The quantity y is partly constant and partly varies inversely as the square of x....

Explanation

(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,

\(y = 3 + \frac{8}{4^{2}} = 3 + \frac{1}{2}\)

= \(3\frac{1}{2}\)