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JEE Main 2019 · 8 Apr · Shift II

JEE Main 2019 Mathematics question on Sets Relations and Functions

Let f(x)=axf(x) = a^x (a>0)(a > 0) be written as f(x)=f1(x)+f2(x)f(x) = f_1(x) + f_2(x), where f1(x)f_1(x) is an even function and f2(x)f_2(x) is an odd function. Then f1(x+y)+f1(x−y)f_1(x + y) + f_1(x - y) equals
  1. A.
    2f1(x+y)⋅f2(x−y)2f_1(x+y) \cdot f_2(x-y)
  2. B.
    2f1(x+y)⋅f1(x−y)2f_1(x+y) \cdot f_1(x-y)
  3. C.
    2f1(x)⋅f2(y)2f_1(x) \cdot f_2(y)
  4. D.
    2f1(x)⋅f1(y)2f_1(x) \cdot f_1(y)
Show answer and solution

Correct option: D

Solution

f₁(x) = (aˣ + a⁻ˣ)/2 (even part) and f₂(x) = (aˣ - a⁻ˣ)/2 (odd part).

f₁(x+y) + f₁(x-y) = (a^(x+y) + a-(x+y))/2 + (a^(x-y) + a-(x-y))/2

= (aˣaʸ + a⁻ˣa⁻ʸ + aˣa⁻ʸ + a⁻ˣaʸ)/2 = (aˣ + a⁻ˣ)(aʸ + a⁻ʸ)/2

= 2 · (aˣ + a⁻ˣ)/2 · (aʸ + a⁻ʸ)/2 = 2f₁(x)·f₁(y).

[JEE Main 2019 · 8 Apr · Shift II]

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