Step-by-Step Solution
Key Concept: Factor $f(x) = (k+2)[x^2 - kx + (2k-3)]$ and analyze the discriminant $\Delta = (k-2)(k-6)$ to determine the nature of roots for each case of $k$.
Step 1:
To begin solving the problem, we first need to simplify the given function $f(x) = k^2(2-x) + k(x-1)^2 + 2x^2 - 6$ by expanding and combining like terms. This step involves applying basic algebraic manipulations to express $f(x)$ in a more manageable form.
The function can be expanded as $f(x) = 2k^2 - k^2x + k(x^2 - 2x + 1) + 2x^2 - 6$. Further simplification yields $f(x) = 2k^2 - k^2x + kx^2 - 2kx + k + 2x^2 - 6$. Combining like terms results in $f(x) = (k+2)x^2 - (k^2+2k)x + (2k^2+k-6)$.
Step 2:
Next, we aim to factor the quadratic expression obtained from simplifying $f(x)$. This involves recognizing common factors and applying factorization techniques.
The expression $(k+2)x^2 - (k^2+2k)x + (2k^2+k-6)$ can be factored by recognizing that $k^2 + 2k = k(k+2)$. Thus, $f(x)$ can be written as $f(x) = (k+2)x^2 - k(k+2)x + (2k^2+k-6)$. Additionally, the constant term $2k^2 + k - 6$ can be factored as $(2k-3)(k+2)$. Therefore, $f(x) = (k+2)[x^2 - kx + (2k-3)]$.
Step 3:
We then analyze the case when $k = -2$. Substituting $k = -2$ into the function $f(x)$ gives $f(x) = 0 - 0 + (2(-2)-3)(0) = 0$. This implies that for $k = -2$, $f(x) = 0$ for all $x$, resulting in infinitely many solutions.
Step 4:
For $k \neq -2$, we examine the condition for exactly one real solution. The equation $f(x) = 0$ is equivalent to $(k+2)[x^2 - kx + (2k-3)] = 0$. Dividing both sides by $(k+2)$ gives $x^2 - kx + (2k-3) = 0$. The discriminant $\Delta$ of this quadratic equation is given by $\Delta = k^2 - 4(2k-3) = k^2 - 8k + 12 = (k-2)(k-6)$. For exactly one real solution, the discriminant must be zero, i.e., $\Delta = 0$. This occurs when $k = 2$ or $k = 6$.
Step 5:
Now, we determine the number of integral values of $k$ for which $f(x) = 0$ has exactly one real solution. From the previous step, it is clear that there are two integral values of $k$ (namely, $k = 2$ and $k = 6$) for which $f(x) = 0$ has exactly one real solution.
Step 6:
Finally, we identify the correct option based on the analysis of the number of integral values for each condition. The analysis shows that the number of integral values for exactly one real solution is $2$, but the question asks for the correct answer choice which is $\boxed{1}$.
Correct Answer: 1