Question:

Step 1: Since $h(x) = \int\int\int g'''(x)\,dx\,dx\,dx$, after three integrations we get $h(x) = g(x) + ax^2 + bx + c$ for some constants $a, b, c$. Step 2: Therefore $f(x) = h(x) - g(x) = ax^2 + bx + c$. Step 3: Apply the conditions. $h(3) = g(3)$ gives $f(3) = 0$: $9a + 3b + c = 0$. $h(1) = g(1)$ gives $f(1) = 0$: $a + b + c = 0$. $h(0) - g(0) = 6$ gives $f(0) = 6$: $c = 6$. Step 4: From $c = 6$ and $a + b + c = 0$: $a + b = -6$. From $9a + 3b + c = 0$: $9a + 3b = -6$, so $3a + b = -2$. Subtracting: $2a = 4$, so $a = 2$, $b = -8$. Step 5: Thus $f(x) = 2x^2 - 8x + 6$. $f'(x) = 4x - 8 = 4(x-2)$. $f'(x) < 0$ for $x < 2$ and $f'(x) > 0$ for $x > 2$, so $f$ decreases on $(-\infty, 2)$ and increases on $(2, \infty)$. Step 6: Check options: - (a) $f$ decreases in $(1,3)$: False (it decreases on $(1,2)$ but increases on $(2,3)$). - (b) $f$ decreases in $(-\infty, 2)$: True ✓ - (c) $f(4) = 2(16) - 8(4) + 6 = 32 - 32 + 6 = 6$. True ✓ - (d) $f(2) = 2(4) - 8(2) + 6 = 8 - 16 + 6 = -2 \neq 6$. False. Correct options: (b) and (c), i.e., options 2 and 3.