Question:

Step 1: Assume $f(x) = ax^n$ (polynomial with non-negative coefficients, $f(0)=0$). Then: $$f(f(x)) = a(ax^n)^n = a^{n+1}x^{n^2}$$ $$x\int_0^x f(t)\,dt = x \cdot \dfrac{ax^{n+1}}{n+1} = \dfrac{a}{n+1}x^{n+2}$$ Step 2: Equate exponents and coefficients: $$n^2 = n+2 \Rightarrow n^2 - n - 2 = 0 \Rightarrow (n-2)(n+1) = 0 \Rightarrow n = 2$$ $$a^{n+1} = \dfrac{a}{n+1} \Rightarrow a^3 = \dfrac{a}{3} \Rightarrow a^2 = \dfrac{1}{3} \Rightarrow a = \dfrac{1}{\sqrt{3}}$$ So $f(x) = \dfrac{x^2}{\sqrt{3}}$. Step 3: Check option (a): $|f(|x|)| = \left|\dfrac{x^2}{\sqrt{3}}\right| = \dfrac{x^2}{\sqrt{3}}$, which is differentiable everywhere. Number of non-derivable points = 0. ✓ Step 4: Check option (b): $f(x) = \dfrac{x^2}{\sqrt{3}} \geq 0$ for all $x$, so $\text{sgn}(f(x)) = 0$ at $x=0$ and $1$ elsewhere. It is discontinuous at $x=0$, not at $x=1$. ✗ Step 5: Check option (c): Find $\dfrac{df}{d(\sin^{-1}(2x/(1+x^2)))}$ at $x=\sqrt{3}$. $$\dfrac{df/dx}{d(\sin^{-1}(2x/(1+x^2)))/dx}$$ $f'(x) = \dfrac{2x}{\sqrt{3}}$. At $x=\sqrt{3}$: $f'(\sqrt{3}) = \dfrac{2\sqrt{3}}{\sqrt{3}} = 2$. Let $\theta = \sin^{-1}\left(\dfrac{2x}{1+x^2}\right)$. For $x > 1$: $\dfrac{d}{dx}\sin^{-1}\left(\dfrac{2x}{1+x^2}\right) = \dfrac{-2}{1+x^2}$... Actually $\dfrac{2x}{1+x^2} = \sin(2\tan^{-1}x)$, so $\sin^{-1}\left(\dfrac{2x}{1+x^2}\right) = \pi - 2\tan^{-1}x$ for $x>1$. Derivative $= \dfrac{-2}{1+x^2}$. At $x=\sqrt{3}$: $= \dfrac{-2}{4} = -\dfrac{1}{2}$. So the required derivative $= \dfrac{2}{-1/2} = -4$. ✓ Step 6: Check option (d): $\displaystyle\lim_{x\to 0^+}\left(\dfrac{\sqrt{3} \cdot x^2/\sqrt{3}}{x}\right)^x = \lim_{x\to 0^+}(x)^x = 1$. ✓