Question

Statement-1 :The function $f(x)=\underset{0}{\overset{x}{\int }}\sqrt{1+t^2}dt$ is an odd function and $g(x)=f^{\prime }(x)$ is an even function
Statement-2: For a differentiable function $f(x)$ if $f^{\prime }(x)$ is an even function then f (x) is an odd function

• A. Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.
• B. Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.
• C. Statement-1 is true, statement-2 is false.
• D. Statement-1 is false, statement-2 is true.

Answer 1

Answer: (C)

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