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JEE Main 20198 AprilEvening Shift
Exam
JEE Main
Year
2019
Date
8 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Functions

JEE Main 2019 · 8 April · Evening Shift · Mathematics · Functions

Let f(x) = a ^ x ( a > 0 ) be written as f(x) = f _ 1 (x) + f _ 2...

Let f(x)=ax(a>0)f\left(x\right) = {a}^{x} ( a > 0 ) be written as f(x)=f1(x)+f2(x),f\left(x\right) = {f}_{1}\left(x\right) + {f}_{2}\left(x\right), 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}\left(x + y\right) + {f}_{1} ( x - y ) equals:

Options

  1. A2f1(x)f1(y)2{f}_{1}\left(x\right){f}_{1}\left(y\right)
  2. B2f1(x+y)f1(x−y)2{f}_{1}\left(x + y\right){f}_{1}\left(x - y\right)
  3. C2f1(x)f2(y)2{f}_{1}\left(x\right){f}_{2}\left(y\right)
  4. D2f1(x+y)f2(x−y)2{f}_{1}\left(x + y\right){f}_{2}\left(x - y\right)

Correct answer

A. 2f1(x)f1(y)2{f}_{1}\left(x\right){f}_{1}\left(y\right)

Step-by-step solution

f(x)=ax,a>0f\left(x\right) = {a}^{x}, a > 0

f(x)=ax+a−x+ax−a−x2f\left(x\right) = \dfrac{{a}^{x} + {a}^{- x} + {a}^{x} - {a}^{- x}}{2}
⇒f1(x)=ax+a−x2⇒ {f}_{1}\left(x\right) = \dfrac{{a}^{x} + {a}^{- x}}{2}
f2(x)=ax−a−x2{f}_{2}\left(x\right) = \dfrac{{a}^{x} - {a}^{- x}}{2}

⇒f1(x+y)+f1(x−y)⇒ {f}_{1}\left(x + y\right) + {f}_{1}\left(x - y\right)

=ax+y+a−x−y2+ax−y+a−x+y2= \dfrac{{a}^{x + y} + {a}^{- x - y}}{2} + \dfrac{{a}^{x - y} + {a}^{- x + y}}{2}
=(ax+a−x)2(ay+a−y)= \dfrac{\left({a}^{x} + {a}^{- x}\right)}{2}\left({a}^{y} + {a}^{- y}\right)

=f1(x)×2f1(y)= {f}_{1}\left(x\right) \times 2{f}_{1}\left(y\right)

=2f1(x)f1(y)= 2{f}_{1}\left(x\right){f}_{1}\left(y\right)

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