Question

Let $f$ be a non-negative continuous and twice differentiable function defined on $R$ such that $f\left(x\right)+f(x+\frac{3}{2})=6,\forall x\in R.$ Then

• A. $f\left(x\right)$ is a periodic function
• B. $f\left(x\right)$ is a non-periodic function
• C. $\underset{0}{\overset{300}{\int }}f\left(x\right)dx=900$
• D. If $f\left(0\right)=3,$ then $f^{\prime \prime }\left(x\right)$ has at least $3$ roots in $(0,6)$

Answer 1

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

If $f\left(0\right)=3,$ then $f^{\prime \prime }\left(x\right)$ has at least $3$ roots in $(0,6)$

$f\left(x\right)+f(x+\frac{3}{2})=6,\forall x\in R\cdots \left(1\right)$
Replacing $x$ by $x+\frac{3}{2},$ we get
$f(x+\frac{3}{2})+f(x+3)=6\cdots \left(2\right)$
$\Rightarrow 6-f\left(x\right)+f(x+3)=6\left[\text{Using}\right(1\left)\right]$
$\Rightarrow f(x+3)=f\left(x\right)$
$\therefore f\left(x\right)$ is periodic.

$I=\underset{0}{\overset{300}{\int }}f\left(x\right)dx$
$\Rightarrow I=100\underset{0}{\overset{3}{\int }}f\left(x\right)dx$
$\Rightarrow I=100\left(\underset{0}{\overset{3/2}{\int }}f\right(x)dx+\underset{3/2}{\overset{3}{\int }}f\left(x\right)dx)$
$\Rightarrow I=100\left(\underset{0}{\overset{3/2}{\int }}f\right(x)dx+\underset{0}{\overset{3/2}{\int }}f(x+\frac{3}{2})dx)$
$\Rightarrow I=100\underset{0}{\overset{3/2}{\int }}6dx=100\times 6\times \frac{3}{2}=900$


From Rolle's theorem,
$f^{\prime \prime }\left(x\right)=0$ has at least $3$ roots.