Let $f(x) = x^2 - px + q$, $p, q \in R$. If $x_1, x_2, x_3, x_4, x_5$ (where $x_i \in I$) are the 5 points where $g(x) = |f(|x|)|$ is non-derivable and $\sum_{i=1}^{5} |x_i| = 10$, then $p + q$ can be:
Step-by-Step Solution
Key Concept: The non-differentiability of $g(x) = |f(|x|)|$ primarily stems from two sources: the point $x=0$ due to the inner $|x|$, and points where $f(|x|)=0$ due to the outer absolute value. For exactly 5 distinct integer non-differentiable points, $x=0$ must be one such point, and $f(t)=0$ must have two distinct positive integer roots $t=\alpha, \beta$ which then generate the points $x=\pm\alpha, \pm\beta$.
Step 1: Analyze non-derivability of $g(x) = |f(|x|)|$. The function $g(x)$ can be non-derivable at: $x = 0$ (due to $|x|$), at points where $f(|x|) = 0$ (due to outer absolute value), and at points where $f$ has a corner.
Step 2: Since $f(x) = x^2 - px + q$, the roots of $f(t) = 0$ are $t = \alpha, \beta$ (say). For $g(x) = |f(|x|)|$, non-derivable points occur at $x = 0$, $x = \pm\alpha$, $x = \pm\beta$ (where $\alpha, \beta > 0$).
Step 3: For exactly 5 non-derivable integer points, one possibility is $x = 0, \pm\alpha, \pm\beta$ where $\alpha \neq \beta$ and both are positive integers. This gives 5 points: $\{-\beta, -\alpha, 0, \alpha, \beta\}$.
Step 4: Compute $\sum_{i=1}^5 |x_i| = 0 + \alpha + \alpha + \beta + \beta = 2(\alpha + \beta) = 10$, so $\alpha + \beta = 5$.
Step 5: Since $\alpha$ and $\beta$ are roots of $f(t) = t^2 - pt + q$:
$$p = \alpha + \beta = 5, \quad q = \alpha\beta$$
So $p + q = 5 + \alpha\beta$.
Step 6: With $\alpha + \beta = 5$ and $\alpha, \beta$ positive integers ($\alpha \neq \beta$): possible pairs are $(1,4)$ and $(2,3)$.
- $(\alpha,\beta) = (1,4)$: $q = 4$, $p+q = 9$
- $(\alpha,\beta) = (2,3)$: $q = 6$, $p+q = 11$
Step 7: Check options: $p+q = 9$ (option b) and $p+q = 11$ (option c) are valid. The correct answers are options (b) and (c), i.e., 9 and 11.
Correct Answer: 1, 3