Let f (x) = a^x (a 0) be written as f( x | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

${f_1}\left( x \right) = \frac{{{a^x} + {a^{ - x}}}}{2}$ and ${f_2}\left( x \right) = \frac{{{a^x} - {a^{ - x}}}}{2}$

${f_1}\left( {x + y} \ri

Vedclass · Accepted Answer

$2{f_1}\left( x \right){f_1}\left( y \right)$

Vedclass · Answer

${f_1}\left( x \right) = \frac{{{a^x} + {a^{ - x}}}}{2}$ and ${f_2}\left( x \right) = \frac{{{a^x} - {a^{ - x}}}}{2}$

${f_1}\left( {x + y} \right) + {f_1}\left( {x - y} \right)$

$ = \frac{1}{2}\left( {{a^{x + y}} + {a^{ - x - y}} + {a^{x - y}} + {a^{ - x + y}}} \right)$

$ = \frac{1}{2}\left( {{a^x}\left( {{a^y} + {a^{ - y}}} \right) + {a^{ - x}}\left( {{a^y} + {a^{ - y}}} \right)} \right)$

$ = 2.\left( {\frac{{{a^x} + {a^{ - x}}}}{2}} \right)\left( {\frac{{{a^y} + {a^{ - y}}}}{2}} \right)$

$ = 2{f_1}\left( x \right){f_1}\left( y \right)$

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 and $f_2( x)$ is an odd function. Then $f_1( x + y) + f_1( x - y )$ equals

Similar Questions

If $x = {\log _2}\left( {\sqrt {56 + \sqrt {56 + \sqrt {56 + .... + \infty } } } } \right)$ then

If $f(x)$ be a polynomial function satisfying $f(x).f (\frac{1}{x}) = f(x) + f (\frac{1}{x})$ and $f(4) = 65$ then value of $f(6)$ is

If the domain of the function $f(x)=\log _e$ $\left(\frac{2 x+3}{4 x^2+x-3}\right)+\cos ^{-1}\left(\frac{2 x-1}{x+2}\right)$ is $(\alpha, \beta]$, then the value of $5 \beta-4 \alpha$ is equal to

The period of the function $f(x) = \log \cos 2x + \sin 4x$ is :-

Consider the identity function $I _{ N }: N \rightarrow N$ defined as $I _{ N }$ $(x)=x$ $\forall $ $x \in N$ Show that although $I _{ N }$ is onto but $I _{ N }+ I _{ N }:$ $ N \rightarrow N$ defined as $\left(I_{N}+I_{N}\right)(x)=$ $I_{N}(x)+I_{N}(x)$ $=x+x=2 x$ is not onto.