Mathematics Questions and Answers

\(9^{y} - 4 \times 3^{y} + 3 = 0\)

\(\equiv (3^{2})^{y} - 4 \times 3^{y} + 3 = 0\)

\((3^{y})^{2} - 4 \times 3^{y} + 3 = 0\)

Let \(3^{y}\) be r. Then,

\(r^{2} - 4r + 3 = 0\)

Solving the equation, 

\(r^{2} - 3r - r + 3 = 0\)

\(r(r - 3) - 1(r - 3) = 0\)

\((r - 3)(r - 1) = 0\)

\(\therefore \text{r = 3 or 1}\)

Recall, \(3^{y} = r\)

\(3^{y} = 3 = 3^{1} \text{ or } 3^{y} = 1 = 3^{0}\)

\(\implies \text{y = 1 or 0}\)

\(\log_{3} p + 3\log_{3} q = 3\)

\(\log_{3} p + \log_{3} q^{3} = 3\)

\(\implies \log_{3} (pq^{3}) = 3\)

\(pq^{3} = 3^{3} = 27\)

\(\therefore p = \frac{27}{q^{3}}\)

= \((\frac{3}{q})^{3}\)

\(4 - \frac{1}{2 - \sqrt{3}} = \frac{4(2 - \sqrt{3}) - 1}{2 - \sqrt{3}}\)

= \(\frac{8 - 4\sqrt{3} - 1}{2 - \sqrt{3}}\)

= \(\frac{7 - 4\sqrt{3}}{2 - \sqrt{3}}\)

Rationalizing,

\((\frac{7 - 4\sqrt{3}}{2 - \sqrt{3}}) (\frac{2 + \sqrt{3}}{2 + \sqrt{3}})\)

= \(\frac{14 + 7\sqrt{3} - 8\sqrt{3} - 12}{4 + 2\sqrt{3} - 2\sqrt{3} - 3}\)

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

Surface Area of Sphere A = 4\(\pi r^2\)

∴ A = 4\(\pi\)\(\frac{(D)^2}{2}\)

= \(\frac{(D)^2}{2}\)

= \(\pi\)D²

When increased by 44% A = \(\frac{144 \pi D^2}{100}\)

\(\pi\)\(\frac{(12D)^2}{10}\) = \(\pi\)\(\frac{(6D)^2}{5}\)

Increase in diameter = \(\frac{6D}{5}\) - D = \(\frac{1}{5}\)D

Percentage increase = \(\frac{1}{5}\) x \(\frac{1}{100}\)%

= 20%

P = N800 (Principal), r = 12\(\frac{1}{2}\)% or 0.125

After sometimes, A = 1.5 x 800 = N1,200

A = P(1 + Tr) = 1200

= 800(1 + 0.125T)

= 1200

= 800(1 + 0.125T)

1200 = 800(1 + 0.125T)

0.125T = 0.5

T = \(\frac{0.5}{0.125}\)

= 4