$\mathrm{Let}{\mathrm{ x}}_1\left(t\right)=\cos \left(2\pi nt\right)\text{and}{\mathrm{ x}}_2\left(t\right)=2\sin \left(4\pi nt\right)\text{represent two sinusoids for a positive integer}n.$

$\text{Which of the following statements about }{x}_1\left(t\right)\text{and}{x}_2\left(t\right)\text{is/are valid?}$

The correct option is:
$x_1\left(t\right)\text{and}{x}_2\left(t\right)\text{are non-orthogonal to each other over}0\le t<\frac{1}{2n}$

Step-by-Step Explanation:

Two real-valued signals, $x_1\left(t\right)$ and $x_2\left(t\right)$ , are orthogonal to each other over a time interval $t_a\le t<t_b$ if their inner product (the integral of their product over the interval) is equal to zero:
${\int }_{t_a}^{t_b}{x}_1\left(t\right)x_2\left(t\right)dt=0$
If the integral is non-zero, the signals are non-orthogonal.

Let us evaluate the product of the two given sinusoids:
$x_1\left(t\right)=\cos \left(2\pi nt\right)$
$x_2\left(t\right)=2\sin \left(4\pi nt\right)$

Using the product-to-sum trigonometric identity $2\sin \left(A\right)\cos \left(B\right)=\sin (A+B)+\sin (A-B)$ , where we let $A=4\pi nt$ and $B=2\pi nt$ , we obtain:
$x_1\left(t\right)x_2\left(t\right)=2\sin \left(4\pi nt\right)\cos \left(2\pi nt\right)=\sin \left(6\pi nt\right)+\sin \left(2\pi nt\right)$

Now, we calculate the integral of this product over the interval $0\le t<\frac{1}{2n}$ :
$I={\int }_0^{\frac{1}{2n}}\left[\sin \left(6\pi nt\right)+\sin \left(2\pi nt\right)\right]dt$

Integrating each term individually gives:
$I={\left[-\frac{\cos \left(6\pi nt\right)}{6\pi n}-\frac{\cos \left(2\pi nt\right)}{2\pi n}\right]}_0^{\frac{1}{2n}}$

Substitute the upper limit $t=\frac{1}{2n}$ :
$\cos \left(6\pi n\cdot \frac{1}{2n}\right)=\cos \left(3\pi \right)=-1$
$\cos \left(2\pi n\cdot \frac{1}{2n}\right)=\cos \left(\pi \right)=-1$

Substitute the lower limit $t=0$ :
$\cos \left(0\right)=1$

Evaluating the integral:
$I=\left(-\frac{-1}{6\pi n}-\frac{-1}{2\pi n}\right)-\left(-\frac{1}{6\pi n}-\frac{1}{2\pi n}\right)$
$I=\left(\frac{1}{6\pi n}+\frac{1}{2\pi n}\right)+\left(\frac{1}{6\pi n}+\frac{1}{2\pi n}\right)$
$I=\frac{2}{6\pi n}+\frac{2}{2\pi n}=\frac{1}{3\pi n}+\frac{1}{\pi n}=\frac{4}{3\pi n}$

Since $n>0$ , the integral value $\frac{4}{3\pi n}$ is strictly non-zero. Therefore, $x_1\left(t\right)$ and $x_2\left(t\right)$ are indeed non-orthogonal to each other over the interval $0\le t<\frac{1}{2n}$ .