1. Tardigrade
  2. Question
  3. Mathematics
  4. Statement-1 :The function f( x )=∫ limits0 x √1+ t 2 dt is an odd function and g ( x )= f prime( x ) is an even function Statement-2: For a differentiable function f (x) if f ' (x) is an even function then f (x) is an odd function

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Q. Statement-1 :The function $f( x )=\int\limits_0^{ x } \sqrt{1+ t ^2} dt$ is an odd function and $g ( x )= f ^{\prime}( x )$ is an even function
Statement-2: For a differentiable function $f (x)$ if $f ' (x)$ is an even function then f (x) is an odd function

Integrals

Solution:

image
$\text { for S-1: } f(x)=\int\limits_0^x \sqrt{1+t^2} d t ; g(x)=\sqrt{1+x^2} $
$f(-x)=\int\limits_0^{-x} \sqrt{1+t^2} d t ; t=-y $
$f(-x)=-\int\limits_0^x \sqrt{1+y^2} d y $
$\therefore f ( x )+ f (- x )=0 \Rightarrow f \text { is odd and } g \text { is obviously even. }$