Limits, Continuity & Differentiability Question - None

Question

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:

Accepted Answer

Step 1: Identify non-derivable points of g(x) = |f(|x|)|.

Non-derivability occurs at:

x = 0 (corner from |x|)
x = ±a where f(a) = 0 and a > 0 (zeros of f when f changes sign)
x = ±b where f(b) = 0 and b > 0 (second zero if it exists)

Step 2: For exactly 5 non-derivable points, f(x) must have exactly 2 positive roots a and b (with a ≠ b). This gives points: -b, -a, 0, a, b.

Step 3: Apply the constraint ∑|xᵢ| = 10.

|-b| + |-a| + |0| + |a| + |b| = b + a + 0 + a + b = 2a + 2b = 10

Therefore: a + b = 5

Step 4: Since a and b are roots of f(x) = x² - px + q:

a + b = p (sum of roots)
ab = q (product of roots)

From a + b = 5: p = 5

Step 5: Determine range of q. Since a, b are positive real roots with a + b = 5 and a ≠ b:

By AM-GM: ab ≤ (a+b)²/4 = 25/4, with equality when a = b (excluded)

So q = ab ∈ (0, 25/4)

Therefore: p + q ∈ (5, 45/4) or p + q ∈ (5, 11.25)

∴ Answer: Values in the range (5, 11.25). Common options like p+q = 6, 7, 8, 9, 10, or 11 are all possible, while p+q ≤ 5 or p+q ≥ 11.25 are impossible.

Question:

Step-by-Step Solution

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