Maths-
General
Easy

Question

Let f : R →R and g : R →R be continuous functions, then the value of integral subscript negative pi divided by 2 end subscript superscript pi text /2 end text end superscript (f(x) + f(-x)) (g(x) – g(-x)) dx, is equal to

  1. – 1    
  2. 0    
  3. 1    
  4. None of these    

The correct answer is: 0


    f space colon space R space rightwards arrow R space a n d space g space colon space R space rightwards arrow R space a r e space c o n t i n u o u s space f u n c t i o n s.
I equals integral subscript negative pi divided by 2 end subscript superscript pi text /2 end text end superscript left parenthesis f left parenthesis x right parenthesis space plus space f left parenthesis negative x right parenthesis right parenthesis space left parenthesis g left parenthesis x right parenthesis space – space g left parenthesis negative x right parenthesis right parenthesis space d x space space space space......... open parentheses 1 close parentheses
N o w comma space i f space w e space t a k e space x rightwards arrow negative x comma
rightwards double arrow integral subscript negative a end subscript superscript a f open parentheses negative x close parentheses d x equals integral subscript negative a end subscript superscript a minus f open parentheses x close parentheses d x
T h e r e f o r e comma
w h e n space x equals straight pi over 2 comma space t h e n space minus x equals fraction numerator negative straight pi over denominator 2 end fraction
w h e n space x equals fraction numerator negative straight pi over denominator 2 end fraction comma space t h e n space minus x equals straight pi over 2

F o r
I equals integral subscript negative pi divided by 2 end subscript superscript pi text /2 end text end superscript left parenthesis f left parenthesis x right parenthesis space plus space f left parenthesis negative x right parenthesis right parenthesis space left parenthesis g left parenthesis x right parenthesis space – space g left parenthesis negative x right parenthesis right parenthesis space d x
I equals integral subscript negative pi divided by 2 end subscript superscript pi text /2 end text end superscript left parenthesis f left parenthesis negative x right parenthesis space plus space f left parenthesis x right parenthesis right parenthesis space left parenthesis g left parenthesis negative x right parenthesis space – space g left parenthesis x right parenthesis right parenthesis space d x
I equals negative integral subscript negative pi divided by 2 end subscript superscript pi text /2 end text end superscript left parenthesis f left parenthesis x right parenthesis space plus space f left parenthesis negative x right parenthesis right parenthesis space left parenthesis g left parenthesis x right parenthesis space – space g left parenthesis negative x right parenthesis right parenthesis space d x space space space space space space space........ open parentheses 2 close parentheses

A d d i n g space u p space e q u a t i o n space open parentheses 1 close parentheses space a n d space open parentheses 2 close parentheses comma
2 I equals integral subscript negative pi divided by 2 end subscript superscript pi text /2 end text end superscript left parenthesis f left parenthesis x right parenthesis space plus space f left parenthesis negative x right parenthesis right parenthesis space left parenthesis g left parenthesis x right parenthesis space – space g left parenthesis negative x right parenthesis right parenthesis space d x space minus integral subscript negative pi divided by 2 end subscript superscript pi text /2 end text end superscript left parenthesis f left parenthesis x right parenthesis space plus space f left parenthesis negative x right parenthesis right parenthesis space left parenthesis g left parenthesis x right parenthesis space – space g left parenthesis negative x right parenthesis right parenthesis space d x
2 I equals 0
therefore I equals 0

    A function f is odd if f(x)=f(x)f(x)=f(x)<br> </br>

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