To solve the problem of determining the product of possible values of [tex]\( a \)[/tex] such that the function [tex]\( f(x) = 2ax - x^2 - 4 \)[/tex] has a positive integral value at exactly one real [tex]\( x \)[/tex], let's go through the steps systematically.
1. Define the Problem:
We have the quadratic function [tex]\( f(x) = 2ax - x^2 - 4 \)[/tex]. We need to determine for which values of [tex]\( a \)[/tex] this function has only one real root.
2. Identify the Condition for One Real Root:
A quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex] has exactly one real root when its discriminant is zero. The discriminant for [tex]\( f(x) \)[/tex] is given by:
[tex]\[
\Delta = b^2 - 4ac
\][/tex]
For [tex]\( f(x) = 2ax - x^2 - 4 \)[/tex], rewrite it in the form [tex]\( -x^2 + 2ax - 4 \)[/tex]:
[tex]\[
a = -1, \quad b = 2a, \quad c = -4
\][/tex]
Thus, the discriminant [tex]\(\Delta\)[/tex] is:
[tex]\[
\Delta = (2a)^2 - 4(-1)(-4) = 4a^2 - 16
\][/tex]
3. Set the Discriminant to Zero:
To ensure there is exactly one real root, set the discriminant to zero:
[tex]\[
4a^2 - 16 = 0
\][/tex]
[tex]\[
4a^2 = 16
\][/tex]
[tex]\[
a^2 = 4
\][/tex]
[tex]\[
a = \pm 2
\][/tex]
So, the possible values of [tex]\(a\)[/tex] are [tex]\(-2\)[/tex] and [tex]\(2\)[/tex].
4. Determine the Product of the Possible Values of [tex]\(a\)[/tex]:
The product of [tex]\(-2\)[/tex] and [tex]\(2\)[/tex] is:
[tex]\[
(-2) \times 2 = -4
\][/tex]
Therefore, the product of the possible values of [tex]\(a\)[/tex] is [tex]\(-4\)[/tex].
The correct answer is:
[tex]\[
(4) -5
\][/tex]
