JEE Main · 2019 · 08 Apr 2019 · Shift 2 · Sets Relations and Functions

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

Answer: (B)

Solution

Split

f (x) = a^{x}

into its even and odd parts,

f_{1} (x) = \frac{a ^{x} + a ^{- x}}{2}

and

f_{2} (x) = \frac{a ^{x} - a ^{- x}}{2}

. Then

f_{1} (x + y) + f_{1} (x - y) = \frac{1}{2} (a^{x + y} + a^{- (x + y)} + a^{x - y} + a^{- (x - y)})

. Grouping, this is

\frac{1}{2} [a^{x} (a^{y} + a^{- y}) + a^{- x} (a^{y} + a^{- y})] = \frac{1}{2} (a^{x} + a^{- x}) (a^{y} + a^{- y})

, which equals

\frac{a ^{x} + a ^{- x}}{2} \cdot \frac{a ^{y} + a ^{- y}}{2} \cdot 2 = 2 f_{1} (x) f_{1} (y)

JEE Main · 2019 · 08 Apr 2019 · Shift 2 · Single correct option