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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:

JEE Main 2019 — Mathematics Algebra

2019mcqmedium

Let f(x)=ax(a>0)f(x)={a}^{x} (a>0)f(x)=ax(a>0) be written as f(x)=f1(x)+f2(x),f(x)={f}_{1}(x)+{f}_{2}(x),f(x)=f1​(x)+f2​(x), where f1(x){f}_{1}(x)f1​(x) is an even function and f2(x){f}_{2}(x)f2​(x) is an odd function. Then f1(x+y)+f1(x−y){f}_{1}(x+y)+{f}_{1}(x-y)f1​(x+y)+f1​(x−y) equals:

Official previous-year question

Held on 8 Apr 2019 · Verified 6 Jul 2026.

Options

  1. A

    2f1(x)f1(y)2{f}_{1}(x){f}_{1}(y)2f1​(x)f1​(y)

  2. B

    2f1(x+y)f1(x−y)2{f}_{1}(x+y){f}_{1}(x-y)2f1​(x+y)f1​(x−y)

  3. C

    2f1(x)f2(y)2{f}_{1}(x){f}_{2}(y)2f1​(x)f2​(y)

  4. D

    2f1(x+y)f2(x−y)2{f}_{1}(x+y){f}_{2}(x-y)2f1​(x+y)f2​(x−y)

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