Question

The period of the function f(n) = cos x$^2$ .
A. 2 $\pi $
B. $\pi $
C. $\dfrac{\pi }{2}$
D. Does not exist

Answer 1

Hint: Generally period of any trigonometric function can be found by substituting (x$\pi $) instead of x and evaluating ($\pi $) verifying with option.

Complete step-by-step answer:
 Explanation-1
F(x) = cosx$^2$
As, x$^2$ $ \geqslant $ 0
For periodic function graph should be plated on both sides of Y-axis
But x$^2$< 0 cannot happen, so that cos x$^2$ is non-periodic function.
Now this question can be solved by the given below method as stated in hint.
Explanation-II
For periodic function f(x)
\[
  {\text{F}}\left( {{\text{x + T}}} \right){\text{ = f}}\left( {\text{x}} \right){\text{ }}\left[ {{\text{T is period of f}}\left( {\text{x}} \right)} \right] \\
  {\text{Let f}}\left( {\text{x}} \right){\text{ = cosx}}{{\text{}}^2} \\
  {\text{cos }}{\left( {{\text{x + T}}} \right)^2}{\text{ = cos}}{{\text{x}}^2} \\
   \Rightarrow {\text{(}}{{\text{x}}^2}{\text{ + }}{{\text{T}}^2}{\text{ + 2xT)=cos }}{\left( {\text{x}} \right)^2} \\
\]
Not possible

Therefore, Period of cosx$^2$ does not exist

Note: This kind of question can also be dealt with brief ideas about graphs. Simplify it further to check the period of the function. You can verify the solution by plotting the graph of the function.