(a) \(y = k + \frac{c}{x^{2}}\)
where c and k are constants.
(b) When x = 1, y = 11
\(11 = k + \frac{c}{1^{2}} \implies 11 = k + c ... (1)\)
When x = 2, y = 5
\(5 = k + \frac{c}{2^{2}} \implies 5 = k + \frac{c}{4}\)
\(\equiv 20 = 4k + c ... (2)\)
(2) - (1) :
\(4k - k = 20 - 11 \implies 3k = 9\)
\(k = 3\)
Put k = 3 in (1), we have
\(11 = 3 + c \implies c = 11 - 3 = 8\)
\(\therefore y = 3 + \frac{8}{x^{2}}\)
When x = 4,
\(y = 3 + \frac{8}{4^{2}} = 3 + \frac{1}{2}\)
= \(3\frac{1}{2}\)