Question

Let $f\left(x\right)$ and $\varphi \left(x\right)$ are two continuous functions on $R$ satisfying $\varphi \left(x\right)=\underset{a}{\overset{x}{\int }}f\left(t\right)dt,a\ne 0$ and another continuous function $g\left(x\right)$ satisfying $g(x+\alpha )+g\left(x\right)=0\forall x\in R,\alpha >0$, and $\underset{b}{\overset{2k}{\int }}g\left(t\right)dt$ is independent of $b$.

If $f\left(x\right)$ is an odd function, then

• A. $\varphi \left(x\right)$ is also an odd function.
• B. $\varphi \left(x\right)$ is an even function
• C. $\varphi \left(x\right)$ is neither an even nor an odd function
• D. for $\varphi \left(x\right)$ to be an even function, it must satisfy $\underset{0}{\overset{a}{\int }}f\left(x\right)dx=0$

Answer 1

The correct option is B $\varphi \left(x\right)$ is an even function

$f\left(x\right)$ is odd function so,
$f\left(x\right)=-f(-x)$
$\varphi (-x)=\underset{a}{\overset{-x}{\int }}f\left(t\right)dt$
Putting $t=-y$
$\Rightarrow \varphi (-x)=\underset{-a}{\overset{x}{\int }}f(-y)(-dy)$
$\Rightarrow \varphi (-x)=\underset{-a}{\overset{x}{\int }}f\left(y\right)dy$
$\Rightarrow \varphi (-x)=\underset{-a}{\overset{x}{\int }}f\left(t\right)dt\Rightarrow \varphi (-x)=\underset{-a}{\overset{a}{\int }}f\left(t\right)dt+\underset{a}{\overset{x}{\int }}f\left(t\right)dt\Rightarrow \varphi (-x)=0+\underset{a}{\overset{x}{\int }}f\left(t\right)dt=\varphi \left(x\right)$

Hence, $\varphi \left(x\right)$ is an even function