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: Analyze non-derivability points of g(x) = |f(|x|)|

The function g(x) is non-derivable at:

• x = 0 (due to |x| and potentially |f(·)|)

• Points where f(|x|) = 0, i.e., where |x| equals a root of f

• Potentially at the vertex of f when it equals zero at some point

Step 2: Identify the 5 non-derivable points

Let f(x) = x² - px + q have roots r₁ and r₂ (real roots required). For g(x) = |f(|x|)|:

If f has two positive roots 0 < r₁ < r₂, the non-derivable points are:

• x = 0 (always one point)

• x = r₁, x = -r₁ (two points where f(|x|) = 0)

• x = r₂, x = -r₂ (two points where f(|x|) = 0)

This gives exactly 5 points: {-r₂, -r₁, 0, r₁, r₂}

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

The sum of absolute values:

|-r₂| + |-r₁| + |0| + |r₁| + |r₂| = r₂ + r₁ + 0 + r₁ + r₂ = 2(r₁ + r₂) = 10

Therefore: r₁ + r₂ = 5

Step 4: Use Vieta's formulas

For f(x) = x² - px + q with roots r₁, r₂:

• Sum of roots: r₁ + r₂ = p = 5

• Product of roots: r₁ · r₂ = q

Step 5: Determine possible values of q

For two distinct positive real roots, we need:

• Discriminant p² - 4q > 0, so 25 - 4q > 0, giving q < 6.25

• Both roots positive: r₁ + r₂ = 5 > 0 ✓ and r₁ · r₂ = q > 0

Thus: 0 < q < 6.25

Step 6: Calculate p + q = 5 + q

Since 0 < q < 6.25:

• p + q = 5 + q where 0 < q < 6.25

• Therefore: 5 < p + q < 11.25

From the options:

• A: 7 ✓ (when q = 2)

• B: 9 ✓ (when q = 4)

• C: 11 ✓ (when q = 6, boundary case)

• D: 13 ✗ (exceeds upper bound)

Checking boundary: If q = 6.25, discriminant = 0 (one repeated root, doesn't give 5 points). If q = 6, discriminant = 1 > 0 (two distinct roots exist). So C is valid.

∴ Answer: A,B,C,D

Question:

Step-by-Step Solution

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