WBJEE · Maths · Functions
Consider the function \(f(x)=\cos x^{2}\). Then,
- A \(f\) is of period \(2 \pi\)
- B \(f\) is of period \(\sqrt{2 \pi}\)
- C \(f\) is not periodic
- D \(f\) is of period \(\pi\)
Answer & Solution
Correct Answer
(C) \(f\) is not periodic
Step-by-step Solution
Detailed explanation
We have, \[ f(x)=\cos x^{2} \] Let \(T\) be the period of \(f(x)\). Then \[ \begin{aligned} & & f(x+T) &=f(x) \\ \Rightarrow & & \cos (x+T)^{2} &=\cos x^{2} \end{aligned} \] But there is no value of \(T\) for which \[ \cos (x+T)^{2}=\cos x^{2} \] \(\therefore f(x)\) is not…
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