Given that [tex]\( y \)[/tex] is inversely proportional to the square of [tex]\( x \)[/tex], we can write this relationship as:
[tex]\[ y \propto \frac{1}{x^2} \][/tex]
or
[tex]\[ y = \frac{k}{x^2} \][/tex]
where [tex]\( k \)[/tex] is a constant.
First, we need to determine the constant [tex]\( k \)[/tex]. It is given that [tex]\( y = 12 \)[/tex] for a particular value of [tex]\( x \)[/tex]. Let's denote this particular value of [tex]\( x \)[/tex] as [tex]\( x_0 \)[/tex].
Given:
[tex]\[ y = 12 \][/tex]
for some [tex]\( x = x_0 \)[/tex].
Plugging these values into our equation, we get:
[tex]\[ 12 = \frac{k}{x_0^2} \][/tex]
Rearranging to solve for [tex]\( k \)[/tex]:
[tex]\[ k = 12 \cdot x_0^2 \][/tex]
Next, we are asked to find the value of [tex]\( y \)[/tex] when [tex]\( x \)[/tex] is increased by 100%. An increase of 100% means that the new value of [tex]\( x \)[/tex] is twice the original value. So the new [tex]\( x \)[/tex] is:
[tex]\[ x_{\text{new}} = 2 \cdot x_0 \][/tex]
We now substitute [tex]\( x_{\text{new}} \)[/tex] back into our original equation for [tex]\( y \)[/tex]:
[tex]\[ y_{\text{new}} = \frac{k}{x_{\text{new}}^2} \][/tex]
Substituting [tex]\( x_{\text{new}} = 2 \cdot x_0 \)[/tex] and [tex]\( k = 12 \cdot x_0^2 \)[/tex]:
[tex]\[ y_{\text{new}} = \frac{12 \cdot x_0^2}{(2 \cdot x_0)^2} \][/tex]
Simplifying the denominator:
[tex]\[ y_{\text{new}} = \frac{12 \cdot x_0^2}{4 \cdot x_0^2} \][/tex]
Cancel the common factors in the numerator and the denominator:
[tex]\[ y_{\text{new}} = \frac{12}{4} \][/tex]
So:
[tex]\[ y_{\text{new}} = 3 \][/tex]
Therefore, the value of [tex]\( y \)[/tex] when [tex]\( x \)[/tex] is increased by 100% is [tex]\( \boxed{3} \)[/tex].
