Let f(x ) = ax (a 0) be written as f( x) = f1(x ) + f2(x) , where f1( x), is an even function of f2(x) is an odd function. Then f1(x + y) + f1(x - y) equals

Question

Let f(x ) = ax (a > 0) be written as f( x) = f1(x ) + f2(x) , where f1( x), is an even function of f2(x) is an odd function. Then f1(x + y) + f1(x - y) equals (A) 2 f1 (x) f1 (y) (B) 2 f1 (x) f2 (y) (C) 2 f1 (x + y) f2 (x - y ) (D) 2 f1 (x + y) f1 (x - y )

Tardigrade Admin · Accepted Answer

f(x) =ax, a >0 f(x) = (ax +a-x+ax -a-x/2) ⇒ f1 (x)=(ax+a-x/2) f2(x)= (ax-a-x/2) ⇒ f1(x+y) + f1(x-y) = (ax+y+ a-x-y/2) + (ax-y+a-x+y/2) = ((ax+a-x)/2) (ay +a-y) =f1(x) ×2f1(y) = 2f1(x)f1(y)

Q. Let $f (x) = a^{x} (a > 0)$ be written as $f (x) = f_{1} (x) + f_{2} (x),$ where $f_{1} (x),$ is an even function of $f_{2} (x)$ is an odd function. Then $f_{1} (x + y) + f_{1} (x - y)$ equals

$2 f_{1} (x) f_{1} (y)$

$2 f_{1} (x) f_{2} (y)$

$2 f_{1} (x + y) f_{2} (x - y)$

$2 f_{1} (x + y) f_{1} (x - y)$

Q. Let f(x)=ax(a>0) be written as f(x)=f1​(x)+f2​(x), where f1​(x), is an even function of f2​(x) is an odd function. Then f1​(x+y)+f1​(x−y) equals

2f1​(x)f1​(y)

2f1​(x)f2​(y)

2f1​(x+y)f2​(x−y)

2f1​(x+y)f1​(x−y)

f(x)=ax,a>0 f(x)=2ax+a−x+ax−a−x​ ⇒f1​(x)=2ax+a−x​ f2​(x)=2ax−a−x​ ⇒f1​(x+y)+f1​(x−y) =2ax+y+a−x−y​+2ax−y+a−x+y​ =2(ax+a−x)​(ay+a−y) =f1​(x)×2f1​(y) =2f1​(x)f1​(y)

Q. Let $f (x) = a^{x} (a > 0)$ be written as $f (x) = f_{1} (x) + f_{2} (x),$ where $f_{1} (x),$ is an even function of $f_{2} (x)$ is an odd function. Then $f_{1} (x + y) + f_{1} (x - y)$ equals

$2 f_{1} (x) f_{1} (y)$

$2 f_{1} (x) f_{2} (y)$

$2 f_{1} (x + y) f_{2} (x - y)$

$2 f_{1} (x + y) f_{1} (x - y)$