Key Concepts and Formulas

• Definition of a Periodic Function: A function $f(x)$ is periodic if there exists a positive real number $T$ (the period) such that $f(x+T) = f(x)$ for all $x$ in the domain of $f$. The smallest such positive $T$ is the fundamental period.
• Period of Basic Trigonometric Functions:
  • The period of $\sin(kx)$ and $\cos(kx)$ is $\frac{2\pi}{|k|}$.
  • The period of $\tan(kx)$ is $\frac{\pi}{|k|}$.
• Period of Sum/Difference of Periodic Functions: If $f_1(x)$ has period $T_1$ and $f_2(x)$ has period $T_2$, then the period of $f_1(x) \pm f_2(x)$ is the least common multiple (LCM) of $T_1$ and $T_2$, provided the LCM exists. If the LCM does not exist (e.g., one function is not periodic), then the sum/difference may not be periodic.
• Period of $|f(x)|$: If $f(x)$ is periodic with period $T$, then $|f(x)|$ is also periodic. Its period is either $T$ or $T/2$.
• Period of $f^2(x)$: If $f(x)$ is periodic with period $T$, then $f^2(x)$ is also periodic. Its period is $T/2$. For example, $\sin^2 x = \frac{1-\cos 2x}{2}$, so its period is $\frac{2\pi}{2} = \pi$.

Step-by-Step Solution

We need to determine which of the given functions is not periodic. We will analyze each option by finding the periods of its constituent parts and then determining the period of the combined function.

Step 1: Analyze Option (A): $f(x) = |\sin 3x| + \sin^2 x$

• Part 1: $|\sin 3x|$
  • The function $\sin 3x$ has a period of $T_1 = \frac{2\pi}{|3|} = \frac{2\pi}{3}$.
  • The function $|\sin 3x|$ will also be periodic. The period of $|\sin kx|$ is $\frac{\pi}{|k|}$. Therefore, the period of $|\sin 3x|$ is $T_{1'} = \frac{\pi}{|3|} = \frac{\pi}{3}$.
• Part 2: $\sin^2 x$
  • We know that $\sin^2 x = \frac{1 - \cos 2x}{2}$.
  • The period of $\cos 2x$ is $T_2 = \frac{2\pi}{|2|} = \pi$.
  • Therefore, the period of $\sin^2 x$ is also $\pi$.
• Combined Function: We need to find the LCM of $T_{1'} = \frac{\pi}{3}$ and $T_2 = \pi$.
  • LCM$(\frac{\pi}{3}, \pi) = \pi \times$ LCM$(\frac{1}{3}, 1) = \pi \times \frac{1}{3} \times \text{LCM}(1, 3) = \pi \times \frac{1}{3} \times 3 = \pi$.
  • Alternatively, we can write $\pi = \frac{3\pi}{3}$. LCM$(\frac{\pi}{3}, \frac{3\pi}{3}) = \frac{\pi}{3} \times$ LCM$(1, 3) = \frac{\pi}{3} \times 3 = \pi$.
  • Since the LCM exists and is a positive real number ($\pi$), the function $|\sin 3x| + \sin^2 x$ is periodic with period $\pi$.

Step 2: Analyze Option (B): $f(x) = \cos \sqrt x + \cos^2 x$

• Part 1: $\cos \sqrt x$
  • Consider the behavior of $\cos \sqrt x$. For the function to be periodic, we need $\cos \sqrt{x+T} = \cos \sqrt{x}$ for all $x$ in the domain.
  • Let's examine the domain. For $\sqrt{x}$ to be defined, we must have $x \ge 0$.
  • If $\cos \sqrt{x+T} = \cos \sqrt{x}$, then $\sqrt{x+T} = \sqrt{x} + 2n\pi$ or $\sqrt{x+T} = -\sqrt{x} + 2n\pi$ for some integer $n$.
  • Since $\sqrt{x+T} \ge 0$ and $\sqrt{x} \ge 0$, the second case $\sqrt{x+T} = -\sqrt{x} + 2n\pi$ is only possible if $n=0$, leading to $\sqrt{x+T} = -\sqrt{x}$, which implies $x+T=x$, so $T=0$, which is not allowed. If $n > 0$, $\sqrt{x+T} = 2n\pi - \sqrt{x}$. Squaring both sides gives $x+T = (2n\pi)^2 - 4n\pi\sqrt{x} + x$, so $T = (2n\pi)^2 - 4n\pi\sqrt{x}$. This implies $T$ depends on $x$, which contradicts the definition of a period.
  • Thus, we must have $\sqrt{x+T} = \sqrt{x} + 2n\pi$.
  • Squaring both sides: $x+T = x + 4n^2\pi^2 + 4n\pi\sqrt{x}$.
  • This gives $T = 4n^2\pi^2 + 4n\pi\sqrt{x}$. For $T$ to be a constant period, this equation must hold for all $x$. This is only possible if $n=0$, which gives $T=0$, but the period must be positive.
  • Therefore, $\cos \sqrt x$ is not periodic for $x \ge 0$.
• Part 2: $\cos^2 x$
  • The function $\cos x$ has a period of $2\pi$.
  • The period of $\cos^2 x = \frac{1+\cos 2x}{2}$ is $\frac{2\pi}{2} = \pi$.
• Combined Function: Since one of the components ($\cos \sqrt x$) is not periodic, the sum of a non-periodic function and a periodic function is generally not periodic. Therefore, $\cos \sqrt x + \cos^2 x$ is not periodic.

Step 3: Analyze Option (C): $f(x) = \cos 4x + \tan^2 x$

• Part 1: $\cos 4x$
  • The period of $\cos 4x$ is $T_1 = \frac{2\pi}{|4|} = \frac{\pi}{2}$.
• Part 2: $\tan^2 x$
  • The function $\tan x$ has a period of $\pi$.
  • The period of $\tan^2 x$ is $\frac{\pi}{1} = \pi$.
• Combined Function: We need to find the LCM of $T_1 = \frac{\pi}{2}$ and $T_2 = \pi$.
  • LCM$(\frac{\pi}{2}, \pi) = \pi \times$ LCM$(\frac{1}{2}, 1) = \pi \times \frac{1}{2} \times \text{LCM}(1, 2) = \pi \times \frac{1}{2} \times 2 = \pi$.
  • Since the LCM exists and is a positive real number ($\pi$), the function $\cos 4x + \tan^2 x$ is periodic with period $\pi$.

Step 4: Analyze Option (D): $f(x) = \cos 2x + \sin x$

• Part 1: $\cos 2x$
  • The period of $\cos 2x$ is $T_1 = \frac{2\pi}{|2|} = \pi$.
• Part 2: $\sin x$
  • The period of $\sin x$ is $T_2 = 2\pi$.
• Combined Function: We need to find the LCM of $T_1 = \pi$ and $T_2 = 2\pi$.
  • LCM$(\pi, 2\pi) = 2\pi$.
  • Since the LCM exists and is a positive real number ($2\pi$), the function $\cos 2x + \sin x$ is periodic with period $2\pi$.

Step 5: Identify the Non-Periodic Function

From our analysis, we found that:

• Option (A) is periodic.
• Option (B) is not periodic because $\cos \sqrt x$ is not periodic.
• Option (C) is periodic.
• Option (D) is periodic.

Therefore, the function that is not periodic is $\cos \sqrt x + \cos^2 x$.

Common Mistakes & Tips

• Domain Considerations: Always check the domain of the functions, especially when square roots are involved. A function with a restricted domain (like $x \ge 0$ for $\sqrt{x}$) might not be periodic, even if its trigonometric part appears to be.
• Period of Squared Trigonometric Functions: Remember that $\sin^2(kx)$ and $\cos^2(kx)$ have a period of $\frac{\pi}{|k|}$, not $\frac{2\pi}{|k|}$.
• Period of Absolute Value Functions: The period of $|f(x)|$ is generally half the period of $f(x)$ if $f(x)$ crosses the x-axis, but it's important to verify this. For $|\sin kx|$, the period is indeed $\frac{\pi}{|k|}$.
• LCM of Periods: For a sum/difference of periodic functions, the period is the LCM of their individual periods. If any component is not periodic, the entire function is likely not periodic.

Summary

To determine which function is not periodic, we analyzed each option by finding the periods of its individual components and then considering the period of the combined function. We used the properties of basic trigonometric functions, their squares, absolute values, and the concept of the least common multiple for periods. Option (B) was identified as non-periodic because the term $\cos \sqrt x$ is not periodic due to the behavior of the square root function and the constant nature required for a period.

Final Answer

The final answer is \boxed{B} which corresponds to option (B).