Question

Given an odd function $f$ that is defined & integrable everywhere & periodic with period $2$.
Let $g\left(x\right)={\int }_0^{x}f\left(t\right)dt$

If $f^{\prime }(-2)=-2$, then $f^{\prime }\left(2\right)$ equals?

• A. $0$
• B. $1$
• C. $2$
• D. $-2$

Answer 1

Answer: (D)

$-2$

To find $f^{\prime }\left(2\right)$ if $f^{\prime }(-2)=-2$

Using Leinuitz Rule

$g^{\prime }\left(x\right)=f\left(x\right)$

Since $f$ is a periodic function with period$2$

$\Rightarrow f(x+2k)=f\left(x\right);k\in I$

Also, f is an odd function

$f(-x)=-f\left(x\right)f^{\prime }(-x)\times (-1)=-f^{\prime }\left(x\right)f^{\prime }\left(x\right)=f^{\prime }(-x)$

Take $x=-2$

$f^{\prime }(-2)=f^{\prime }\left(2\right)$

Since $f^{\prime }(-2)=-2$

$f^{\prime }\left(2\right)=-2$

Hence the correct answer is $f^{\prime }\left(2\right)=-2$