(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).