Question:

Step 1: Apply the condition for continuity at $x=0$ For $f(x)$ to be continuous at $x=0$, the limit of $f(x)$ as $x \to 0$ must be equal to $f(0)$. Given $f(0) = 2$, we must have: $$ \lim_{x \to 0} \frac{ax^3 + bx^2 + cx + d}{x} = 2 $$ For this limit to exist, the numerator must approach $0$ as $x \to 0$. This implies that the constant term $d$ must be $0$. $$ \lim_{x \to 0} (ax^2 + bx + c) = 2 $$ Substituting $x=0$ into the expression, we get: $$ a(0)^2 + b(0) + c = 2 $$ $$ c = 2 $$ So, we have found $d=0$ and $c=2$. Step 2: Utilize the A.P. condition to find $a$ and $b$ We are given that $a, b, c, d$ are in an arithmetic progression (A.P.). We know $c=2$ and $d=0$. Let the common difference of the A.P. be $k$. Then $d = c + k$ is incorrect. It should be $c = b+k$ and $d = c+k$. Alternatively, $k = d-c$. $$ k = d - c = 0 - 2 = -2 $$ Now we can find $b$ and $a$: $$ c = b + k \implies 2 = b + (-2) \implies b = 4 $$ $$ b = a + k \implies 4 = a + (-2) \implies a = 6 $$ Thus, the sequence $a, b, c, d$ is $6, 4, 2, 0$, which is an A.P. with a common difference of $-2$. Step 3: Simplify the function $f(x)$ Substitute the values of $a, b, c, d$ into the definition of $f(x)$ for $x \neq 0$: $$ f(x) = \frac{6x^3 + 4x^2 + 2x + 0}{x} $$ For $x \neq 0$, we can divide each term by $x$: $$ f(x) = 6x^2 + 4x + 2 \quad \text{for } x \neq 0 $$ Combining this with $f(0)=2$, the function $f(x)$ can be written as: $$ f(x) = \begin{cases} 6x^2 + 4x + 2, & x \neq 0 \\ 2, & x = 0 \end{cases} $$ Since $6(0)^2 + 4(0) + 2 = 2$, this definition can be simplified to: $$ f(x) = 6x^2 + 4x + 2 \quad \text{for all } x \in \mathbb{R} $$ Step 4: Analyze the function $y = |f(|x|)|$ First, let's find $f(|x|)$ by replacing $x$ with $|x|$ in $f(x)$: $$ f(|x|) = 6(|x|)^2 + 4|x| + 2 $$ Since $(|x|)^2 = x^2$, this simplifies to: $$ f(|x|) = 6x^2 + 4|x| + 2 $$ Now we need to consider $y = |f(|x|)|$. To remove the absolute value, we check the sign of $f(|x|)$. Let $t = |x|$. Then $f(|x|) = 6t^2 + 4t + 2$. The discriminant of the quadratic $6t^2 + 4t + 2$ is $\Delta = (4)^2 - 4(6)(2) = 16 - 48 = -32$. Since the discriminant is negative and the leading coefficient ($6$) is positive, the quadratic $6t^2 + 4t + 2$ is always positive for all real values of $t$. Therefore, $f(|x|) = 6x^2 + 4|x| + 2 > 0$ for all $x \in \mathbb{R}$. This means that $|f(|x|)| = f(|x|)$. So, the function we need to analyze is: $$ y = 6x^2 + 4|x| + 2 $$ Step 5: Find the number of local minima of $y$ The function $y = 6x^2 + 4|x| + 2$ is an even function because $y(-x) = 6(-x)^2 + 4|-x| + 2 = 6x^2 + 4|x| + 2 = y(x)$. Let's analyze the function for $x \ge 0$. In this domain, $|x|=x$, so: $$ y = 6x^2 + 4x + 2 \quad \text{for } x \ge 0 $$ To find local minima, we compute the derivative with respect to $x$: $$ \frac{dy}{dx} = 12x + 4 $$ Set the derivative to zero to find critical points: $$ 12x + 4 = 0 \implies 12x = -4 \implies x = -\frac{1}{3} $$ This critical point $x = -1/3$ is not in the domain $x \ge 0$. For $x > 0$, $\frac{dy}{dx} = 12x + 4 > 0$. This means that $y(x)$ is strictly increasing on $(0, \infty)$. Therefore, for $x \ge 0$, the minimum value must occur at $x=0$. At $x=0$, $y(0) = 6(0)^2 + 4|0| + 2 = 2$. Since $y(x)$ is decreasing for $x<0$ (due to $y = 6x^2 - 4x + 2$, giving $\frac{dy}{dx} = 12x-4 < 0$ for $x<0$) and increasing for $x>0$, the point $x=0$ is a local minimum. Since $y(0)=2$ is also the global minimum (as $6x^2+4|x|+2 \ge 2$), it is the only local minimum for the function. The final answer is $\boxed{1}$.