Quadratic Equations Question - None

Question

Let $f(x) = x^2 - 2px + p^2 - 1$, where $p \in R - \{-1, 1\}$. If $\alpha$ and $\beta$ are distinct real roots of the equation $f(x) = 0$ such that $\left|\dfrac{\alpha^2 + \beta^2 + 3\alpha\beta}{\alpha\beta}\right| \leq 5$, then set of values of $p \in [a, b]$. The value of $[2(a^2 + b^2)]$ is:

[Note: $[k]$ denotes greatest integer less than or equal to $k$.]

Accepted Answer

Step 1: Find conditions for distinct real roots.

For f(x) = x² - 2px + p² - 1 = 0, the discriminant must be positive:

Δ = 4p² - 4(p² - 1) = 4p² - 4p² + 4 = 4 > 0

This is always true, so distinct real roots always exist.

Step 2: Apply Vieta's formulas.

From f(x) = 0:

α + β = 2p

αβ = p² - 1

Step 3: Simplify the given expression.

|α² + β² + 3αβ|/|αβ| ≤ 5

Since α² + β² = (α + β)² - 2αβ = 4p² - 2(p² - 1) = 2p² + 2:

|(2p² + 2) + 3(p² - 1)|/|p² - 1| ≤ 5

|2p² + 2 + 3p² - 3|/|p² - 1| ≤ 5

|5p² - 1|/|p² - 1| ≤ 5

Step 4: Solve the absolute value inequality.

-5|p² - 1| ≤ 5p² - 1 ≤ 5|p² - 1|

Case 1: p² > 1 (so p < -1 or p > 1)

|p² - 1| = p² - 1

-5(p² - 1) ≤ 5p² - 1 ≤ 5(p² - 1)

From -5p² + 5 ≤ 5p² - 1: -10p² ≤ -6, so p² ≥ 3/5 (always satisfied when p² > 1)

From 5p² - 1 ≤ 5p² - 5: -1 ≤ -5 (false, contradiction)

Case 2: p² < 1 (so -1 < p < 1)

|p² - 1| = 1 - p²

-5(1 - p²) ≤ 5p² - 1 ≤ 5(1 - p²)

From -5 + 5p² ≤ 5p² - 1: -5 ≤ -1 (true)

From 5p² - 1 ≤ 5 - 5p²: 10p² ≤ 6, so p² ≤ 3/5, thus |p| ≤ √(3/5)

Step 5: Combine conditions.

We need -1 < p < 1 and |p| ≤ √(3/5) = √15/5 ≈ 0.7746

Therefore: p ∈ [-√(3/5), √(3/5)] = [-√15/5, √15/5]

So a = -√15/5 and b = √15/5

Step 6: Calculate 2(a² + b²).

a² = b² = 3/5

2(a² + b²) = 2(3/5 + 3/5) = 2(6/5) = 12/5 = 2.4

[2.4] = 2

∴ Answer: 2

Question:

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