Mathematics Questions and Answers

3 + 2 + \(\frac{4}{3}\) + \(\frac{8}{9}\) + \(\frac{16}{17}\) + .....

a = 3

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

s \(\alpha\) = \(\frac{a}{1 - r}\) = \(\frac{3}{1 - \frac{2}{3}}\)

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

= 3 x 3

= 9

To solve -11 \(\leq\) 4 - 3x \(\leq\) 28

-11 \(\leq\) 4 - 3x also 4 -3x \(\leq\) 28

15 \(\leq\) -3x \(\leq\) 24 = 15 \(\geq\) 3x - 3x \(\geq\) -24

-5 \(\geq\) x, x \(\geq\) -8

i.e. x \(\leq\) 5

∴ -8 \(\leq\) x \(\leq\) 5

\(\frac{3}{x^2 + x - 2}\) = \(\frac{3}{(x - 1)(x + 2)}\)

\(\frac{A}{x - 1}\) + \(\frac{B}{x + 2}\)

A(x + 2) + B(x - 1) = 3

when x = 1, 3A = 3 \(\to\) a = 1

when x = -2, -3B = 3 \(\to\) B = -1

= \(\frac{1}{x - 1} - \frac{1}{x + 2}\)

x + 1 is a factor of x³ + 3x² + kx + 4

Let f(x) = x³ + 3x² + kx + 4

∴ f(-1) = (-1)³ + 3(-1)² + k(-1) + 4 = 0

-1 + 3 - k + 4 = 0

∴ k = 6

Let the amounts invested at 4% and 5% respectively be x and y.

\(\therefore x + y = 280 ... (i)\)

Interest on x = \(\frac{x \times 4 \times 1}{100} = 0.04x\)

Interest on y = \(\frac{y \times 5 \times 1}{100} = 0.05y\)

\(\therefore 0.04x + 0.05y = 12.80\)

\(\implies 4x + 5y = 1280 ... (ii)\)

From (i), \(x = 280 - y\).

Put into (ii), \(4(280 - y) + 5y = 1280\)

\(1120 - 4y + 5y = 1280\)

\(1120 + y = 1280 \implies y = 1280 - 1120 = N160\)

\(\therefore\) N160 was invested at the rate of 5% per annum.