Question

Assertion :If $f\left(x\right)$ is a non-negative continuous function such that $f\left(x\right)+f(x+\frac{1}{2})=1$ then ${\int }_0^{2}f\left(x\right)dx=1$ Reason: $f\left(x\right)$ is a periodic function having period $1$.

• 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,

$f\left(x\right)+f(x+\frac{1}{2})=1$ ...(1)

Replacing $x$ by $x+\frac{1}{2}$, equation (1) reduces to

$f(x+\frac{1}{2})+f(x+1)=1$ ...(2)

Subtracting equation (1) from equation (2), we get

$f(x+1)-f\left(x\right)=0$ i.e., $f(x+1)=f\left(x\right)$

$\Rightarrow f$ is periodic having period $1$

$\therefore I={\int }_0^{2}f\left(x\right)dx={\int }_0^{1}f\left(x\right)dx+{\int }_0^{1}f\left(x\right)dx$

$={\int }_{-1/2}^{1/2}f\left(x\right)dx+{\int }_{-1/2}^{1/2}f(t+\frac{1}{2})dt$

$[\because f$ has period $1,\therefore {\int }_0^{1}f\left(x\right)dx={\int }_{-1/2}^{1/2}f\left(x\right)dx$ and putting $x=t+\frac{1}{2}$ in the second integral$]$

$={\int }_{-1/2}^{1/2}[f\left(t\right)+f(t+\frac{1}{2})]dt={\int }_{-1/2}^{1/2}1dt$

$={\left[t\right]}_{-1/2}^{1/2}=1$