Question

The period of $f\left(x\right)={10}^{\left\{{\cos }^2\left(\mathrm{\pi x}\right)+x-\left\{x\right\}\right\}}+{\sin }^2\left(\mathrm{\pi x}\right)$ is

• A. $1$
• B. $2$
• C. $3$
• D. $\mathrm{\pi }$

Answer 1

The correct option is A

$1$

Explanation for the correct option:

Given function is $f\left(x\right)={10}^{\left\{{\cos }^2\left(\mathrm{\pi x}\right)+x-\left\{x\right\}\right\}}+{\sin }^2\left(\mathrm{\pi x}\right)$

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 $c=1$,

$\begin{array}{cccll}& f\left(x+1\right)& =& {10}^{\left\{{\cos }^2\left(\pi \left(x+1\right)\right)+\left(x+1\right)-\left\{x+1\right\}\right\}}+{\sin }^2\left(\pi \left(x+1\right)\right)& \\ & & =& {10}^{\left\{{\cos }^2\left(\pi x+\mathrm{\pi }\right)+x-\left\{x\right\}\right\}}+{\sin }^2\left(\pi x+\mathrm{\pi }\right)& \left[\because \left(x+1\right)-\left\{x+1\right\}=x-\left\{x\right\}\right]\\ & & =& {10}^{\left\{{\left(-\cos \left(\pi x\right)\right)}^2+x-\left\{x\right\}\right\}}+{\left(-\sin \left(\pi x\right)\right)}^2& \left[\because \cos \left(\mathrm{\pi }+\mathrm{\theta }\right)=-\cos \left(\theta \right),\sin \left(\mathrm{\pi }+\mathrm{\theta }\right)=-\sin \left(\theta \right)\right]\\ & & =& {10}^{\left\{{\cos }^2\left(\pi x\right)+x-\left\{x\right\}\right\}}+{\sin }^2\left(\pi x\right)& \\ \Rightarrow & f\left(x+1\right)& =& f\left(x\right)& \end{array}$

$\begin{array}{cccll}& f\left(x+2\right)& =& {10}^{\left\{{\cos }^2\left(\pi \left(x+2\right)\right)+\left(x+2\right)-\left\{x+2\right\}\right\}}+{\sin }^2\left(\pi \left(x+2\right)\right)& \\ & & =& {10}^{\left\{{\cos }^2\left(\pi x+2\mathrm{\pi }\right)+x-\left\{x\right\}\right\}}+{\sin }^2\left(\pi x+2\mathrm{\pi }\right)& \left[\because \left(x+2\right)-\left\{x+2\right\}=x-\left\{x\right\}\right]\\ & & =& {10}^{\left\{{\left(\cos \left(\pi x\right)\right)}^2+x-\left\{x\right\}\right\}}+{\left(\sin \left(\pi x\right)\right)}^2& \left[\because \cos \left(2\mathrm{\pi }+\mathrm{\theta }\right)=\cos \left(\theta \right),\sin \left(2\mathrm{\pi }+\mathrm{\theta }\right)=\sin \left(\theta \right)\right]\\ & & =& {10}^{\left\{{\cos }^2\left(\pi x\right)+x-\left\{x\right\}\right\}}+{\sin }^2\left(\pi x\right)& \\ \Rightarrow & f\left(x+2\right)& =& f\left(x\right)& \end{array}$

$\begin{array}{cccll}& f\left(x+3\right)& =& {10}^{\left\{{\cos }^2\left(\pi \left(x+3\right)\right)+\left(x+3\right)-\left\{x+3\right\}\right\}}+{\sin }^2\left(\pi \left(x+3\right)\right)& \\ & & =& {10}^{\left\{{\cos }^2\left(\pi x+3\mathrm{\pi }\right)+x-\left\{x\right\}\right\}}+{\sin }^2\left(\pi x+3\mathrm{\pi }\right)& \left[\because \left(x+3\right)-\left\{x+3\right\}=x-\left\{x\right\}\right]\\ & & =& {10}^{\left\{{\left(-\cos \left(\pi x\right)\right)}^2+x-\left\{x\right\}\right\}}+{\left(-\sin \left(\pi x\right)\right)}^2& \left[\because \cos \left(3\mathrm{\pi }+\mathrm{\theta }\right)=-\cos \left(\theta \right),\sin \left(3\mathrm{\pi }+\mathrm{\theta }\right)=-\sin \left(\theta \right)\right]\\ & & =& {10}^{\left\{{\cos }^2\left(\pi x\right)+x-\left\{x\right\}\right\}}+{\sin }^2\left(\pi x\right)& \\ \Rightarrow & f\left(x+3\right)& =& f\left(x\right)& \end{array}$

From the above we can observe that $c=1,2,3$ satisfies the condition $f\left(x+c\right)=f\left(x\right)$

To determine the period of function consider the least possible value of $c$ i.e. $1$

Thus the period of the given function $f\left(x\right)$ is $1$

Hence the correct option is option $\left(A\right)$ i.e. $1$