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Functions question

2004 · Shift 0 · Q67
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Functions question

2004 · Shift 0 · Q67

JEE MainMathematicsFunctionsMCQ+4 / −1
The graph of the function y = f(x) is symmetrical about the line x = 2, then
  1. A
    f(x)=−f(−x)f\left( x \right) = - f\left( { - x} \right)f(x)=−f(−x)
  2. B
    f(2+x)=f(2−x)f\left( {2 + x} \right) = f\left( {2 - x} \right)f(2+x)=f(2−x)
  3. C
    f(x)=f(−x)f\left( x \right) = f\left( { - x} \right)f(x)=f(−x)
  4. D
    f(x+2)=f(x−2)f\left( {x + 2} \right) = f\left( {x - 2} \right)f(x+2)=f(x−2)
View written solutionFree

Correct answer: B

  1. Meaning of symmetry about the line x=2x=2x=2

If the graph of y=f(x)y=f(x)y=f(x) is symmetric about the vertical line x=2x=2x=2, then points at equal horizontal distances from x=2x=2x=2 have the same yyy-value.

So, for any real number xxx, f(2+x)=f(2−x).f(2+x)=f(2-x).f(2+x)=f(2−x).

This is the standard condition for symmetry about the line x=ax=ax=a: f(a+h)=f(a−h).f(a+h)=f(a-h).f(a+h)=f(a−h). Here a=2a=2a=2.

  1. Check the options
  • Option A: f(x)=−f(−x)f(x)=-f(-x)f(x)=−f(−x)

    This represents an odd-function type symmetry about the origin, not symmetry about x=2x=2x=2.

    So, A is incorrect.

  • Option B: f(2+x)=f(2−x)f(2+x)=f(2-x)f(2+x)=f(2−x)

    This exactly matches the condition for symmetry about the line x=2x=2x=2.

    So, B is correct.

  • Option C: f(x)=f(−x)f(x)=f(-x)f(x)=f(−x)

    This represents an even-function type symmetry about the yyy-axis, i.e. symmetry about x=0x=0x=0, not x=2x=2x=2.

    So, C is incorrect.

  • Option D: f(x+2)=f(x−2)f(x+2)=f(x-2)f(x+2)=f(x−2)

    This does not represent reflection symmetry about x=2x=2x=2. It is a different relation and is not the required symmetry condition.

    So, D is incorrect.

  1. Final answer

Therefore, the correct option is B.\boxed{B}.B​.

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