Further Mathematics Questions and Answers

Let the original volume be V with radius r.

\(V = \frac{4}{3}\pi r^{3}\)

45% increased volume = 145%V. 

Let the %age increase in radius = m%r

\(\frac{145}{100}V = \frac{4}{3}\pi (\frac{mr}{100})^{3}\)

\(1.45V = (\frac{4}{3}\pi r^{3})(\frac{m}{100})^{3}\)

\(1.45V = V(\frac{m}{100})^{3}\)

\(\implies 1.45 \times 10^{6} = m^{3}\)

\(m = \sqrt[3]{1.45 \times 10^{6}} = 113.2%\)

\(\therefore \text{%age increase =}  113.2 - 100 = 13.2%\)

\(\approxeq 13%\)

\(T_{n} = ar^{n - 1}\) (Exponential sequence)

\(T_{4} = ar^{3} = 2 .... (1)\)

\(T_{6} = ar^{5} = 8 ......(2)\)

Divide (2) by (1),

\(\frac{ar^{5}}{ar^{3}} = \frac{8}{2} \)

\(r^{2} = 4\)

\(r = \pm 2\)

\(\begin{pmatrix} 3 & 2 \\ 7 & x \end{pmatrix} \begin{pmatrix} 2 \\ 3 \end{pmatrix} = \begin{pmatrix} 12 \\ 29 \end{pmatrix}\)

\(\begin{pmatrix} 3 \times 2 + 2 \times 3 \\ 7 \times 2 + x \times 3 \end{pmatrix} = \begin{pmatrix} 12 \\ 29 \end{pmatrix}\)

\(\implies 14 + 3x = 29 \)

\(3x = 29 - 14 = 15\)

\(x = 5\)

The inverse of the inverse of a function gives the function 

i.e \(f^{-1}(f^{-1}(x)) = f(x)\)

\(f^{-1}(x) = \frac{x + 1}{4}\)

Take y = x, so

\(f^{-1}(y) = \frac{y + 1}{4}\)

Let \(x = f^{-1}(y)\),

\(x = \frac{y + 1}{4} \implies 4x = y + 1\)

\(y = f(x) = 4x - 1\)

\(9^{2x + 1} = 81^{3x + 2}\)

\(9^{2x + 1} = (9^{2})^{3x + 2}\)

\(9^{2x + 1} = 9^{6x + 4}\)

Equating powers, 

\(2x + 1 = 6x + 4 \implies -3 = 4x\)

\(\therefore x = \frac{-3}{4}\)