 0 & x = 0 \end{cases}$. This function is discontinuous only at $x=0$. It is continuous at $x=1$. This property is incorrect. **Property 3: Derivative of $f(x)$ with respect to $\sin^{-1}\left(\frac{2x}{1+x^2}\right)$ at $x = \sqrt{3}$ is $-4$.** Let $u = f(x) = \frac{1}{\sqrt{3}} x^2$. Let $v = \sin^{-1}\left(\frac{2x}{1+x^2}\right)$. We need to find $\frac{du}{dv} = \frac{du/dx}{dv/dx}$ at $x = \sqrt{3}$. First, $\frac{du}{dx} = \frac{d}{dx}\left(\frac{1}{\sqrt{3}} x^2\right) = \frac{2}{\sqrt{3}} x$. At $x = \sqrt{3}$, $\frac{du}{dx} = \frac{2}{\sqrt{3}} (\sqrt{3}) = 2$. For $v = \sin^{-1}\left(\frac{2x}{1+x^2}\right)$: We use the identity $\sin^{-1}\left(\frac{2x}{1+x^2}\right) = \pi - 2 \tan^{-1}(x)$ for $x > 1$. Since $x = \sqrt{3} > 1$, we use this form. $\frac{dv}{dx} = \frac{d}{dx}(\pi - 2 \tan^{-1}(x)) = 0 - 2 \cdot \frac{1}{1+x^2} = -\frac{2}{1+x^2}$. At $x = \sqrt{3}$, $\frac{dv}{dx} = -\frac{2}{1+(\sqrt{3})^2} = -\frac{2}{1+3} = -\frac{2}{4} = -\frac{1}{2}$. Therefore, $\frac{du}{dv} = \frac{du/dx}{dv/dx} = \frac{2}{-1/2} = -4$. This property is correct. **Property 4: $\displaystyle\lim_{x \to 0^+}\left(\dfrac{\sqrt{3}f(x)}{x}\right)^x = 1$.** Substitute $f(x) = \frac{1}{\sqrt{3}} x^2$: $$\frac{\sqrt{3}f(x)}{x} = \frac{\sqrt{3} \left(\frac{1}{\sqrt{3}} x^2\right)}{x} = \frac{x^2}{x} = x$$ So we need to evaluate $\displaystyle\lim_{x \to 0^+} (x)^x$. This is an indeterminate form $0^0$. Let $L = \displaystyle\lim_{x \to 0^+} x^x$. Take the natural logarithm of both sides: $$\ln L = \displaystyle\lim_{x \to 0^+} \ln(x^x) = \displaystyle\lim_{x \to 0^+} x \ln x$$ This is an indeterminate form $0 \cdot (-\infty)$. Rewrite it as a fraction: $$\ln L = \displaystyle\lim_{x \to 0^+} \frac{\ln x}{1/x}$$ This is an indeterminate form $\frac{-\infty}{\infty}$, so we apply L'Hopital's Rule: $$\ln L = \displaystyle\lim_{x \to 0^+} \frac{d/dx(\ln x)}{d/dx(1/x)} = \displaystyle\lim_{x \to 0^+} \frac{1/x}{-1/x^2} = \displaystyle\lim_{x \to 0^+} (-x) = 0$$ Since $\ln L = 0$, we have $L = e^0 = 1$. This property is correct.