Periodic functions Questions in English

(A) Let the given expression be $f(\theta) = \left(\tan \theta - \frac{1}{3} \tan^3 \theta\right) \left(\frac{1}{3} - \tan^2 \theta\right)^{-1}$.
$f(\theta) = \frac{\tan \theta - \frac{1}{3} \tan^3 \theta}{\frac{1}{3} - \tan^2 \theta} = \frac{3 \tan \theta - \tan^3 \theta}{3(1 - 3 \tan^2 \theta)} \times 3 = \frac{3 \tan \theta - \tan^3 \theta}{1 - 3 \tan^2 \theta}$.
Using the identity $\tan 3\theta = \frac{3 \tan \theta - \tan^3 \theta}{1 - 3 \tan^2 \theta}$,we get $f(\theta) = \tan 3\theta$.
The period of $\tan x$ is $\pi$. Therefore,the period of $\tan 3\theta$ is $\frac{\pi}{|3|} = \frac{\pi}{3}$.

(C) Given the function $f(x) = e^{\log(\sin x)} + (\tan x)^3 - \operatorname{cosec}(3x - 5)$.
First,simplify the expression: $f(x) = \sin x + (\tan x)^3 - \operatorname{cosec}(3x - 5)$.
Let $f_1(x) = \sin x$,$f_2(x) = (\tan x)^3$,and $f_3(x) = -\operatorname{cosec}(3x - 5)$.
The period of $f_1(x) = \sin x$ is $T_1 = 2\pi$.
The period of $f_2(x) = (\tan x)^3$ is $T_2 = \pi$.
The period of $f_3(x) = -\operatorname{cosec}(3x - 5)$ is $T_3 = \frac{2\pi}{|3|} = \frac{2\pi}{3}$.
The period of $f(x)$ is the $\text{LCM}$ of $(T_1, T_2, T_3) = \text{LCM}(2\pi, \pi, \frac{2\pi}{3})$.
To find the $\text{LCM}$ of fractions,we use $\frac{\text{LCM of numerators}}{\text{HCF of denominators}} = \frac{\text{LCM}(2\pi, \pi, 2\pi)}{\text{HCF}(1, 1, 3)} = \frac{2\pi}{1} = 2\pi$.

(B) The given function is $f(x) = \cos(3x + 5) + 7$.
We know that the period of the function $\cos(ax + b) + c$ is given by the formula $T = \frac{2 \pi}{|a|}$.
In the given expression,$a = 3$.
Therefore,the period $T = \frac{2 \pi}{|3|} = \frac{2 \pi}{3}$.
Thus,the correct option is $B$.

(C) We have the expression $\cos(x + 8x + 27x + \ldots + n^3x)$.
This can be written as $\cos\left(\sum_{k=1}^{n} k^3 x\right) = \cos\left(x \sum_{k=1}^{n} k^3\right)$.
Using the formula for the sum of cubes of the first $n$ natural numbers,$\sum_{k=1}^{n} k^3 = \left(\frac{n(n+1)}{2}\right)^2 = \frac{n^2(n+1)^2}{4}$.
Thus,the expression becomes $\cos\left(\frac{n^2(n+1)^2}{4} x\right)$.
The period of $\cos(kx)$ is $\frac{2\pi}{|k|}$.
Here,$k = \frac{n^2(n+1)^2}{4}$.
Therefore,the period is $\frac{2\pi}{\frac{n^2(n+1)^2}{4}} = \frac{8\pi}{n^2(n+1)^2}$.

(D) Given the function $f(x) = \sin 5x \cos 3x$.
Using the trigonometric identity $2 \sin A \cos B = \sin(A+B) + \sin(A-B)$,we can rewrite the function as:
$f(x) = \frac{1}{2} [\sin(5x+3x) + \sin(5x-3x)] = \frac{1}{2} (\sin 8x + \sin 2x)$.
The period of $\sin 8x$ is $T_1 = \frac{2\pi}{8} = \frac{\pi}{4}$.
The period of $\sin 2x$ is $T_2 = \frac{2\pi}{2} = \pi$.
The period of the sum of two periodic functions is the Least Common Multiple ($L$.$C$.$M$.) of their individual periods.
$\alpha = \text{L.C.M.}\left(\frac{\pi}{4}, \pi\right) = \frac{\text{L.C.M.}(\pi, \pi)}{\text{H.C.F.}(4, 1)} = \frac{\pi}{1} = \pi$.
Thus,$\alpha = \pi$.
Finally,$\cos \alpha = \cos \pi = -1$.

(D) Given function is $f(x) = \cos \left(\frac{x}{3}\right) + \sin \left(\frac{x}{2}\right)$.
We know that the period of $\cos(ax)$ is $\frac{2 \pi}{|a|}$ and the period of $\sin(bx)$ is $\frac{2 \pi}{|b|}$.
For the first term $\cos \left(\frac{x}{3}\right)$, the period $T_1 = \frac{2 \pi}{1/3} = 6 \pi$.
For the second term $\sin \left(\frac{x}{2}\right)$, the period $T_2 = \frac{2 \pi}{1/2} = 4 \pi$.
The period of the sum of two periodic functions is the Least Common Multiple $(LCM)$ of their individual periods.
Therefore, the period of $f(x) = \text{LCM}(6 \pi, 4 \pi)$.
Since $6 \pi = 2 \times 3 \pi$ and $4 \pi = 2 \times 2 \pi$, the $\text{LCM}(6 \pi, 4 \pi) = 12 \pi$.
Thus, the period of $f(x)$ is $12 \pi$.

(D) The function is given by $f(x) = 7 + \cos(5x + 3)$.
We know that the fundamental period of the function $\cos(ax + b)$ is given by $\frac{2\pi}{|a|}$.
Here,$a = 5$.
Therefore,the period of $\cos(5x + 3)$ is $\frac{2\pi}{|5|} = \frac{2\pi}{5}$.
Since adding a constant $7$ to the function does not change its period,the period of $f(x)$ is $\frac{2\pi}{5}$.

(D) Let $f(x) = \sin ^4 x + \cos ^4 x$.
We know that $\sin ^4 x + \cos ^4 x = (\sin ^2 x + \cos ^2 x)^2 - 2 \sin ^2 x \cos ^2 x$.
Since $\sin ^2 x + \cos ^2 x = 1$,we have $f(x) = 1 - 2 \sin ^2 x \cos ^2 x$.
Using the identity $\sin 2x = 2 \sin x \cos x$,we can write $2 \sin ^2 x \cos ^2 x = \frac{1}{2} (2 \sin x \cos x)^2 = \frac{1}{2} \sin ^2 2x$.
Thus,$f(x) = 1 - \frac{1}{2} \sin ^2 2x$.
Using the identity $\sin ^2 \theta = \frac{1 - \cos 2\theta}{2}$,we get $f(x) = 1 - \frac{1}{2} \left( \frac{1 - \cos 4x}{2} \right) = 1 - \frac{1}{4} + \frac{\cos 4x}{4} = \frac{3}{4} + \frac{1}{4} \cos 4x$.
The period of $\cos(kx)$ is $\frac{2\pi}{|k|}$.
Therefore,the period of $f(x) = \frac{3}{4} + \frac{1}{4} \cos 4x$ is $\frac{2\pi}{4} = \frac{\pi}{2}$.

(B) function $f(x)$ is periodic if there exists a positive constant $T$ such that $f(x+T) = f(x)$ for all $x$ in the domain.
For $f(x) = x + \sin 2x$:
As $x \to \infty$,$f(x) \to \infty$,because $x$ is a strictly increasing function and $\sin 2x$ is bounded between $-1$ and $1$. Thus,$f(x+T) = x + T + \sin(2(x+T)) = x + \sin 2x$ cannot hold for any $T > 0$. Therefore,$x + \sin 2x$ is not a periodic function.
For $g(x) = \cos (\sqrt{x}+1)$:
As $x$ increases,the argument $\sqrt{x}+1$ increases,but the rate of change of the argument decreases. For a function to be periodic,the values must repeat at regular intervals. Since the distance between consecutive zeros of $\cos (\sqrt{x}+1)$ is not constant,it is not a periodic function.
Thus,both statements $B$ and $D$ are correct. Given the structure,we identify that $x + \sin 2x$ is not periodic and $\cos (\sqrt{x}+1)$ is not periodic.

(D) function $f$ is periodic with period $T$ if $f(x+T) = f(x)$ for all $x$.
For the sum $f+g$ to be periodic,there must exist a constant $T > 0$ such that $(f+g)(x+T) = (f+g)(x)$.
This implies $f(x+T) + g(x+T) = f(x) + g(x)$.
If $\frac{T_{1}}{T_{2}}$ is a rational number,say $\frac{T_{1}}{T_{2}} = \frac{p}{q}$ where $p, q \in \mathbb{Z}^{+}$,then $qT_{1} = pT_{2} = T$.
In this case,$f(x+T) = f(x)$ and $g(x+T) = g(x)$,so $(f+g)(x+T) = f(x) + g(x)$.
Thus,$f+g$ is periodic if the ratio of their periods is a rational number.

(C) function $f(x) = \sin(px) + \cos(qx)$ is periodic if and only if the ratio of the periods of the two components is a rational number.
The period of $\sin x$ is $T_1 = 2\pi$.
The period of $\cos(ax)$ is $T_2 = \frac{2\pi}{|a|}$ (for $a \neq 0$).
For the sum to be periodic,the ratio $\frac{T_1}{T_2}$ must be a rational number.
$\frac{T_1}{T_2} = \frac{2\pi}{2\pi / |a|} = |a|$.
Thus,$|a|$ must be a rational number.
Therefore,$a$ must be a rational number.

(C) We have the function $f(x) = \cos(x^2)$.
For a function to be periodic with period $T > 0$,it must satisfy $f(x+T) = f(x)$ for all $x$ in its domain.
Substituting the function,we get $\cos((x+T)^2) = \cos(x^2)$.
This implies $(x+T)^2 = x^2 + 2n\pi$ or $(x+T)^2 = -(x^2) + 2n\pi$ for some integer $n$.
Expanding the left side,$x^2 + 2xT + T^2 = x^2 + 2n\pi$,which simplifies to $2xT + T^2 = 2n\pi$.
For this to hold for all $x$,the coefficient of $x$ must be zero,which implies $T=0$. However,the period $T$ must be a positive constant.
Since no such $T > 0$ exists,the function $f(x) = \cos(x^2)$ is not periodic.

(A) The period of $\cos(ax)$ is $\frac{2\pi}{|a|}$. Thus,the period of $\cos 4x$ is $T_1 = \frac{2\pi}{4} = \frac{\pi}{2}$.
The period of $\tan(bx)$ is $\frac{\pi}{|b|}$. Thus,the period of $\tan 3x$ is $T_2 = \frac{\pi}{3}$.
The period of the sum of two periodic functions is the least common multiple $(LCM)$ of their individual periods.
We need to find $\text{LCM}\left(\frac{\pi}{2}, \frac{\pi}{3}\right)$.
To find the $LCM$ of fractions $\frac{a}{b}$ and $\frac{c}{d}$,we use the formula $\frac{\text{LCM}(a, c)}{\text{HCF}(b, d)}$.
Here,$\text{LCM}(\pi, \pi) = \pi$ and $\text{HCF}(2, 3) = 1$.
Therefore,the period is $\frac{\pi}{1} = \pi$.