Question

Assertion :If $f\left(x\right)={\int }_0^{x}g\left(t\right)dt$, where $g$ is an even function and $f(x+5)=g\left(x\right)$, then $g\left(0\right)-g\left(x\right)={\int }_0^{x}f\left(t\right)dt$ Reason: $f$ is an odd function.

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Both Assertion and Reason are incorrect

Answer 1

The correct option is A Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

We have,

$f\left(x\right)={\int }_0^{x}g\left(t\right)dt={\int }_0^{x}g(-t)dt$ $(g$ is even$)$

$=-{\int }_0^{-x}g\left(u\right)du$ $[$Putting $-t=u]$

$=-f(-x)$

$\therefore f$ is an odd function

Now, $g\left(0\right)-g\left(x\right)=f\left(5\right)-f(x+5)$ $[$using $f(x+5)=g\left(x\right)]$

$={\int }_0^{5}g\left(t\right)dt-{\int }_0^{x+5}g\left(t\right)dt$$={\int }_{x+5}^{5}g\left(t\right)dt={\int }_{x+5}^{5}g(-t)dt$ $(g$ is even$)$

$={\int }_{x+5}^{5}f(-t+5)dt$ $[$using $f(x+5)=g\left(x\right)]$

$=-{\int }_{-x}^{0}g\left(z\right)dz$ $[$Putting $-t+5=z]$

$={\int }_{-x}^{0}f(-z)dz$ $[f$ is odd$]$

$=-{\int }_x^{0}f\left(y\right)dy$ $[$Putting $-z=y]$

$={\int }_0^{x}f\left(y\right)dy={\int }_0^{x}f\left(t\right)dt$