Explanation

Where A is the assumed mean = 13

substituting, mean = 13 + \(\frac{18}{100}\) = 13 + 1.85

= 14.85 years

Explanation

(a) \(S_{n} = \frac{5n^{2}}{2} + \frac{5n}{2}\)

\(S_{1} = \frac{5(1^{2})}{2} + \frac{5(1)}{2} = 5 \implies T_{1} = 5\)

\(S_{2} = \frac{5(2^{2})}{2} + \frac{5(2)}{2} = 10 + 5 = 15\)

\(S_{3} = \frac{5(3^{2})}{2} + \frac{5(3)}{2} = 30\)

\(S_{4} = \frac{5(4^{2})}{2} + \frac{5(4)}{2} = 50\)

\(T_{2} = S_{2} - S_{1} = 15 - 5 = 10\)

\(T_{3} = S_{3} - S_{2} = 30 - 15 = 15\)

\(T_{4} = 50 - 30 = 20\)

The first four terms are 5, 10, 15 and 20.

This is a linear sequence with d = 5.

\(T_{n} = a + (n - 1)d\)

= \(5 + (n - 1)5 \implies T_{n} = 5 + 5n - 5 = 5n\)

(b) \(x^{2} + y^{2} - 10x - 8y + 25 = 0\)

For circle to touch x- axis, the distance from centre of circle to the x- axis must be equal to the radius.

\(x^{2} + y^{2} - 10x - 8y + 25 = 0\)

\(2g = -10, g = -5\)

\(2f = -8 , f = -4\)

Centre = (-g, -f) = (5, 4).

\(Radius = \sqrt{g^{2} + f^{2} - c} = \sqrt{(-5)^{2} + (-4)^{2} - 25} = 4\)

\(|PQ| = \sqrt{4^{2}} = 4\)

Since PQ = radius, circle touches x- axis.

Correct Answer: C. (1,4)

Explanation

\(f(x) = 5x^{2} - 4x + 3\)

\(f'(x) = 10x - 4 = 6 \implies 10x = 10; x=1\)

When x = 1, f(x) = y = \(5(1^{2}) - 4(1) + 3 = 4\)

The coordinates are (1,4)

Correct Answer: C. -9i - 44j

Explanation

= 3(5i+4j)?8(3i+7j)=15i+12j?24i?56j = ?9i?44j

Correct Answer: D. 128 + 64 + 32 + 16 + ...

Correct Answer: C. \(\frac{2}{3}\)

Explanation

variance = \(\frac{\sum(x - \bar{x})^2}{n}\)

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

Correct Answer: B. \(\frac{27}{8}\)y\(^2\)

Explanation

(2x + \(\frac{3y}{4}\))\(^3\)

= 3(2x) (\(\frac{3y}{4}\))\(^2\)

= \(\frac{27}{8}\)y\(^2\)

Correct Answer: C. \(81\sqrt{2}\)

Explanation

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

\(a = \sqrt{6}, r = \frac{T_{2}}{T_{1}} = \frac{3\sqrt{2}}{\sqrt{6}} \)

\(r = \frac{\sqrt{18}}{\sqrt{6}} = \sqrt{3}\)

\(\therefore T_{8} = (\sqrt{6})(\sqrt{3})^{8 - 1} \)

= \((\sqrt{6})(27\sqrt{3}) = 27\sqrt{18} = 81\sqrt{2}\)

Correct Answer: A. \(5\frac{2}{5}\)

Explanation

Sum to infinity of a G.P when /r/ < 1 = \(\frac{a}{1 - r}\)

a = \(\frac{9}{2},r = \frac{T_2}{T_1} = \frac{3}{4} + \frac{9}{2}\)

r = \(\frac{3}{4} \times \frac{2}{9} = \frac{1}{6}\)

\(S_∞ = \frac{\frac{9}{2}}{1-\frac{1}{6}} = \frac{\frac{9}{2}}{\frac{5}{6}} = \frac{27}{5}\)

\(\therefore S_∞ = 5\frac{2}{5}\)

Explanation

\(3x^{2} + 2y^{2} + xy + x - 7 = 0\)

We differentiate y implicitly with respect to x.

\(6x + 2(2y)\frac{\mathrm d y}{\mathrm d x} + y + x\frac{\mathrm d y}{\mathrm d x} + 1 = 0\)

\(\frac{\mathrm d y}{\mathrm d x} = \frac{- y - 1 - 6x}{4y + x}\)

= \(\frac{-(6x + y + 1)}{4y + x}\).

At the point (-2, 1),

\(\frac{\mathrm d y}{\mathrm d x} = \frac{-(-12 + 1 + 1)}{4(1) - 2}\)

= \(\frac{-(-10)}{2} = 5\)