Critical Point

Step 1: Understand the Condition for 3 Distinct Real Roots

Let the given function be:

$$f(x) = 2x^3 - 3x^2 - 12x + k$$

For a cubic polynomial with a positive leading coefficient ($$2 > 0$$) to intersect the x-axis at three distinct points, its graph must rise to a local maximum above the x-axis ($$f(x_{\text{max}}) > 0$$) and dip to a local minimum below the x-axis ($$f(x_{\text{min}}) < 0$$).

Mathematically, this translates to the condition:

$$f(x_{\text{max}}) \cdot f(x_{\text{min}}) < 0$$

Step 2: Find the Critical Points

To find the turning points, differentiate $$f(x)$$ with respect to $$x$$ and equate it to zero:

$$f'(x) = 6x^2 - 6x - 12$$

Set $$f'(x) = 0$$:

$$6(x^2 - x - 2) = 0$$ $$(x - 2)(x + 1) = 0$$

The critical points are $$x = -1$$ and $$x = 2$$.

• By the first derivative test (or looking at the sign change of $$f'(x)$$), $$x = -1$$ is the point of local maxima.
• $$x = 2$$ is the point of local minima.

Step 3: Evaluate the Function at Critical Points

Substitute these critical points back into original function $f(x)$:

1. At $$x = -1$$ (Local Maxima):$$f(-1) = 2(-1)^3 - 3(-1)^2 - 12(-1) + k$$ $$f(-1) = -2 - 3 + 12 + k = k + 7$$
2. At $$x = 2$$ (Local Minima):$$f(2) = 2(2)^3 - 3(2)^2 - 12(2) + k$$ $$f(2) = 16 - 12 - 24 + k = k - 20$$

Step 4: Apply the Root Condition

For three distinct real roots:

$$(k + 7)(k - 20) < 0$$

Using the method of intervals (wavy curve method), the range of $$k$$ is:

$$-7 < k < 20$$

Step 5: Count the Integral Values

We need to find the number of integers strictly between $$-7$$ and $$20$$:

$$\text{Number of integers} = 20 - (-7) - 1 = 26$$

The specific integers are from $$-6$$ to $$19$$ inclusive.

Correct Option: B. 26

$$f(x) = x^3 - 3x + 1$$
$$f'(x) = 3x^2 - 3 = 3(x-1)(x+1)$$
The critical points are $$x = 1$$ and $$x = -1$$.
• Local maximum value: $$f(-1) = (-1)^3 - 3(-1) + 1 = 3$$
• Local minimum value: $$f(1) = (1)^3 - 3(1) + 1 = -1$$
Since $$f(-1) > 0$$ and $$f(1) < 0$$, the cubic equation $$f(x) = 0$$ has 3 distinct real roots. 
Let's denote them as $$\alpha, \beta, \gamma$$.).
• $$f(-2) = -8 + 6 + 1 = -1 < 0$$ and $$f(-1) = 3 > 0 \implies \alpha \in (-2, -1)$$
• $$f(0) = 1 > 0$$ and $$f(1) = -1 < 0 \implies \beta \in (0, 1)$$
• $$f(1) = -1 < 0$$ and $$f(2) = 8 - 6 + 1 = 3 > 0 \implies \gamma \in (1, 2)$$
So, $$-2 < \alpha < -1$$, $$0 < \beta < 1$$, and $$1 < \gamma < 2$$.
We want to find the number of real roots for $$f(f(x)) = 0$$.

$$f(x) = \alpha \quad \text{or} \quad f(x) = \beta \quad \text{or} \quad f(x) = \gamma$$
Now, we analyze how many times the horizontal lines $$y = \alpha$$, $$y = \beta$$, and $$y = \gamma$$ intersect the graph of $$y = f(x)$$. Remember that the local maximum of $$f(x)$$ is $$3$$ and the local minimum is $$-1$$.
• For $$f(x) = \alpha$$: Since $$\alpha \in (-2, -1)$$, the line $$y = \alpha$$ lies entirely below the local minimum ($$y = -1$$). A horizontal line below the local minimum intersects the cubic curve exactly 1 time.
• For $$f(x) = \beta$$: Since $$\beta \in (0, 1)$$, the line $$y = \beta$$ lies strictly between the local minimum ($$-1$$) and the local maximum ($$3$$). A horizontal line in this range intersects the cubic curve exactly 3 times.
• For $$f(x) = \gamma$$: Since $$\gamma \in (1, 2)$$, the line $$y = \gamma$$ also lies strictly between the local minimum ($$-1$$) and the local maximum ($$3$$). Therefore, it also intersects the curve exactly 3 times.
$$\text{Total roots} = 1 (\text{from } \alpha) + 3 (\text{from } \beta) + 3 (\text{from } \gamma) = 7 \text{ roots}$$

Total intersections = $$1 + 3 + 3 = 7$$.