Question

Suppose f(x) is an even function on the symmetric interval x 6 [-A, A] and g(x) is an odd function defined on the same interval. Which of the following must be true? A/3 A/3 84(3x) + 1 dx = 2 84(3x) + 1 dx -A/3 0 f(x) is not an odd function. A/2 A/2 ✓ f(x) dx = 2 ✓ f(x) dx -A/2 A | f(x)g?(x) dx = 0 -A

Answer 1

2 Solution - There are two correct options, which are as follows: A/3 (1) All (98 (3x) + 1) dx -A/3 A/2 I f(x) dx Violes) +1) de A/2 S f(x) dx 2 -1/2 Explanation - First observe that [44] is also a symmetric interval To verify that above two options are correct, first we prove the second option i.e. [ f(x) dx I f(x) dx. -1/2 A/2 Al2 = 2 0 A/2 f(x) da f(x) dx = I, + I₂ (soy). -A/2 if(x) dx + y los da Off(x) dx 9 ft-t) (-at) { putting -1/2 Now, I -A/2 +A A/2 If that A/2 0 Since, f is functon, so fl-t=flt) o A/2 - I floo) da A12 Hence, I f(x) dx = 2 2 A/2 f(x) dx. -Al2 O How to prove the first option, it is sufficient to replace limit value appropriately and to prove that (94(3x)+1) even function. is on

Since product of two odd functions is even, therefore g4 (3x) = g(3) 9 (3x) 9 (3x) g(x).is even function. an Note : 96) = -9(x) => 943x) - - 9(390). These fore 9 (3x) is also odd function is also even and even, there fore. Net, the constant function 1 um of two even functions is 94(5x) +1 even function. IS Hence, Al A/3 5 ( 4 (3x) +1 ) dx 2 Jogʻ(30) +1) olx da. -A/ (Answon)