Question

The period of a function $f\left(x\right)=a^{\left\{\tan \left(\pi x\right)+x-\left|x\right|\right\}}$ where $a>0$, $\mathrm{\pi }$ denotes the greatest integer function and $x$ is a real number, is defined as

• A. $\mathrm{\pi }$
• B. $\frac{\mathrm{\pi }}{2}$
• C. $\frac{\mathrm{\pi }}{4}$
• D. $2\mathrm{\pi }$
• E. $1$

Answer 1

The correct option is E

$1$

Step 1: Explanation for the correct option:

Option(E): Given function is $f\left(x\right)=a^{\left\{\tan \left(\pi x\right)+x-\left|x\right|\right\}}$

The period of a function is that the value or interval at which the function repeats its values and is defined as the smallest constant for which $f\left(x+c\right)=f\left(x\right)$

Here for $c=1$,

$\begin{array}{cccll}& f\left(x+1\right)& =& a^{\left\{\tan \left(\pi \left(x+1\right)\right)+\left(x+1\right)-\left|x+1\right|\right\}}& \\ & & =& a^{\left\{\tan \left(\pi x+\mathrm{\pi }\right)+x-\left|x\right|\right\}}& \left[\because \left(x+1\right)-\left|x+1\right|=x-\left|x\right|\right]\\ & & =& a^{\left\{\tan \left(\pi x\right)+x-\left|x\right|\right\}}& \left[\because \tan \left(\mathrm{\pi }+\mathrm{\theta }\right)=\tan \left(\theta \right)\right]\\ \Rightarrow & f\left(x+1\right)& =& f\left(x\right)& \end{array}$

Thus option $\left(E\right)$ is correct

Step 2: Explanation for the incorrect options

Option(A): Here for $c=\mathrm{\pi }$

$\begin{array}{lllll}& f\left(x+\mathrm{\pi }\right)& =& a^{\left\{\tan \left(\pi \left(x+\mathrm{\pi }\right)\right)+\left(x+\mathrm{\pi }\right)-\left|x+\mathrm{\pi }\right|\right\}}& \\ & & =& a^{\left\{\tan \left(\pi x+{\mathrm{\pi }}^2\right)+x-\left|x\right|\right\}}& \left[\because \left(x+\mathrm{\pi }\right)-\left|x+\mathrm{\pi }\right|=x-\left|x\right|\right]\\ \Rightarrow & f\left(x+\mathrm{\pi }\right)& \ne & f\left(x\right)& \end{array}$

Thus option $\left(A\right)$ is incorrect

Option(B): Here for $c=\frac{\mathrm{\pi }}{2}$

$\begin{array}{lllll}& f\left(x+\frac{\mathrm{\pi }}{2}\right)& =& a^{\left\{\tan \left(\pi \left(x+\frac{\mathrm{\pi }}{2}\right)\right)+\left(x+\frac{\mathrm{\pi }}{2}\right)-\left|x+\frac{\mathrm{\pi }}{2}\right|\right\}}& \\ & & =& a^{\left\{\tan \left(\pi x+\frac{{\mathrm{\pi }}^2}{2}\right)+x-\left|x\right|\right\}}& \left[\because \left(x+\frac{\mathrm{\pi }}{2}\right)-\left|x+\frac{\mathrm{\pi }}{2}\right|=x-\left|x\right|\right]\\ \Rightarrow & f\left(x+\frac{\mathrm{\pi }}{2}\right)& \ne & f\left(x\right)& \end{array}$

Thus option $\left(B\right)$ is incorrect

Option(C): Here for $c=\frac{\mathrm{\pi }}{4}$

$\begin{array}{lllll}& f\left(x+\frac{\mathrm{\pi }}{4}\right)& =& a^{\left\{\tan \left(\pi \left(x+\frac{\mathrm{\pi }}{4}\right)\right)+\left(x+\frac{\mathrm{\pi }}{4}\right)-\left|x+\frac{\mathrm{\pi }}{4}\right|\right\}}& \\ & & =& a^{\left\{\tan \left(\pi x+\frac{{\mathrm{\pi }}^2}{4}\right)+x-\left|x\right|\right\}}& \left[\because \left(x+\frac{\mathrm{\pi }}{4}\right)-\left|x+\frac{\mathrm{\pi }}{4}\right|=x-\left|x\right|\right]\\ \Rightarrow & f\left(x+\frac{\mathrm{\pi }}{4}\right)& \ne & f\left(x\right)& \end{array}$

Thus option $\left(C\right)$ is incorrect

Option(D): Here for $c=2\mathrm{\pi }$

$\begin{array}{lllll}& f\left(x+2\mathrm{\pi }\right)& =& a^{\left\{\tan \left(\pi \left(x+2\mathrm{\pi }\right)\right)+\left(x+2\mathrm{\pi }\right)-\left|x+2\mathrm{\pi }\right|\right\}}& \\ & & =& a^{\left\{\tan \left(\pi x+2{\mathrm{\pi }}^2\right)+x-\left|x\right|\right\}}& \left[\because \left(x+2\mathrm{\pi }\right)-\left|x+2\mathrm{\pi }\right|=x-\left|x\right|\right]\\ \Rightarrow & f\left(x+2\mathrm{\pi }\right)& \ne & f\left(x\right)& \end{array}$

Thus option $\left(D\right)$ is incorrect

Hence, the correct option is option(E) i.e. $1$