Question

Statement$-1:$ If the function $f\left(x\right)=a{x}^2–2bx\left[x\right]+c[x{]}^2$, where $a,b,c\in N$ is periodic with period $1$, then $a,b,c$ are in A.P., G.P. and H.P.

Statement$-2:$ Three non-zero numbers are in A.P., G.P. and H.P. if and only if they are equal.
$($ Here, $[.]$ denotes the greatest integer function$)$

• A. Statement$-1$ is true, Statement$-2$ is true and Statement$-2$ is correct explanation for Statement$-1$.
• B. Statement$-1$ is true, Statement$-2$ is true and Statement$-2$ is NOT the correct explanation for Statement$-1$.
• C. Statement$-1$ is true, Statement$-2$ is false.
• D. Statement$-1$ is false, Statement$-2$ is true.

Answer 1

The correct option is A Statement$-1$ is true, Statement$-2$ is true and Statement$-2$ is correct explanation for Statement$-1$.

$f\left(x\right)=a{x}^2–2bx\left[x\right]+c[x{]}^2$
Put $\left[x\right]=x-\left\{x\right\}$
$\Rightarrow f\left(x\right)=a{x}^2-2bx(x-\{x\left\}\right)+c(x-\{x\}{)}^2\Rightarrow f(x)=(a-2b+c)x^2+(2b-2c\left)x\right\{x\}+c\{x{\}}^2$
We know that $\left\{x\right\}$ is periodic with period $1$, so $f\left(x\right)$ is periodic if
$(a-2b+c)=0,(2b-2c)=0\Rightarrow b=c=a$
Hence, three non-zero equal numbers are in A.P., G.P. and H.P.