Let F(x) = \(\big[\frac{g(x)-g(-x)}{f(x)+f(-x)}\big]^m\) such that m

1.	Let F(x) = \(\big[\frac{g(x)-g(-x)}{f(x)+f(-x)}\big]^m\) such that m = 2n, n ∈ N and f (– x) ≠ – f (x). Then, F (x) is(a) an odd function (b) an even function (c) constant function (d) None of these
Answer» Answer : (b) an even function F(x) = \(\big[\frac{g(x)-g(-x)}{f(x)+f(-x)}\big]^m\) ∴ F(-x) = \(\big[\frac{g(-x)-g(x)}{f(-x)+f(x)}\big]^m =\big[\frac{-g(x)-g(-x)}{f(-x)+f(x)}\big]^m\) = \(\big[-\big(\frac{g(x)-g(-x)}{f(-x)+f(x)}\big)\big]^{2n}\) = \(\big[\frac{g(x)-g(-x)}{f(-x)+f(x)}\big]\) (∴ 2n ⇒ even power) = F (x) ⇒ F is an even function.

Let F(x) = \(\big[\frac{g(x)-g(-x)}{f(x)+f(-x)}\big]^m\) such that m = 2n, n ∈ N and f (– x) ≠ – f (x). Then, F (x) is(a) an odd function (b) an even function (c) constant function (d) None of these

Answer»

Answer : (b) an even function

F(x) = \(\big[\frac{g(x)-g(-x)}{f(x)+f(-x)}\big]^m\)

∴ F(-x) = \(\big[\frac{g(-x)-g(x)}{f(-x)+f(x)}\big]^m =\big[\frac{-g(x)-g(-x)}{f(-x)+f(x)}\big]^m\)

= \(\big[-\big(\frac{g(x)-g(-x)}{f(-x)+f(x)}\big)\big]^{2n}\)

= \(\big[\frac{g(x)-g(-x)}{f(-x)+f(x)}\big]\) (∴ 2n ⇒ even power)

= F (x)

⇒ F is an even function.