Question

Match $\text{List I}$ with the $\text{List II}$ and select the correct answer using the code given below the lists :
Let $[.]$ denote the greatest integer function. Let $f\left(x\right)=\left(\ln \right(a^2-3a-3\left)\right)|\sin x|+\left[\frac{a^2}{9}\right]\cos \pi x$ for all $x\in R$ and $a\in [-4,4]$

$\begin{array}{cccc}& \text{List I}& & \text{List II}\\ \text{(A)}& \text{If}f\left(x\right)\text{is periodic, then the number of integral values of}a\text{is}& \text{(P)}& 1\\ \text{(B)}& \text{If}f\left(x\right)\text{is periodic with period to be a rational number, then}& \text{(Q)}& 2\\ & \text{the number of integral values of}a\text{is}& & \\ \text{(C)}& \text{If}f\left(x\right)\text{is periodic with period to be an irrational number, then}& \text{(R)}& 3\\ & \text{the number of integral values of}a\text{is}& & \\ \text{(D)}& \text{If}f\left(x\right)\text{is non-periodic function, then the number of integral}& \text{(S)}& 4\\ & \text{values of}a\text{is}& & \end{array}$

Which of the following is a CORRECT combination?

• A. $\left(A\right)\to \left(R\right),\left(B\right)\to \left(P\right)$
• B. $\left(A\right)\to \left(R\right),\left(B\right)\to \left(Q\right)$
• C. $\left(A\right)\to \left(P\right),\left(B\right)\to \left(Q\right)$
• D. $\left(A\right)\to \left(S\right),\left(B\right)\to \left(P\right)$

Answer 1

The correct option is A $\left(A\right)\to \left(R\right),\left(B\right)\to \left(P\right)$

$\left(A\right)$
Clearly, $a^2-3a-3>0$
$\Rightarrow a<\frac{3-\sqrt{21}}{2}$ or $a>\frac{3+\sqrt{21}}{2}$

For $f\left(x\right)$ to be periodic, there are three cases possible.
Case $1:$
$\ln (a^2-3a-3)=0$ and $\left[\frac{a^2}{9}\right]\ne 0$
$\Rightarrow a^2-3a-3=1$
$\Rightarrow a^2-3a-4=0$
$\Rightarrow (a-4)(a+1)=0$
$\Rightarrow a=4,a=-1$
But at $a=-1,\left[\frac{a^2}{9}\right]=\left[\frac{1}{9}\right]=0$
At $a=4,\left[\frac{a^2}{9}\right]=\left[\frac{16}{9}\right]=1\ne 0$
$\therefore a=4$

Case $2:$
$\ln (a^2-3a-3)\ne 0$ and $\left[\frac{a^2}{9}\right]=0$
If $\left[\frac{a^2}{9}\right]=0,$
then $0\le \frac{a^2}{9}<1$
$\Rightarrow -3<a<3$
and $a^2-3a-3>0$
$\Rightarrow a<\frac{3-\sqrt{21}}{2}$ or $a>\frac{3+\sqrt{21}}{2}$
$\therefore a\in (-3,\frac{3-\sqrt{21}}{2})$
But at $a=-1,$ $\ln (a^2-3a-3)=0$
$\therefore a=-2$

Case $3:$ $\ln (a^2-3a-3)=0$ and $\left[\frac{a^2}{9}\right]=0$
$\Rightarrow a=-1$
$\therefore$ For $f\left(x\right)$ to be periodic,
Integral values $=-1,-2,4$
$\therefore$ Number of integral values $=3$

$\left(B\right)$
For period to be a rational number, Case $1$ is feasible.
$\therefore$ Number of integral values $=1$