(a) Two functions, f and g, are defined by \(f : x \to 2x^{2} -
(a) Two functions, f and g, are defined by \(f : x \to 2x^{2} - 1\) and \(g : x \to 3x + 2\) where x is a real number.
(i) If \(f(x - 1) - 7 = 0\), find the values of x.
(ii) Evaluate : \(\frac{f(-\frac{1}{2}) . g(3)}{f(4) - g(5)}\).
(b) An operation, \((\ast)\) is defined on the set R, of real numbers, by \(m \ast n = \frac{-n}{m^{2} + 1}\), where \(m, n \in R\). If \(-3, -10 \in R\), show whether or not \(\ast\) is commutative.
Explanation
(a) \(f : x \to 2x^{2} - 1 ; g : x \to 3x + 2\)
(i) \(f(x - 1) - 7 = 0\)
\(f(x - 1) = 2(x - 1)^{2} - 1 = 2(x^{2} - 2x + 1) - 1\)
= \(2x^{2} - 4x + 2 - 1\)
\(f(x - 1) - 7 = 2x^{2} - 4x + 1 - 7 = 2x^{2} - 4x - 6 = 0\)
\(2x^{2} - 6x + 2x - 6 = 0 \implies 2x(x - 3) + 2(x - 3) = 0\)
\((2x + 2)(x - 3) = 0 \implies 2x = -2; x = 3\)
\(x = -1 ; 3\)
(ii) \(\frac{f(-\frac{1}{2}) . g(3)}{f(4) - g(5)}\)
\(f(-\frac{1}{2}) = 2(-\frac{1}{2})^{2} - 1 = \frac{1}{2} - 1 = -\frac{1}{2}\)
\(g(3) = 3(3) + 2 = 9 + 2 = 11\)
\(f(4) = 2(4^{2}) - 1 = 32 - 1 = 31\)
\(g(5) = 3(5) + 2 = 15 + 2 = 17\)
\(\frac{f(-\frac{1}{2}) . g(3)}{f(4) - g(5)} = \frac{(-\frac{1}{2}) . (11)}{31 - 17}\)
= \(\frac{-\frac{11}{2}}{14}\)
= \(-\frac{11}{28}\)
(b) \(m \ast n = \frac{-n}{m^{2} + 1}; m, n \in R\)
\(-3 \ast -10 = \frac{-(-10)}{(-3)^{2} + 1}\)
= \(\frac{10}{10} = 1\)
\(-10 \ast -3 = \frac{-(-3)}{(-10)^{2} + 1}\)
= \(\frac{3}{101}\)
Hence, \(\ast\) is not commutative as \(m \ast n \neq n \ast m\).