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  • I have a doubt, kindly clarify. - Sets, Relations and Functions - JEE Main-3

 A  real valued function f\left ( x \right ) satisfies the functional equation  f\left ( x -y\right )=f(x)f(y)-f(a-x)f(a+y) where a  is a given constant and f(0)=1.f(2a-x)  is equal to

  • Option 1)

    f(x)\;

  • Option 2)

    -f(x)\; \;

  • Option 3)

    \; f(-x)\;

  • Option 4)

    \; f(a)+f(a-x)

 

Answers (2)

best_answer

As we learnt in

FUNCTIONS -

A relation f from a set A to a set B is said to be a function if every element of set A has one and only one image in set B.

-

 

 f(x - y) = f(x) f(y) - f(a - x) f(a + y)                        (i)

f(0) =1,   f(2a - 2) = ?

Put x = y = 0 

f(0)=f(0)\times f(0)-f(a)\times f(a)

1 = | x | - f2(a) 

f2(a) = 0    f(a) = 0 

Now  f(2a - x) = f(a + a - x) = f(a - (x - a))

Where x\rightarrow a

    y\rightarrow x-a

\therefore    f(a) f(x - a) - f(a - a) f(a + x - a) = 0 - 1 \times f(x) = - f(x)

Correct option is 2.


Option 1)

f(x)\;

This is an incorrect option.

Option 2)

-f(x)\; \;

This is the correct option.

Option 3)

\; f(-x)\;

This is an incorrect option.

Option 4)

\; f(a)+f(a-x)

This is an incorrect option.

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