Further Mathematics Questions and Answers

\(Mean = \frac{\text{sum of ages}}{\text{no of pupils}}\)

Let the sum of the ages of the 15 pupils be x and the sum of the 16 pupils be y.

\(\frac{x}{15} = 14.2 \implies x = 14.2 \times 15 = 213\)

\(\frac{y}{16} = 14.1 \implies y = 14.1 \times 16 = 225.6\)

\(\text{Age of new pupil} = 225.6 - 213 = 12.6\)

\(y = 4x^{2} - 12x + 7\)

\(\frac{\mathrm d y}{\mathrm d x} = 8x - 12\)

At x = 2, y = 8(2) - 12 = 4

Equation of the tangent to the curve: \(y - (-1) = 4(x - 2)\)

\(y + 1 = 4x - 8 \implies y - 4x + 1 + 8 = y - 4x + 9 = 0\)

The vertical line \(x = \frac{-b}{2a}\) is the axis of symmetry of the curve.

\(y = x^{2} - 4x - 12\)

\(\text{Axis of symmetry} = x = \frac{-(-4)}{2(1)} = \frac{4}{2} = 2\)

\(T_{n} = ar^{n - 1}\) ( Geometric Progression)

\(T_{3} = ar^{3 - 1} = ar^{2} = 10 .... (1)\)

\(T_{6} = ar^{6 - 1} = ar^{5} = 80 .....(2)\)

Divide (2) by (1)

\(r^{5 - 2} = r^{3} = 8 \)

\(r = \sqrt[3]{8} = 2\)

\(P = {x : \text{x is a factor of 6}} \implies P = {1, 2, 3, 6}\)

\(g(x) = x^{2} + 3x - 5\)

\(g(1) = 1^{2} + 3(1) - 5 = 1 + 3 - 5 = -1\)

\(g(2) = 2^{2} + 3(2) - 5 = 4 + 6 - 5 = 5\)

\(g(3) = 3^{2} + 3(3) - 5 = 9 + 9 - 5 = 13\)

\(g(6) = 6^{2} + 3(6) - 5 = 36 + 18 - 5 = 49\)

\(\therefore Range(g(x)) = {-1, 5, 13, 49}\)