Mathematics Questions and Answers

The identity element (e) under an operation, say \(\otimes\), is the element such that for any given element under the operation, say a,

\(a \otimes e = e \otimes a = a\)

From the table, q is the identity element.

\(p \otimes q = q \otimes p = p\)

Same as all through.

Let the G.p be a, ar, ar², S³ = \(\frac{1}{2}\)S

a + ar + ar² = \(\frac{1}{2}\)(\(\frac{a}{1 - r}\))

2(1 + r + r)(r - 1) = 1

= 2r³ = 3

= r³ = \(\frac{3}{2}\)

r(\(\frac{3}{2}\))^{\(\frac{1}{3}\)} = \(\sqrt{\frac{3}{2}}\)

2p - 10 = \(\frac{p + 1 + 1 - 4P^2}{2}\) (Arithmetic mean)

= 2(2p - 100 = p + 2 - 4P²)

= 4p - 20 = p + 2 - 4p²

= 4p² + 3p - 22 = 0

= (p - 2)(4p + 11) = 0

∴ p = 2 or -\(\frac{4}{11}\)

\(\frac{1}{3x}\) + \(\frac{1}{2}\) > \(\frac{1}{4x}\)

= \(\frac{2 + 3x}{6x}\) > \(\frac{1}{4x}\)

= 4(2 + 3x) > 6x = 12x² - 2x = 0

= 2x(6x - 1) > 0 = x(6x - 1) > 0

Case 1 (-, -) = x < 0, 6x -1 < 0

= x < 0, x < \(\frac{1}{6}\) = x < \(\frac{1}{6}\) (solution)

Case 2 (+, +) = x > 0, 6x -1 > 0 = x > 0, x > \(\frac{1}{6}\)

Combining solutions in cases(1) and (2)

= x > 0, x < \(\frac{1}{6}\) = 0 < x < \(\frac{1}{6}\)

\(\frac{11x + 2}{6x^2 - x - 1}\) = \(\frac{11x + 2}{(3x + 1)(2x - 1)}\)

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

11x + 2 = A(2x - 1) + B(3x + 1)

put x = \(\frac{1}{2}\)

\(\frac{15}{2} = \frac{5}{2}B\)

B = 3.

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

\(-\frac{5}{3} = \frac{-5}{3}\)A \(\implies\) A = 1

∴ \(\frac{11x +2}{6x^2 - x - 1}\) = \(\frac{1}{3x + 1}\) + \(\frac{3}{2x - 1}\)