Question

Let $f$ be a non-negative continuous function defined on $R$ such that $f\left(x\right)+f(x+\frac{1}{2})=6$ and $N=\underset{0}{\overset{6}{\int }}f\left(x\right)dx.$ Then

• A. number of divisors of $N$ is $6.$
• B. number of divisors of $N$ is $9.$
• C. sum of the digits in $N$ is $9.$
• D. sum of the digits in $N$ is $6.$

Answer 1

Answer: (A), (C)

sum of the digits in $N$ is $9.$

Given, $f\left(x\right)+f(x+\frac{1}{2})=6\cdots \left(1\right)$
$x\to x+\frac{1}{2}$
$f(x+\frac{1}{2})+f(x+1)=6\cdots \left(2\right)$

Equation $\left(1\right)-\left(2\right),$ we get
$f\left(x\right)-f(x+1)=0$
$\Rightarrow f\left(x\right)=f(x+1)$

Now, $N=\underset{0}{\overset{6}{\int }}f\left(x\right)dx$
$\Rightarrow N=6\underset{0}{\overset{1}{\int }}f\left(x\right)dx$
$\Rightarrow N=6\left(\underset{0}{\overset{1/2}{\int }}f\left(x\right)dx+\underset{1/2}{\overset{1}{\int }}f\left(x\right)dx\right)$
Put $x=y+\frac{1}{2}\Rightarrow dx=dy$ in the second integral
$\Rightarrow N=6\left(\underset{0}{\overset{1/2}{\int }}f\left(x\right)dx+\underset{0}{\overset{1/2}{\int }}f(y+\frac{1}{2})dy\right)$
$\Rightarrow N=6\underset{0}{\overset{1/2}{\int }}\left(f\right(x)+(x+\frac{1}{2}))dx$
$\Rightarrow N=6\underset{0}{\overset{1/2}{\int }}6dx=6\times 6\times \frac{1}{2}=18$
$\Rightarrow N=18=2^1\times 3^2$
Hence, number of divisors of $N$ is $2\times 3=6$
Sum of digits in $N$ is $1+8=9$