Verified Solution • 2025–26 Session moderate

QJEEM19Maeaf4: JEE Main 2019 Mathematics - Class 11 JEEM19Maeaf4 Solution: Let ƒ(x) = a x (a > 0) be written as ƒ(x) = ƒ 1 (x) + ƒ 2 (x),…

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

Multiple Choice Options

  • A 2ƒ1 (x)ƒ1 (y)
  • B 2ƒ1 (x + y)ƒ1 (x – y)
  • C 2ƒ1 (x)ƒ2 (y)
  • D 2ƒ1 (x + y)ƒ2 (x – y)

Step-by-Step Verified Solution

f(x) = ax

As f1(x) is even function then

f1(x) = f(x)+f(−x)2{{{f\left( x \right) + f\left( { - x} \right)} \over 2}}

= ax+a−x2{{{a^x} + {a^{ - x}}} \over 2}

As f2(x) is odd function then

f2(x) = f(x)−f(−x)2{{{f\left( x \right) - f\left( { - x} \right)} \over 2}}

= ax−a−x2{{{a^x} - {a^{ - x}}} \over 2}

Now,

ƒ1 (x + y) + ƒ1 (x – y)

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

Also ƒ1 (x)ƒ1 (y) = (ax+a−x2)(ay+a−y2)\left( {{{{a^x} + {a^{ - x}}} \over 2}} \right)\left( {{{{a^y} + {a^{ - y}}} \over 2}} \right)

= ax+y+ax−y+a−x+y+a−x−y4{{{a^{x + y}} + {a^{x - y}} + {a^{ - x + y}} + {a^{ - x - y}}} \over 4}

= f1(x+y)+f2(x−y)2{{{{f_1}\left( {x + y} \right) + {f_2}\left( {x - y} \right)} \over 2}}

∴\therefore ƒ1 (x + y) + ƒ1 (x – y) = 2ƒ1 (x)ƒ1 (y)

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