Question:

Step 1: Let $g'(1) = a$ and $g''(2) = b$. Then the function $f(x)$ is given by $f(x) = x^2 + ax + b$. Step 2: Compute the necessary derivatives and values of $f(x)$: $f(1) = 1^2 + a(1) + b = 1 + a + b$. $f'(x) = \frac{d}{dx}(x^2 + ax + b) = 2x + a$. $f''(x) = \frac{d}{dx}(2x + a) = 2$. Step 3: Substitute these expressions into the definition of $g(x)$: $g(x) = f(1)x^2 + xf'(x) + f''(x)$ $g(x) = (1+a+b)x^2 + x(2x+a) + 2$ $g(x) = (1+a+b)x^2 + 2x^2 + ax + 2$ $g(x) = (3+a+b)x^2 + ax + 2$. Step 4: Determine $g'(x)$ and use the condition $g'(1)=a$: $g'(x) = \frac{d}{dx}((3+a+b)x^2 + ax + 2) = 2(3+a+b)x + a$. Using $g'(1)=a$: $2(3+a+b)(1) + a = a$ $2(3+a+b) = 0$ $3+a+b = 0$ $a+b = -3$. Step 5: Determine $g''(x)$ and use the condition $g''(2)=b$: $g''(x) = \frac{d}{dx}(2(3+a+b)x + a) = 2(3+a+b)$. Using $g''(2)=b$: $2(3+a+b) = b$. Substitute $a+b=-3$ from Step 4: $2(3+(-3)) = b$ $2(0) = b$ $b = 0$. Now, substitute $b=0$ back into $a+b=-3$: $a+0 = -3$ $a = -3$. Step 6: Substitute the values of $a$ and $b$ into $f(x)$ and $g(x)$: $f(x) = x^2 + (-3)x + 0 = x^2 - 3x$. $g(x) = (3+a+b)x^2 + ax + 2 = (3+(-3)+0)x^2 + (-3)x + 2 = 0x^2 - 3x + 2 = -3x + 2$. Step 7: Determine the minimum value of $f(x)$: $f(x) = x^2 - 3x$. This is a parabola opening upwards. Its minimum occurs at the vertex. The x-coordinate of the vertex is $x = -\frac{-3}{2(1)} = \frac{3}{2}$. The minimum value is $f\left(\frac{3}{2}\right) = \left(\frac{3}{2}\right)^2 - 3\left(\frac{3}{2}\right) = \frac{9}{4} - \frac{9}{2} = \frac{9}{4} - \frac{18}{4} = -\frac{9}{4} = -2.25$. Step 8: Evaluate the definite integral $\displaystyle\int_2^3 \frac{dx}{f(x)-x+5}$: First, simplify the denominator: $f(x)-x+5 = (x^2-3x) - x + 5 = x^2 - 4x + 5$. Complete the square for the denominator: $x^2 - 4x + 5 = (x^2 - 4x + 4) + 1 = (x-2)^2 + 1$. The integral becomes: $\displaystyle\int_2^3 \frac{dx}{(x-2)^2+1}$. Let $u = x-2$. Then $du = dx$. When $x=2$, $u=2-2=0$. When $x=3$, $u=3-2=1$. The integral transforms to: $\displaystyle\int_0^1 \frac{du}{u^2+1} = [\arctan u]_0^1 = \arctan(1) - \arctan(0) = \frac{\pi}{4} - 0 = \frac{\pi}{4}$. Step 9: Determine the number of positive integral values in the domain of $\sqrt{\dfrac{f(x)}{g(x)}}$: For $\sqrt{\dfrac{f(x)}{g(x)}}$ to be defined, we must have $\dfrac{f(x)}{g(x)} \geq 0$. $\dfrac{x^2-3x}{-3x+2} \geq 0$ $\dfrac{x(x-3)}{-(3x-2)} \geq 0$ $\dfrac{x(x-3)}{3x-2} \leq 0$. The critical points are $x=0$ (from $x$), $x=3$ (from $x-3$), and $x=2/3$ (from $3x-2$, where the denominator is zero). We perform a sign analysis for the expression $\dfrac{x(x-3)}{3x-2}$: \begin{itemize} \item For $x < 0$: Let $x=-1$. $\dfrac{(-1)(-4)}{-5} = -\frac{4}{5} < 0$. \item For $0 < x < 2/3$: Let $x=0.5$. $\dfrac{(0.5)(-2.5)}{-0.5} = 2.5 > 0$. \item For $2/3 < x < 3$: Let $x=1$. $\dfrac{(1)(-2)}{1} = -2 < 0$. \item For $x > 3$: Let $x=4$. $\dfrac{(4)(1)}{10} = \frac{4}{10} > 0$. \end{itemize} The inequality $\dfrac{x(x-3)}{3x-2} \leq 0$ holds for $x \in (-\infty, 0] \cup (2/3, 3]$. Note that $x=2/3$ is excluded because it makes the denominator zero. $x=0$ and $x=3$ are included because they make the numerator zero, satisfying $0 \leq 0$. The positive integral values in this domain are $x=1, x=2, x=3$. There are 3 such positive integral values. Step 10: Determine the number of points where $g(|x|)$ is non-derivable: $g(x) = -3x+2$. Then $g(|x|) = -3|x|+2$. The function $y = |x|$ is non-differentiable at $x=0$. For $x>0$, $g(|x|) = -3x+2$, and its derivative is $-3$. For $x<0$, $g(|x|) = -3(-x)+2 = 3x+2$, and its derivative is $3$. At $x=0$: The left-hand derivative is $\lim_{h \to 0^-} \frac{g(|h|) - g(0)}{h} = \lim_{h \to 0^-} \frac{3h+2 - 2}{h} = \lim_{h \to 0^-} \frac{3h}{h} = 3$. The right-hand derivative is $\lim_{h \to 0^+} \frac{g(|h|) - g(0)}{h} = \lim_{h \to 0^+} \frac{-3h+2 - 2}{h} = \lim_{h \to 0^+} \frac{-3h}{h} = -3$. Since the left-hand derivative ($3$) is not equal to the right-hand derivative ($-3$), $g(|x|)$ is not differentiable at $x=0$. There is 1 point where $g(|x|)$ is non-derivable.