Question

Let $f$ and $g$ be two differentiable functions such that $f\left(x\right)$ is odd and $g\left(x\right)$ is even. If $f\left(5\right)=7,$ $f\left(0\right)=0,$ $g\left(x\right)=f(x+5)$ and $f\left(x\right)=\underset{0}{\overset{x}{\int }}g\left(t\right)dt\forall x\in R,$ then which of the following is/are CORRECT?

• A. $f(x-5)=-g\left(x\right)\forall x\in R$
• B. $\underset{0}{\overset{5}{\int }}f\left(t\right)dt=7$
• C. $\underset{0}{\overset{x}{\int }}f\left(t\right)dt=g\left(0\right)-g\left(x\right)$
• D. $\underset{0}{\overset{5}{\int }}f\left(t\right)dt=\underset{0}{\overset{5}{\int }}f(5-t)dt$

Answer 1

Answer: (A), (B), (C), (D)

$\underset{0}{\overset{5}{\int }}f\left(t\right)dt=\underset{0}{\overset{5}{\int }}f(5-t)dt$

We have $g\left(x\right)=f(x+5)$
$\Rightarrow g(-x)=f(-x+5)=f(-(x-5\left)\right)$
$\Rightarrow f(x-5)=-g\left(x\right)$
( $\because f\left(x\right)$ is odd function and $g\left(x\right)$ is even function.)


$\underset{0}{\overset{5}{\int }}f\left(t\right)dt$
$=\underset{0}{\overset{5}{\int }}f(5-t)dt=\underset{0}{\overset{5}{\int }}g\left(t\right)dt$
$=f\left(5\right)=7$


$I=\underset{0}{\overset{x}{\int }}f\left(t\right)dt$
Put $t=u-5\Rightarrow dt=du$
$I=\underset{5}{\overset{5+x}{\int }}f(u-5)du$
$=-\underset{5}{\overset{5+x}{\int }}g\left(u\right)du[\because f(x-5)=-g(x\left)\right]$
Also, $f\left(x\right)=\underset{0}{\overset{x}{\int }}g\left(t\right)dt\Rightarrow f^{\prime }\left(x\right)=g\left(x\right)$
Then, $I=-\underset{5}{\overset{5+x}{\int }}{f}^{\prime }\left(u\right)du$
$=[f\left(u\right){]}_{5+x}^5=f\left(5\right)-f(5+x)=g\left(0\right)-g\left(x\right)$
$\therefore I=\underset{0}{\overset{x}{\int }}f\left(t\right)dt=g\left(0\right)-g\left(x\right)$