Mathematics · 2019
JEE Main · 8 April 2019, Shift 2 · Q61
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:
Let $\displaystyle f(x)=\mathrm{a}^x(\mathrm{a}>0)$ be written as $\displaystyle f(x)=f_1(x)+f_2(x)$, where $\displaystyle f_1(x)$ is an even function and $\displaystyle f_2(x)$ is an odd function. Then $\displaystyle f_1(x+y)+f_1(x-y)$ equals :
Official answer
(1)
$\displaystyle 2 f_1(x) f_1(y)$
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JEE Main 2019 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.