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Functions question

2019 · 8 Apr · Shift 2 · Q38
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Functions question

2019 · 8 Apr · Shift 2 · Q38

JEE MainMathematicsFunctionsMCQ+4 / −1
Let ƒ(x) = ax (a > 0) be written as ƒ(x) = ƒ1 (x) + ƒ2 (x), where ƒ1 (x) is an even function of ƒ2 (x) is an odd function. Then ƒ1 (x + y) + ƒ1 (x – y) equals
  1. A
    2ƒ1 (x)ƒ1 (y)
  2. B
    2ƒ1 (x + y)ƒ1 (x – y)
  3. C
    2ƒ1 (x)ƒ2 (y)
  4. D
    2ƒ1 (x + y)ƒ2 (x – y)
View written solutionFree

Correct answer: A

  1. We are given f(x)=ax(a>0)f(x)=a^x \quad (a>0)f(x)=ax(a>0) and it is decomposed as f(x)=f1(x)+f2(x),f(x)=f_1(x)+f_2(x),f(x)=f1​(x)+f2​(x), where f1f_1f1​ is even and f2f_2f2​ is odd.

  2. For any function f(x)f(x)f(x), its even and odd parts are: f1(x)=f(x)+f(−x)2,f2(x)=f(x)−f(−x)2.f_1(x)=\frac{f(x)+f(-x)}{2}, \qquad f_2(x)=\frac{f(x)-f(-x)}{2}.f1​(x)=2f(x)+f(−x)​,f2​(x)=2f(x)−f(−x)​.

Since f(x)=axf(x)=a^xf(x)=ax, we have f(−x)=a−x.f(-x)=a^{-x}.f(−x)=a−x. Therefore, f1(x)=ax+a−x2,f_1(x)=\frac{a^x+a^{-x}}{2},f1​(x)=2ax+a−x​, f2(x)=ax−a−x2.f_2(x)=\frac{a^x-a^{-x}}{2}.f2​(x)=2ax−a−x​.

  1. Now compute f1(x+y)+f1(x−y)f_1(x+y)+f_1(x-y)f1​(x+y)+f1​(x−y): f1(x+y)=ax+y+a−(x+y)2,f_1(x+y)=\frac{a^{x+y}+a^{-(x+y)}}{2},f1​(x+y)=2ax+y+a−(x+y)​, f1(x−y)=ax−y+a−(x−y)2.f_1(x-y)=\frac{a^{x-y}+a^{-(x-y)}}{2}.f1​(x−y)=2ax−y+a−(x−y)​. So, f1(x+y)+f1(x−y)f_1(x+y)+f_1(x-y)f1​(x+y)+f1​(x−y) =12(ax+y+a−x−y+ax−y+a−x+y).=\frac{1}{2}\left(a^{x+y}+a^{-x-y}+a^{x-y}+a^{-x+y}\right).=21​(ax+y+a−x−y+ax−y+a−x+y).

  2. Rearrange terms: =12[ax(ay+a−y)+a−x(a−y+ay)]=\frac{1}{2}\left[a^x(a^y+a^{-y})+a^{-x}(a^{-y}+a^y)\right]=21​[ax(ay+a−y)+a−x(a−y+ay)] =12(ax+a−x)(ay+a−y).=\frac{1}{2}(a^x+a^{-x})(a^y+a^{-y}).=21​(ax+a−x)(ay+a−y).

  3. Now use f1(x)=ax+a−x2,f1(y)=ay+a−y2.f_1(x)=\frac{a^x+a^{-x}}{2}, \qquad f_1(y)=\frac{a^y+a^{-y}}{2}.f1​(x)=2ax+a−x​,f1​(y)=2ay+a−y​. Thus, (ax+a−x)(ay+a−y)=4f1(x)f1(y).(a^x+a^{-x})(a^y+a^{-y})=4f_1(x)f_1(y).(ax+a−x)(ay+a−y)=4f1​(x)f1​(y). Therefore, f1(x+y)+f1(x−y)=12⋅4f1(x)f1(y)=2f1(x)f1(y).f_1(x+y)+f_1(x-y)=\frac{1}{2}\cdot 4f_1(x)f_1(y)=2f_1(x)f_1(y).f1​(x+y)+f1​(x−y)=21​⋅4f1​(x)f1​(y)=2f1​(x)f1​(y).

  4. Hence the correct option is A: 2f1(x)f1(y).\boxed{A:~2f_1(x)f_1(y)}.A: 2f1​(x)f1​(y)​.

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