JEE Main 2019 · 8 April · Evening Shift · Mathematics · Functions
A. 2f1(x)f1(y)2{f}_{1}\left(x\right){f}_{1}\left(y\right)2f1(x)f1(y)
f(x)=ax,a>0f\left(x\right) = {a}^{x}, a > 0f(x)=ax,a>0
f(x)=ax+a−x+ax−a−x2f\left(x\right) = \dfrac{{a}^{x} + {a}^{- x} + {a}^{x} - {a}^{- x}}{2}f(x)=2ax+a−x+ax−a−x⇒f1(x)=ax+a−x2⇒ {f}_{1}\left(x\right) = \dfrac{{a}^{x} + {a}^{- x}}{2}⇒f1(x)=2ax+a−xf2(x)=ax−a−x2{f}_{2}\left(x\right) = \dfrac{{a}^{x} - {a}^{- x}}{2}f2(x)=2ax−a−x
⇒f1(x+y)+f1(x−y)⇒ {f}_{1}\left(x + y\right) + {f}_{1}\left(x - y\right)⇒f1(x+y)+f1(x−y)
=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}=2ax+y+a−x−y+2ax−y+a−x+y=(ax+a−x)2(ay+a−y)= \dfrac{\left({a}^{x} + {a}^{- x}\right)}{2}\left({a}^{y} + {a}^{- y}\right)=2(ax+a−x)(ay+a−y)
=f1(x)×2f1(y)= {f}_{1}\left(x\right) \times 2{f}_{1}\left(y\right)=f1(x)×2f1(y)
=2f1(x)f1(y)= 2{f}_{1}\left(x\right){f}_{1}\left(y\right)=2f1(x)f1(y)