The values of 'a' for which one root of the equation $$x^2 - (a+1)x + a^2 + a - 8 = 0$$ exceeds 2 and the other is lesser than 2, are given by :

We are given the quadratic equation $$ x^2 - (a+1)x + a^2 + a - 8 = 0 $$. We need to find the values of $$ a $$ such that one root is greater than 2 and the other root is less than 2. Since the coefficient of $$ x^2 $$ is positive, the parabola opens upwards. For one root to be greater than 2 and the other less than 2, the point $$ x = 2 $$ must lie between the roots. This means the function value at $$ x = 2 $$ must be negative.

Define the quadratic function as $$ f(x) = x^2 - (a+1)x + a^2 + a - 8 $$. Evaluate $$ f(2) $$:

$$ f(2) = (2)^2 - (a+1)(2) + a^2 + a - 8 $$

$$ = 4 - 2(a+1) + a^2 + a - 8 $$

$$ = 4 - 2a - 2 + a^2 + a - 8 $$

$$ = a^2 - a - 6 $$

Set $$ f(2) < 0 $$:

$$ a^2 - a - 6 < 0 $$

Factor the quadratic expression:

$$ a^2 - a - 6 = (a - 3)(a + 2) $$

So,

$$ (a - 3)(a + 2) < 0 $$

The roots are $$ a = 3 $$ and $$ a = -2 $$. These divide the real line into intervals: $$ (-\infty, -2) $$, $$ (-2, 3) $$, and $$ (3, \infty) $$. Test the sign in each interval:

• For $$ a < -2 $$, say $$ a = -3 $$: $$ (-3-3)(-3+2) = (-6)(-1) = 6 > 0 $$ → positive.
• For $$ -2 < a < 3 $$, say $$ a = 0 $$: $$ (0-3)(0+2) = (-3)(2) = -6 < 0 $$ → negative.
• For $$ a > 3 $$, say $$ a = 4 $$: $$ (4-3)(4+2) = (1)(6) = 6 > 0 $$ → positive.

The expression is negative only when $$ -2 < a < 3 $$. However, we must ensure the quadratic has two distinct real roots, so the discriminant must be positive.

The discriminant $$ D $$ is:

$$ D = [-(a+1)]^2 - 4(1)(a^2 + a - 8) $$

$$ = (a+1)^2 - 4(a^2 + a - 8) $$

$$ = a^2 + 2a + 1 - 4a^2 - 4a + 32 $$

$$ = -3a^2 - 2a + 33 $$

Set $$ D > 0 $$:

$$ -3a^2 - 2a + 33 > 0 $$

Multiply both sides by -1 (reverse the inequality):

$$ 3a^2 + 2a - 33 < 0 $$

Solve $$ 3a^2 + 2a - 33 = 0 $$ using the quadratic formula:

$$ a = \frac{ -2 \pm \sqrt{ (2)^2 - 4(3)(-33) } }{2(3)} $$

$$ = \frac{ -2 \pm \sqrt{4 + 396} }{6} $$

$$ = \frac{ -2 \pm \sqrt{400} }{6} $$

$$ = \frac{ -2 \pm 20 }{6} $$

So, the roots are:

$$ a = \frac{18}{6} = 3 \quad \text{and} \quad a = \frac{-22}{6} = -\frac{11}{3} $$

Factor the quadratic:

$$ 3a^2 + 2a - 33 = (a - 3)(3a + 11) $$

Thus,

$$ (a - 3)(3a + 11) < 0 $$

The roots are $$ a = 3 $$ and $$ a = -\frac{11}{3} \approx -3.6667 $$. Since the parabola opens upwards, the expression is negative between the roots:

$$ -\frac{11}{3} < a < 3 $$

Now, combine both conditions:

1. From $$ f(2) < 0 $$: $$ -2 < a < 3 $$
2. From $$ D > 0 $$: $$ -\frac{11}{3} < a < 3 $$

The intersection is $$ \max(-2, -\frac{11}{3}) < a < \min(3, 3) $$, which simplifies to $$ -2 < a < 3 $$ because $$ -2 > -\frac{11}{3} $$.

Check the endpoints:

• At $$ a = -2 $$, $$ f(2) = (-2)^2 - (-2) - 6 = 4 + 2 - 6 = 0 $$, which is not less than 0. The roots are $$ x = 2 $$ and $$ x = -3 $$, so one root is exactly 2, not exceeding it.
• At $$ a = 3 $$, $$ f(2) = (3)^2 - 3 - 6 = 9 - 3 - 6 = 0 $$, again not less than 0. The roots are both 2 (double root), so no root exceeds or is less than 2.

Thus, $$ a = -2 $$ and $$ a = 3 $$ are excluded, and the strict inequality $$ -2 < a < 3 $$ holds.

Verify with a test point in $$ (-2, 3) $$, say $$ a = 0 $$:

$$ f(x) = x^2 - x - 8 = 0 $$

Roots: $$ \frac{1 \pm \sqrt{1 + 32}}{2} = \frac{1 \pm \sqrt{33}}{2} \approx \frac{1 \pm 5.744}{2} $$, so approximately $$ 3.372 $$ and $$ -2.372 $$. One root exceeds 2, the other is less than 2.

Comparing with the options:

A. $$ 3 < a < 10 $$

B. $$ a \geq 10 $$

C. $$ -2 < a < 3 $$

D. $$ a \leq -2 $$

The condition $$ -2 < a < 3 $$ matches option C.

Hence, the correct answer is Option C.