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Functions question

2005 · Shift 0 · Q61
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Functions question

2005 · Shift 0 · Q61

JEE MainMathematicsFunctionsMCQ+4 / −1
A real valued function f(x) satisfies the functional equation f(x - y) = f(x)f(y) - f(a - x)f(a + y) where a is given constant and f(0) = 1, f(2a - x) is equal to
  1. A
    - f(x)
  2. B
    f(x)
  3. C
    f(a) + f(a - x)
  4. D
    f(- x)
View written solutionFree

Correct answer: A

  1. We are given f(x−y)=f(x)f(y)−f(a−x)f(a+y)f(x-y)=f(x)f(y)-f(a-x)f(a+y)f(x−y)=f(x)f(y)−f(a−x)f(a+y) for all real x,yx,yx,y, with f(0)=1.f(0)=1.f(0)=1. We need to find f(2a−x)f(2a-x)f(2a−x).

  2. Use the functional equation at convenient values.

Take x=0,y=0x=0, y=0x=0,y=0: f(0)=f(0)f(0)−f(a)f(a).f(0)=f(0)f(0)-f(a)f(a).f(0)=f(0)f(0)−f(a)f(a). Since f(0)=1f(0)=1f(0)=1, 1=1−f(a)2 ⇒ f(a)2=0⇒f(a)=0.1=1-f(a)^2 \,\Rightarrow\, f(a)^2=0 \Rightarrow f(a)=0.1=1−f(a)2⇒f(a)2=0⇒f(a)=0.

  1. Now put y=xy=xy=x in the functional equation: f(0)=f(x)f(x)−f(a−x)f(a+x).f(0)=f(x)f(x)-f(a-x)f(a+x).f(0)=f(x)f(x)−f(a−x)f(a+x). Using f(0)=1f(0)=1f(0)=1, 1=f(x)^2-f(a-x)f(a+x). \tag{1}

  2. Put x=ax=ax=a in the functional equation: f(a−y)=f(a)f(y)−f(0)f(a+y).f(a-y)=f(a)f(y)-f(0)f(a+y).f(a−y)=f(a)f(y)−f(0)f(a+y). Since f(a)=0f(a)=0f(a)=0 and f(0)=1f(0)=1f(0)=1, f(a−y)=−f(a+y).f(a-y)=-f(a+y).f(a−y)=−f(a+y). Now replace yyy by x−ax-ax−a so that a+y=xa+y=xa+y=x: f(2a−x)=−f(x).f(2a-x)=-f(x).f(2a−x)=−f(x). Thus, f(2a−x)=−f(x).\boxed{f(2a-x)=-f(x)}.f(2a−x)=−f(x)​.

  3. Check options:

  • A: −f(x)-f(x)−f(x) ✓
  • B: f(x)f(x)f(x) ✗
  • C: f(a)+f(a−x)f(a)+f(a-x)f(a)+f(a−x); since f(a)=0f(a)=0f(a)=0, this becomes f(a−x)f(a-x)f(a−x), not generally equal ✗
  • D: f(−x)f(-x)f(−x) not implied ✗

Hence the correct option is A.

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