Find the set of all values of a for which both roots of the equation x² - 2ax + a² - 1 = 0 lie between -2 and 4 — JEE Mathematics

Let $f(x) = x^2 - 2ax + a^2 - 1$. For both roots to lie in the interval $(-2, 4)$, three conditions must be met:

1. Discriminant $D \ge 0$:

$$(-2a)^2 - 4(1)(a^2 - 1) \ge 0 \implies 4a^2 - 4a^2 + 4 \ge 0 \implies 4 \ge 0$$

This is always true for all real $a$.
2. Vertex position $-2 < -\frac{B}{2A} < 4$:

$$-2 < \frac{2a}{2} < 4 \implies -2 < a < 4$$

3. Endpoint signs $f(-2) > 0$ and $f(4) > 0$:

$$f(-2) = (-2)^2 - 2a(-2) + a^2 - 1 > 0 \implies 4 + 4a + a^2 - 1 > 0 \implies a^2 + 4a + 3 > 0$$

$$(a+1)(a+3) > 0 \implies a \in (-\infty, -3) \cup (-1, \infty)$$

$$f(4) = 4^2 - 2a(4) + a^2 - 1 > 0 \implies 16 - 8a + a^2 - 1 > 0 \implies a^2 - 8a + 15 > 0$$

$$(a-3)(a-5) > 0 \implies a \in (-\infty, 3) \cup (5, \infty)$$

Taking the intersection of all intervals:

• $a \in (-2, 4)$
• $a \in (-\infty, -3) \cup (-1, \infty)$
• $a \in (-\infty, 3) \cup (5, \infty)$

Intersection: $a \in (-1, 3)$.

Answer: $a \in (-1, 3)$