Let A = [– 2, 2] – {0}. Define f : A → R and g : A → R byf(x)

1.	Let A = [– 2, 2] – {0}. Define f : A → R and g : A → R byf(x) = \(\frac{1}{x^3+2x\|x\|}\) and g(x) = \(\sqrt{x^4 +\frac{3}{x^2}}\) then,(a) f and g are odd (b) f is odd, g is even (c) f is even, g is odd (d) f and g are even
Answer» Answer: (b) f is odd g is even \(f(x) = \frac{1}{x^3+2x\|x\|}\) ∴ f(-x) = \(f(x) = \frac{1}{(-x)^3+2(-x)\|-x\|}\) = \(f(x) = \frac{1}{-x^3-2x\|x\|}\) (∴ \| – x \| = \| x \|) = \(f(x) =- \frac{1}{x^3+2x\|x\|}\) = - f(x) ⇒ f is odd. g(x) = \(\sqrt{x^4+\frac{3}{x^2}}\) ∴ g(- x) = \(\sqrt{(-x)^4+\frac{3}{(-x)^2}} = \sqrt{x^4+\frac{3}{x^2}}\) = g(x) ⇒ g is even.

Let A = [– 2, 2] – {0}. Define f : A → R and g : A → R byf(x) = \(\frac{1}{x^3+2x|x|}\) and g(x) = \(\sqrt{x^4 +\frac{3}{x^2}}\) then,(a) f and g are odd (b) f is odd, g is even (c) f is even, g is odd (d) f and g are even

Answer»

Answer: (b) f is odd g is even

\(f(x) = \frac{1}{x^3+2x|x|}\)

∴ f(-x) = \(f(x) = \frac{1}{(-x)^3+2(-x)|-x|}\)

= \(f(x) = \frac{1}{-x^3-2x|x|}\) (∴ | – x | = | x |)

= \(f(x) =- \frac{1}{x^3+2x|x|}\) = - f(x) ⇒ f is odd.

g(x) = \(\sqrt{x^4+\frac{3}{x^2}}\)

∴ g(- x) = \(\sqrt{(-x)^4+\frac{3}{(-x)^2}} = \sqrt{x^4+\frac{3}{x^2}}\)

= g(x)

⇒ g is even.