Mathematics Questions and Answers

a2 - b2 = (\(\frac{2x}{1 - x}\))2 - (\(\frac{1 + x}{1 - x}\))2

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

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

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

< B is the largest since the side facing it is the largest, i.e. (x + 4)cm

Cosine B = \(\frac{1}{5}\)

= 0.2 given

b² - a² + c² - 2a Cos B

Cos B = \(\frac{a^2 + c^2 - b^2}{2ac}\)

\(\frac{1}{5}\) = \(\frac{x^2 + ?(x - 4)^2 - (x + 4)^2}{2x (x - 4)}\)

\(\frac{1}{5}\)= \(\frac{x(x - 16)}{2x(x - 4)}\)

\(\frac{1}{5}\) = \(\frac{x - 16}{2x - 8}\)

= 5(x - 16)

= 2x - 8

3x = 72

x = \(\frac{72}{3}\)

= 24

If diameter of the circle = 12cm; radius of the circle(L) = \(\frac{12}{2}\)

= 6cm

\(\frac{\theta}{360}\) = \(\frac{r}{L}\) where \(\theta\) = 210\(\theta\), L = 6cm

\(\frac{210}{360}\) = \(\frac{r}{6}\)

where r = radius of the base of the cone

V = \(\frac{1260}{360}\)

= 3.50cm

x³ + px² + qx + 1 = (x - 1) Q(x) + R

x - 2 = 0, x = 2, R = 0,

4p + 2p = -9........(i)

x³ + px² + qx + 1 = (x - 1)Q(x) + R

-1 + p - q + 1 = 0

p - q = 0.......(ii)

Solve the equation simultaneously

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

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

p + q = \(\frac{3}{2}\) - \(\frac{3}{2}\)

= \(\frac{-6}{2}\)

= -3

4a2 - 9b2, 8a3 + 27b3, (4a + 6b)2 = (2a + 3b)(2a - 3b)

8a3 + 27b3 = (2a)3 + (3b)3

= (2a + 3b)(4a - 6ab = 9a2)

(4a + 6b)2 = 2(2a + 3b)2