Periodic functions Questions in English

(B) The period of $\sin(ax)$ is $\frac{2\pi}{|a|}$.
So,the period of $\sin 2x$ is $\frac{2\pi}{2} = \pi$.
For a function $f(x) = |\sin(ax)|$,the period is $\frac{\pi}{|a|}$.
Therefore,the period of $|\sin 2x|$ is $\frac{\pi}{2}$.

(B) Given expression is $f(\theta) = \sin \theta \cos \theta$.
Using the identity $\sin 2\theta = 2 \sin \theta \cos \theta$,we can write:
$f(\theta) = \frac{1}{2} \sin 2\theta$.
The period of $\sin(k\theta)$ is given by $\frac{2\pi}{|k|}$.
Here,$k = 2$,so the period is $\frac{2\pi}{2} = \pi$.
Thus,the correct option is $B$.

(C) Let $f(\theta) = \frac{\sin \theta + \sin 2\theta}{\cos \theta + \cos 2\theta}$.
Using the sum-to-product formulas $\sin A + \sin B = 2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)$ and $\cos A + \cos B = 2 \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)$:
$f(\theta) = \frac{2 \sin \left(\frac{3\theta}{2}\right) \cos \left(\frac{\theta}{2}\right)}{2 \cos \left(\frac{3\theta}{2}\right) \cos \left(\frac{\theta}{2}\right)} = \tan \left(\frac{3\theta}{2}\right)$.
The period of $\tan(k\theta)$ is $\frac{\pi}{|k|}$.
Here,$k = \frac{3}{2}$,so the period is $\frac{\pi}{3/2} = \frac{2\pi}{3}$.

(C) The period of a trigonometric function of the form $\cos(ax + b)$ is given by $T = \frac{2\pi}{|a|}$.
Here,$a = 7$ and $b = -5$.
Therefore,the period $T = \frac{2\pi}{|7|} = \frac{2\pi}{7}$.

(D) We can rewrite the expression as: $\sin \theta - \sqrt{3} \cos \theta = 2 \left( \frac{1}{2} \sin \theta - \frac{\sqrt{3}}{2} \cos \theta \right)$.
Using the identity $\sin(A - B) = \sin A \cos B - \cos A \sin B$,we get: $2 \sin \left( \theta - \frac{\pi}{3} \right)$.
The period of $\sin(k\theta)$ is $\frac{2\pi}{|k|}$.
Here,$k = 1$,so the period is $\frac{2\pi}{1} = 2\pi$.

(D) The period of $\sin(ax)$ is given by $\frac{2\pi}{|a|}$.
For $\sin \frac{x}{2}$, the period is $\frac{2\pi}{1/2} = 4\pi$.
For $\cos \frac{x}{3}$, the period is $\frac{2\pi}{1/3} = 6\pi$.
The period of the expression $\sin \frac{x}{2} - \cos \frac{x}{3}$ is the least common multiple $(LCM)$ of $4\pi$ and $6\pi$.
$LCM(4\pi, 6\pi) = 12\pi$.

(C) The period of $\cot 3x$ is $T_1 = \frac{\pi }{3}$.
The period of $\cos (4x + 3)$ is $T_2 = \frac{2\pi }{4} = \frac{\pi }{2}$.
The period of the function $f(x) = \cot 3x - \cos (4x + 3)$ is the $L.C.M.$ of $T_1$ and $T_2$.
$L.C.M. \left( \frac{\pi }{3}, \frac{\pi }{2} \right) = \frac{L.C.M. (\pi, \pi)}{H.C.F. (3, 2)} = \frac{\pi }{1} = \pi$.
Thus,the period is $\pi$.

(D) Let $f(\theta) = 2\sin 3\theta + 4\cos 3\theta$.
We can write this as $f(\theta) = R\sin(3\theta + \alpha)$,where $R = \sqrt{2^2 + 4^2} = \sqrt{20} = 2\sqrt{5}$.
So,$f(\theta) = 2\sqrt{5} \sin(3\theta + \alpha)$.
The function given is $g(\theta) = |f(\theta)| = |2\sqrt{5} \sin(3\theta + \alpha)|$.
The period of $\sin(3\theta + \alpha)$ is $T = \frac{2\pi}{3}$.
The period of $|\sin(3\theta + \alpha)|$ is $\frac{T}{2} = \frac{2\pi/3}{2} = \frac{\pi}{3}$.
Thus,the period of $|2\sin 3\theta + 4\cos 3\theta |$ is $\frac{\pi}{3}$.

(A) Let $f(x) = \sin^4 x + \cos^4 x$.
Using the identity $a^2 + b^2 = (a+b)^2 - 2ab$,we have:
$f(x) = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x$
$f(x) = 1 - 2(\sin x \cos x)^2$
$f(x) = 1 - 2(\frac{\sin 2x}{2})^2$
$f(x) = 1 - \frac{\sin^2 2x}{2}$
Using the identity $\sin^2 \theta = \frac{1 - \cos 2\theta}{2}$:
$f(x) = 1 - \frac{1}{2} \left( \frac{1 - \cos 4x}{2} \right)$
$f(x) = 1 - \frac{1}{4} + \frac{1}{4} \cos 4x$
$f(x) = \frac{3}{4} + \frac{1}{4} \cos 4x$
The period of $\cos(kx)$ is $\frac{2\pi}{|k|}$.
Therefore,the period of $f(x)$ is $\frac{2\pi}{4} = \frac{\pi}{2}$.

(D) The period of the function $f(x) = \sin(kx)$ is given by $\frac{2\pi}{|k|}$.
Here,$k = \frac{1}{n}$.
Therefore,the period of $f(x) = \sin \left( \frac{x}{n} \right)$ is $\frac{2\pi}{|1/n|} = 2n\pi$ (assuming $n > 0$).
Given that the period is $4\pi$,we have $2n\pi = 4\pi$.
Solving for $n$,we get $n = 2$.

(A) We know that $\sin^2 x = \frac{1 - \cos(2x)}{2}$.
The period of $\cos(kx)$ is given by $\frac{2\pi}{|k|}$.
Here,$k = 2$,so the period is $\frac{2\pi}{2} = \pi$.
Thus,the period of $\sin^2 x$ is $\pi$.

(B) The period of a function of the form $y = \sin(ax + b)$ is given by the formula $T = \frac{2\pi}{|a|}$.
Given the function $y = \sin 2x$,here $a = 2$.
Therefore,the period $T = \frac{2\pi}{|2|} = \pi$.

(C) function $y = f(t)$ is periodic with period $T$ if $f(t + T) = f(t)$.
For option $(a)$,the terms have frequencies $2\pi, 3\pi, 5\pi$. The periods are $T_1 = \frac{2\pi}{2\pi} = 1$,$T_2 = \frac{2\pi}{3\pi} = \frac{2}{3}$,$T_3 = \frac{2\pi}{5\pi} = \frac{2}{5}$. The $LCM$ of $(1, \frac{2}{3}, \frac{2}{5})$ is $2$.
For option $(b)$,the periods are $T_1 = \frac{2\pi}{\pi/3} = 6$ and $T_2 = \frac{2\pi}{\pi/4} = 8$. The $LCM$ of $(6, 8)$ is $24$.
For option $(c)$,the period of $\sin t$ is $2\pi$ and the period of $\cos 2t$ is $\frac{2\pi}{2} = \pi$. The $LCM$ of $(2\pi, \pi)$ is $2\pi$. Thus,the function $y = \sin t + \cos 2t$ has a period of $2\pi$.

(D) The period of $\sin \left( \frac{2x}{3} \right)$ is $T_1 = \frac{2\pi}{2/3} = 3\pi$.
The period of $\sin \left( \frac{3x}{2} \right)$ is $T_2 = \frac{2\pi}{3/2} = \frac{4\pi}{3}$.
The period of the sum of two periodic functions is the $L.C.M.$ of their individual periods.
We need to find the $L.C.M.$ of $3\pi$ and $\frac{4\pi}{3}$.
$L.C.M. \left( \frac{a}{b}, \frac{c}{d} \right) = \frac{L.C.M.(a, c)}{H.C.F.(b, d)}$.
$L.C.M. \left( \frac{3\pi}{1}, \frac{4\pi}{3} \right) = \frac{L.C.M.(3\pi, 4\pi)}{H.C.F.(1, 3)} = \frac{12\pi}{1} = 12\pi$.
Thus, the period is $12\pi$.

(A) function $f(x) = \cos px + \sin x$ is periodic if there exists a period $\lambda > 0$ such that $f(x + \lambda) = f(x)$ for all $x \in R$.
This implies $\sin(x + \lambda) + \cos p(x + \lambda) = \sin x + \cos px$.
For this to hold,both $\sin x$ and $\cos px$ must be periodic with a common period $\lambda$.
The period of $\sin x$ is $2\pi$,and the period of $\cos px$ is $\frac{2\pi}{|p|}$.
For the sum to be periodic,the ratio of the periods must be a rational number:
$\frac{2\pi}{2\pi/|p|} = |p| \in Q$.
Thus,$p$ must be a rational number.

(A) The period of a function $f(x) = \sin(ax)$ is given by $T = \frac{2\pi}{|a|}$.
For the term $\sin \left( \frac{\pi x}{2} \right)$,the period $T_1 = \frac{2\pi}{\pi/2} = 4$.
For the term $\cos \left( \frac{\pi x}{2} \right)$,the period $T_2 = \frac{2\pi}{\pi/2} = 4$.
The period of the sum of two periodic functions is the least common multiple ($L$.$C$.$M$.) of their individual periods.
Therefore,the period of $\sin \left( \frac{\pi x}{2} \right) + \cos \left( \frac{\pi x}{2} \right) = \text{L.C.M. of } (4, 4) = 4$.

(D) The period of $\sin \frac{\pi x}{2}$ is $T_1 = \frac{2\pi}{\pi/2} = 4$.
The period of $\cos \frac{\pi x}{3}$ is $T_2 = \frac{2\pi}{\pi/3} = 6$.
The period of $\tan \frac{\pi x}{4}$ is $T_3 = \frac{\pi}{\pi/4} = 4$.
The period of the function $f(x)$ is the $L.C.M.$ of $(T_1, T_2, T_3) = L.C.M.(4, 6, 4) = 12$.

(D) The function is given by $f(x) = |\sin(\pi x)|$.
We know that the period of $\sin(ax)$ is $\frac{2\pi}{|a|}$.
For the function $|\sin(ax)|$,the period is $\frac{\pi}{|a|}$.
Here,$a = \pi$.
Therefore,the period is $\frac{\pi}{\pi} = 1$.

(C) The given function is $f(x) = \sin \left( \frac{\pi x}{n - 1} \right) + \cos \left( \frac{\pi x}{n} \right)$.
The period of $\sin \left( \frac{\pi x}{n - 1} \right)$ is $T_1 = \frac{2\pi}{\left( \frac{\pi}{n - 1} \right)} = 2(n - 1)$.
The period of $\cos \left( \frac{\pi x}{n} \right)$ is $T_2 = \frac{2\pi}{\left( \frac{\pi}{n} \right)} = 2n$.
The period of $f(x)$ is the $LCM$ of $T_1$ and $T_2$,which is the $LCM$ of $2(n - 1)$ and $2n$.
Since $n$ and $n-1$ are coprime,the $LCM(2(n-1), 2n) = 2n(n - 1)$.
Thus,the period is $2n(n - 1)$.

(A) For $n = 2$,the function is $f(x) = \frac{\sin(2x)}{\sin(x/2)}$.
Using the identity $\sin(2\theta) = 2\sin(\theta)\cos(\theta)$,we have $\sin(2x) = 2\sin(x)\cos(x) = 2(2\sin(x/2)\cos(x/2))\cos(x) = 4\sin(x/2)\cos(x/2)\cos(x)$.
Thus,$f(x) = \frac{4\sin(x/2)\cos(x/2)\cos(x)}{\sin(x/2)} = 4\cos(x/2)\cos(x)$.
The period of $\cos(x)$ is $2\pi$ and the period of $\cos(x/2)$ is $\frac{2\pi}{1/2} = 4\pi$.
The period of the product of two functions is the least common multiple of their individual periods. The $LCM$ of $2\pi$ and $4\pi$ is $4\pi$.
For $n = 3$,$f(x) = \frac{\sin(3x)}{\sin(x/3)}$. The period of $\sin(3x)$ is $2\pi/3$ and the period of $\sin(x/3)$ is $6\pi$. The period of $f(x)$ is $LCM(2\pi/3, 6\pi) = 6\pi \neq 4\pi$.
Similarly,for other values of $n$,the period will not be $4\pi$.
Therefore,the correct value is $n = 2$.

(C) Let $f(x)$ be a periodic function with period $T > 0$.
Then,by definition,$f(x + T) = f(x)$ for all $x \in \mathbb{R}$.
Substituting the given function $f(x) = x - [x]$:
$(x + T) - [x + T] = x - [x]$
$x + T - [x + T] = x - [x]$
$T = [x + T] - [x]$
For this equation to hold for all $x \in \mathbb{R}$,$T$ must be an integer.
The smallest positive value of $T$ that satisfies this condition is $T = 1$.
Therefore,the period of the function $f(x) = x - [x]$ (which is the fractional part function ${x}$) is $1$.

(D) function $f(x)$ is periodic with period $T$ if $f(x + T) = f(x)$ for all $x$.
Let $g(x) = f(ax + b)$.
We want to find the period $T'$ such that $g(x + T') = g(x)$.
$g(x + T') = f(a(x + T') + b) = f(ax + aT' + b)$.
For this to be equal to $f(ax + b)$,we must have $aT' = T$.
Therefore,$T' = T/a$.
Thus,the function $f(ax + b)$ is periodic with period $T/a$.

(A) Given that $f(x)$ is an odd function,we have $f(-x) = -f(x)$.
For any odd function,$f(0) = -f(0)$,which implies $2f(0) = 0$,so $f(0) = 0$.
Given that $f(x)$ is a periodic function with period $T = 1$,we have $f(x + 1) = f(x)$ for all $x$.
By the property of periodicity,$f(x + nT) = f(x)$ for any integer $n$.
Therefore,$f(2) = f(0 + 2 \times 1) = f(0)$.
Since $f(0) = 0$,it follows that $f(2) = 0$.

(D) function $f(x)$ is periodic if there exists a constant $T > 0$ such that $f(x + T) = f(x)$ for all $x$ in the domain.
For $f(x) = \cos \sqrt{x}$,we check if $\cos \sqrt{x + T} = \cos \sqrt{x}$.
This implies $\sqrt{x + T} = \sqrt{x} + 2n\pi$ or $\sqrt{x + T} = 2n\pi - \sqrt{x}$.
Squaring both sides,we get $x + T = x + 4n\pi\sqrt{x} + 4n^2\pi^2$.
Since $T$ must be a constant independent of $x$,and the expression contains $\sqrt{x}$,no such constant $T$ exists.
Therefore,$f(x) = \cos \sqrt{x}$ is not a periodic function.

(A) Let $f(x) = \frac{|\sin x| + |\cos x|}{|\sin x - \cos x|}$.
Square the function to simplify: $f(x)^2 = \frac{(|\sin x| + |\cos x|)^2}{(|\sin x - \cos x|)^2} = \frac{\sin^2 x + \cos^2 x + 2|\sin x \cos x|}{\sin^2 x + \cos^2 x - 2\sin x \cos x} = \frac{1 + |\sin 2x|}{1 - \sin 2x}$.
Alternatively,observe the symmetry: $f(x + \pi/2) = \frac{|\sin(x + \pi/2)| + |\cos(x + \pi/2)|}{|\sin(x + \pi/2) - \cos(x + \pi/2)|} = \frac{|\cos x| + |\sin x|}{|\cos x - (-\sin x)|} = \frac{|\sin x| + |\cos x|}{|\cos x + \sin x|} = f(x)$.
Since $f(x + \pi/2) = f(x)$,the period is $\pi/2$.

(B) Given the function $f(x) = nx + n - [nx + n]$.
We know that the fractional part function is defined as $\{y\} = y - [y]$.
Therefore,$f(x) = \{nx + n\}$.
Since $n$ is an integer,$nx + n = n(x + 1)$.
Thus,$f(x) = \{n(x + 1)\}$.
The fractional part function $\{u\}$ has a fundamental period of $1$.
For a function $f(x) = g(ax)$,the period is $\frac{T}{|a|}$,where $T$ is the period of $g(x)$.
Here,the period of $\{u\}$ is $1$,so the period of $\{n(x + 1)\}$ is $\frac{1}{|n|}$.
Since $n \in N$,$n > 0$,so the period is $\frac{1}{n}$.

(D) function $f(x)$ is periodic if there exists a positive real number $T$ such that $f(x + T) = f(x)$ for all $x$ in the domain.
$1$. For $f(x) = x - [x]$,this is the fractional part function,denoted as $\{x\}$. It is periodic with a fundamental period $T = 1$,because $\{x + 1\} = (x + 1) - [x + 1] = x + 1 - ([x] + 1) = x - [x] = \{x\}$. Thus,$(A)$ is periodic.
$2$. For $w(x) = \sin^{-1} (\sin x)$,this function is periodic with a fundamental period $T = 2\pi$. Since $\sin(x + 2\pi) = \sin x$,it follows that $\sin^{-1}(\sin(x + 2\pi)) = \sin^{-1}(\sin x)$. Thus,$(B)$ is periodic.
$3$. For $h(x) = x \cos x$,as $x \to \infty$,the amplitude of the oscillations increases linearly,so it does not repeat its values at regular intervals. Thus,$(C)$ is not periodic.
Therefore,both $(A)$ and $(B)$ are periodic functions.

(C) $f(x) = \sin^2 x + \cos^4 x + 2$
$= \sin^2 x + (1 - \sin^2 x)^2 + 2$
$= \sin^2 x + 1 - 2\sin^2 x + \sin^4 x + 2$
$= \sin^4 x - \sin^2 x + 3$
$= (\sin^2 x - \frac{1}{2})^2 + 3 - \frac{1}{4} = (\sin^2 x - \frac{1}{2})^2 + \frac{11}{4}$
Using $\sin^2 x = \frac{1 - \cos 2x}{2}$,we get $f(x) = (\frac{1 - \cos 2x}{2} - \frac{1}{2})^2 + \frac{11}{4} = (-\frac{\cos 2x}{2})^2 + \frac{11}{4} = \frac{1}{4} \cos^2 2x + \frac{11}{4} = \frac{1}{4} (\frac{1 + \cos 4x}{2}) + \frac{11}{4} = \frac{1}{8} \cos 4x + \frac{23}{8}$.
The period $T_1$ of $\cos 4x$ is $\frac{2\pi}{4} = \frac{\pi}{2}$.
For $g(x) = \cos(\cos x) + \cos(\sin x)$,
$g(x + \frac{\pi}{2}) = \cos(\cos(x + \frac{\pi}{2})) + \cos(\sin(x + \frac{\pi}{2})) = \cos(-\sin x) + \cos(\cos x) = \cos(\sin x) + \cos(\cos x) = g(x)$.
Thus,the period $T_2 = \frac{\pi}{2}$.
Therefore,$T_1 = T_2$.

(B) We know that the fractional part function is defined as $\{x\} = x - [x]$.
Given the function $f(x) = \left( \frac{x}{3} - \left[ \frac{x}{3} - 5 \right] \right) + \left( \frac{x}{4} - \left[ \frac{x}{4} - 5 \right] \right) + \left( \frac{x}{5} - \left[ \frac{x}{5} - 5 \right] \right)$.
Since $[x - n] = [x] - n$ for any integer $n$,we have $\left[ \frac{x}{k} - 5 \right] = \left[ \frac{x}{k} \right] - 5$.
Substituting this into the function:
$f(x) = \left( \frac{x}{3} - \left[ \frac{x}{3} \right] + 5 \right) + \left( \frac{x}{4} - \left[ \frac{x}{4} \right] + 5 \right) + \left( \frac{x}{5} - \left[ \frac{x}{5} \right] + 5 \right)$.
$f(x) = \left\{ \frac{x}{3} \right\} + \left\{ \frac{x}{4} \right\} + \left\{ \frac{x}{5} \right\} + 15$.
The period of $\{ax\}$ is $\frac{1}{|a|}$.
So,the periods of the individual terms are $T_1 = \frac{1}{1/3} = 3$,$T_2 = \frac{1}{1/4} = 4$,and $T_3 = \frac{1}{1/5} = 5$.
The period of the sum of these functions is the $L.C.M.$ of their individual periods.
$T = L.C.M.(3, 4, 5) = 60$.

(C) Given the function $f(x) = \cos^2(\sin x) + \sin^2(\cos x)$.
We check for the period $T$ such that $f(x+T) = f(x)$.
Let us test $T = \pi$:
$f(x + \pi) = \cos^2(\sin(x + \pi)) + \sin^2(\cos(x + \pi))$
Since $\sin(x + \pi) = -\sin x$ and $\cos(x + \pi) = -\cos x$,we have:
$f(x + \pi) = \cos^2(-\sin x) + \sin^2(-\cos x)$
Using the properties $\cos(-\theta) = \cos \theta$ and $\sin(-\theta) = -\sin \theta$,we get:
$f(x + \pi) = \cos^2(\sin x) + (-\sin(\cos x))^2 = \cos^2(\sin x) + \sin^2(\cos x) = f(x)$.
Since $f(x + \pi) = f(x)$,the period is $\pi$.

(A) To find the period of the function $f(x) = \log(\cos 2x) + \sin 4x$,we analyze the individual components.
For the term $\log(\cos 2x)$,the function is defined only when $\cos 2x > 0$.
The period of $\cos 2x$ is $\frac{2\pi}{2} = \pi$.
For the term $\sin 4x$,the period is $\frac{2\pi}{4} = \frac{\pi}{2}$.
The period of the sum of two functions is the least common multiple $(LCM)$ of their individual periods.
The $LCM$ of $\pi$ and $\frac{\pi}{2}$ is $\pi$.
Thus,the period of the function $f(x)$ is $\pi$.

(A) The function is given by $f(x) = e^{\{x\} + |\cos \pi x| + |\cos 2\pi x| + \dots + |\cos n\pi x|}$,where $\{x\} = x - [x]$ is the fractional part function.
$1$. The period of the fractional part function $\{x\}$ is $T_0 = 1$.
$2$. For each term of the form $|\cos(k\pi x)|$,the period $T_k$ is given by $\frac{\pi}{k\pi} = \frac{1}{k}$ for $k = 1, 2, \dots, n$.
$3$. The period of the function $f(x)$ is the least common multiple $(LCM)$ of the periods of its individual components: $T = \text{LCM}(T_0, T_1, T_2, \dots, T_n) = \text{LCM}(1, 1, \frac{1}{2}, \frac{1}{3}, \dots, \frac{1}{n})$.
$4$. The $LCM$ of a set of fractions $\frac{a}{b}, \frac{c}{d}, \dots$ is given by $\frac{\text{LCM}(a, c, \dots)}{\text{GCD}(b, d, \dots)}$.
$5$. Thus,$T = \frac{\text{LCM}(1, 1, 1, \dots, 1)}{\text{GCD}(1, 2, 3, \dots, n)} = \frac{1}{1} = 1$.
Therefore,the period of the function is $1$.

(B) Let $f(x) = |\sin 4x| + |\cos 2x|$.
We know that the period of $|\sin ax|$ is $\frac{\pi}{|a|}$ and the period of $|\cos ax|$ is $\frac{\pi}{|a|}$.
For $|\sin 4x|$,the period is $T_1 = \frac{\pi}{4}$.
For $|\cos 2x|$,the period is $T_2 = \frac{\pi}{2}$.
The period of the sum of two periodic functions is the least common multiple $(LCM)$ of their individual periods.
$T = \text{LCM}\left(\frac{\pi}{4}, \frac{\pi}{2}\right) = \frac{\text{LCM}(\pi, \pi)}{\text{HCF}(4, 2)} = \frac{\pi}{2}$.
Thus,the period of $f(x)$ is $\frac{\pi}{2}$.

(C) Let $f(x) = |\cos x|$.
We know that the function $\cos x$ has a period of $2\pi$.
However,when we take the absolute value,the negative parts of the graph are reflected above the $x$-axis.
Specifically,$|\cos(x + \pi)| = |-\cos x| = |\cos x|$.
Since $f(x + \pi) = f(x)$,the fundamental period of the function $f(x) = |\cos x|$ is $\pi$.
As shown in the graph,the pattern of the function repeats every $\pi$ units.

(D) The period of the function $\cos (kx)$ is given by $T = \frac{2 \pi}{|k|}$.
We check each option:
$A) \cos \left( \frac{\pi}{3} x \right) \implies T = \frac{2 \pi}{\pi/3} = 6$.
$B) \cos \left( \frac{\pi}{2} x \right) \implies T = \frac{2 \pi}{\pi/2} = 4$.
$C) \cos (2 \pi x) \implies T = \frac{2 \pi}{2 \pi} = 1$.
$D) \cos (\pi x) \implies T = \frac{2 \pi}{\pi} = 2$.
Therefore,the function with period $2$ is $\cos (\pi x)$.

(C) The period of a function $f(x) = \frac{g(x)}{h(x)}$ is the least common multiple of the periods of $g(x)$ and $h(x)$.
Given $g(x) = \cos(nx)$,its period is $T_1 = \frac{2\pi}{n}$.
Given $h(x) = \sin\left(\frac{x}{n}\right)$,its period is $T_2 = \frac{2\pi}{1/n} = 2n\pi$.
The period of the quotient is the $LCM$ of $T_1$ and $T_2$.
Since $T_2$ is a multiple of $T_1$ (as $2n\pi = n^2 \cdot \frac{2\pi}{n}$),the period of the function is $T_2 = 2n\pi$.
Given $2n\pi = 4\pi$,we get $n = 2$.

(D) We use the identity $\sin^2 A - \sin^2 B = \sin(A + B) \sin(A - B)$.
Here,$A = \frac{\pi}{8} + \frac{x}{2}$ and $B = \frac{\pi}{8} - \frac{x}{2}$.
Then $A + B = \frac{\pi}{8} + \frac{x}{2} + \frac{\pi}{8} - \frac{x}{2} = \frac{\pi}{4}$.
And $A - B = \frac{\pi}{8} + \frac{x}{2} - \left(\frac{\pi}{8} - \frac{x}{2}\right) = x$.
So,$f(x) = \sin\left(\frac{\pi}{4}\right) \sin(x) = \frac{1}{\sqrt{2}} \sin(x)$.
The period of $\sin(x)$ is $2\pi$.
Therefore,the period of $f(x)$ is $2\pi$.

(C) The periods of $\tan 5x$,$\cos 3x$,and $\sin 6x$ are $\frac{\pi}{5}$,$\frac{2\pi}{3}$,and $\frac{\pi}{3}$ respectively.
To find the period $\alpha$ of $f(x) = \frac{\tan 5x \cos 3x}{\sin 6x}$,we simplify the expression:
$f(x) = \frac{\tan 5x \cos 3x}{2 \sin 3x \cos 3x} = \frac{\tan 5x}{2 \sin 3x}$.
The period of $\tan 5x$ is $\frac{\pi}{5}$ and the period of $\sin 3x$ is $\frac{2\pi}{3}$.
The period $\alpha$ of $f(x)$ is the least common multiple of $\frac{\pi}{5}$ and $\frac{2\pi}{3}$,which is $\frac{2\pi}{\gcd(1, 1)} = 2\pi$.
Thus,$\alpha = 2\pi$.
We need to evaluate $f\left(\frac{\alpha}{8}\right) = f\left(\frac{2\pi}{8}\right) = f\left(\frac{\pi}{4}\right)$.
$f\left(\frac{\pi}{4}\right) = \frac{\tan(5\pi/4)}{2 \sin(3\pi/4)} = \frac{1}{2 \times (1/\sqrt{2})} = \frac{1}{\sqrt{2}}$.

(B) function $f(x) = f_1(x) + f_2(x)$ is periodic if both $f_1(x)$ and $f_2(x)$ are periodic and the ratio of their periods is a rational number.
Here,$f_1(x) = \cos(ax)$ has a period $T_1 = \frac{2\pi}{|a|}$ and $f_2(x) = \sin(x)$ has a period $T_2 = 2\pi$.
For $f(x)$ to be periodic,the ratio $\frac{T_1}{T_2} = \frac{2\pi/|a|}{2\pi} = \frac{1}{|a|}$ must be a rational number.
If $\frac{1}{|a|}$ is a rational number,then $|a|$ must be a rational number,which implies $a$ is a rational number.
Therefore,$a$ must be a rational number. Hence,option $B$ is correct.

(C) The period of $\cos(nx)$ is $T_1 = \frac{2\pi}{n}$.
The period of $\sin(x/n)$ is $T_2 = \frac{2\pi}{1/n} = 2n\pi$.
The period of the quotient $\frac{\cos(nx)}{\sin(x/n)}$ is the least common multiple of the periods of the numerator and the denominator.
Given that the period is $4\pi$,we have $2n\pi = 4\pi$.
Solving for $n$,we get $n = 2$.

(B) The value of $k$ is the sum of the squares of the first $20$ natural numbers:
$k = \sum_{n=1}^{20} n^2 = \frac{n(n+1)(2n+1)}{6}$ for $n=20$.
$k = \frac{20 \times 21 \times 41}{6} = 70 \times 41 = 2870$.
We need the period of $f(y) = \tan(2870y) + \sin(2870y)$.
The period of $\tan(ky)$ is $T_1 = \frac{\pi}{k} = \frac{\pi}{2870}$.
The period of $\sin(ky)$ is $T_2 = \frac{2\pi}{k} = \frac{2\pi}{2870} = \frac{\pi}{1435}$.
The period of the sum of two functions is the Least Common Multiple $(LCM)$ of their individual periods.
$LCM\left(\frac{\pi}{2870}, \frac{2\pi}{2870}\right) = \frac{LCM(\pi, 2\pi)}{GCD(2870, 2870)} = \frac{2\pi}{2870} = \frac{\pi}{1435}$.

(A) The period of $f(x) = 3 \sin \frac{\pi x}{3} - \cos \frac{\pi x}{2} + \tan \frac{\pi x}{4}$ is the $LCM$ of the periods of its components. The periods are $T_1 = \frac{2\pi}{\pi/3} = 6$,$T_2 = \frac{2\pi}{\pi/2} = 4$,and $T_3 = \frac{\pi}{\pi/4} = 4$. Thus,$\alpha = \text{LCM}(6, 4, 4) = 12$.
For $\beta$,use the identity $\sin^2 A - \sin^2 B = \sin(A+B)\sin(A-B)$. Here,$\sin^2 \left( \frac{\pi}{7} + \frac{x}{4} \right) - \sin^2 \left( \frac{\pi}{7} - \frac{x}{4} \right) = \sin \left( \frac{2\pi}{7} \right) \sin \left( \frac{x}{2} \right)$. The period is $\beta = \frac{2\pi}{1/2} = 4\pi$.
For $\gamma$,$\cos^4 x + \sin^4 x = 1 - 2 \sin^2 x \cos^2 x = 1 - \frac{1}{2} \sin^2(2x) = 1 - \frac{1}{2} \left( \frac{1 - \cos(4x)}{2} \right) = \frac{3}{4} + \frac{1}{4} \cos(4x)$. The period is $\gamma = \frac{2\pi}{4} = \frac{\pi}{2}$.
Finally,$\frac{\alpha \gamma}{\beta} = \frac{12 \times \frac{\pi}{2}}{4\pi} = \frac{6\pi}{4\pi} = \frac{3}{2}$.

(D) The period of $\sin(a\theta)$ is $\frac{2\pi}{|a|}$ and the period of $\cos(b\theta)$ is $\frac{2\pi}{|b|}$.
For $f(\theta) = \sin \frac{\theta}{3} + \cos \frac{\theta}{2}$, the period of $\sin \frac{\theta}{3}$ is $T_1 = \frac{2\pi}{1/3} = 6\pi$.
The period of $\cos \frac{\theta}{2}$ is $T_2 = \frac{2\pi}{1/2} = 4\pi$.
The period of the sum of two periodic functions is the Least Common Multiple $(LCM)$ of their individual periods.
We need to find the $LCM(6\pi, 4\pi)$.
$LCM(6, 4) = 12$.
Therefore, the period of $f(\theta)$ is $12\pi$.

(C) The period of $\cos(3x + 4)$ is $T_1 = \frac{2\pi}{3}$.
The period of $\tan(2x - 3)$ is $T_2 = \frac{\pi}{2}$.
The period of $\sin(5x)$ is $T_3 = \frac{2\pi}{5}$.
The period $k$ of $f(x)$ is the $L$.$C$.$M$. of $T_1, T_2, T_3$,which is $\text{L.C.M.}\left(\frac{2\pi}{3}, \frac{\pi}{2}, \frac{2\pi}{5}\right) = \frac{\text{L.C.M.}(2\pi, \pi, 2\pi)}{\text{H.C.F.}(3, 2, 5)} = \frac{2\pi}{1} = 2\pi$.
Thus,$k = 2\pi$.
Checking option $C$: $\tan \frac{k}{3} = \tan \frac{2\pi}{3} = -\sqrt{3}$.

(D) Let $f(x) = \frac{\sin x}{\cos 3x} + \frac{\sin 3x}{\cos 9x} + \frac{\sin 9x}{\cos 27x} + \frac{\sin 27x}{\cos 81x}$.
Using the identity $\tan \theta - \tan \phi = \frac{\sin(\theta - \phi)}{\cos \theta \cos \phi}$,we can rewrite each term:
$\frac{\sin x}{\cos 3x} = \frac{1}{2} \frac{\sin(3x - x)}{\cos 3x \cos x} = \frac{1}{2}(\tan 3x - \tan x)$.
Similarly,$\frac{\sin 3x}{\cos 9x} = \frac{1}{2}(\tan 9x - \tan 3x)$,
$\frac{\sin 9x}{\cos 27x} = \frac{1}{2}(\tan 27x - \tan 9x)$,
$\frac{\sin 27x}{\cos 81x} = \frac{1}{2}(\tan 81x - \tan 27x)$.
Summing these,we get $f(x) = \frac{1}{2}(\tan 81x - \tan x)$.
The period of $\tan(kx)$ is $\frac{\pi}{|k|}$.
The period of $\tan 81x$ is $\frac{\pi}{81}$ and the period of $\tan x$ is $\pi$.
The period of the sum is the $L$.$C$.$M$. of the individual periods: $\text{L.C.M.}\left(\frac{\pi}{81}, \pi\right) = \pi$.

(A) To find the period of $f(x)$,we find the periods of the individual trigonometric functions in the numerator and denominator.
Numerator: $2 \sin \left(\frac{\pi x}{3}\right) \cos \left(\frac{2 \pi x}{5}\right) = \sin \left(\frac{\pi x}{3} + \frac{2 \pi x}{5}\right) + \sin \left(\frac{\pi x}{3} - \frac{2 \pi x}{5}\right) = \sin \left(\frac{11 \pi x}{15}\right) - \sin \left(\frac{\pi x}{15}\right)$.
The period of $\sin \left(\frac{11 \pi x}{15}\right)$ is $T_1 = \frac{2 \pi}{11 \pi / 15} = \frac{30}{11}$.
The period of $\sin \left(\frac{\pi x}{15}\right)$ is $T_2 = \frac{2 \pi}{\pi / 15} = 30$.
The period of the numerator is $\text{LCM} \left(\frac{30}{11}, 30\right) = \frac{\text{LCM}(30, 30)}{\text{HCF}(11, 1)} = 30$.
Denominator: The period of $\tan \left(\frac{7 \pi x}{2}\right)$ is $T_3 = \frac{\pi}{7 \pi / 2} = \frac{2}{7}$.
The period of $\sec \left(\frac{5 \pi x}{3}\right)$ is $T_4 = \frac{2 \pi}{5 \pi / 3} = \frac{6}{5}$.
The period of the denominator is $\text{LCM} \left(\frac{2}{7}, \frac{6}{5}\right) = \frac{\text{LCM}(2, 6)}{\text{HCF}(7, 5)} = \frac{6}{1} = 6$.
The period of $f(x)$ is $\text{LCM}(30, 6) = 30$.