Main answer:The wavefunction of a particle is normalized if the probability of finding the particle within the region of space that the wavefunction describes is equal to 1. We will begin by demonstrating that the wavefunction is normalized, as requested. The given wavefunction is \[\psi(x) = \frac{1}{\sqrt{a}}\cos\frac{\pi x}{a}.\]Since the wavefunction is real, the integral to be solved is as follows:\[\int_{-\infty}^\infty \psi(x)^2 \, dx = \int_{-a/2}^{a/2} \psi(x)^2 \, dx,\]where we used the symmetry of the wavefunction to limit the integration region to [-a/2, a/2]. So, the integral is:\[\int_{-a/2}^{a/2} \psi(x)^2 \, dx = \int_{-a/2}^{a/2} \frac{1}{a} \cos^2\frac{\pi x}{a} \, dx.\]We know that \[\cos^2\theta = \frac{1}{2}\left(1+\cos 2\theta\right),\]so we can use this identity to simplify the integrand, which results in\[\int_{-a/2}^{a/2} \psi(x)^2 \, dx = \frac{1}{2}+\frac{1}{2}\int_{-a/2}^{a/2} \cos\frac{2\pi x}{a} \, dx.\]By taking the integral from -a/2 to a/2 of the cos function, we can get\[\int_{-a/2}^{a/2} \cos\frac{2\pi x}{a} \, dx = \frac{a}{2\pi}\left[\sin\frac{2\pi x}{a}\right]_{-a/2}^{a/2} = 0.\]Thus, we obtain\[\int_{-a/2}^{a/2} \psi(x)^2 \, dx = \frac{1}{2}+\frac{1}{2}(0) = 1.\]So, the wavefunction is indeed normalized. To find the value of x for the particle, we need to find the maximum of the probability density, which is given by\[\rho(x) = \psi(x)^2 = \frac{1}{a}\cos^2\frac{\pi x}{a}.\]
The maximum occurs at x = a/4 and x = 3a/4, so the particle is equally likely to be found at either of these points. Finally, to calculate the uncertainty in the position of the particle, we need to evaluate\[\Delta x = \sqrt{\langle x^2\rangle - \langle x\rangle^2},\]where\[\langle x^2\rangle = \int_{-\infty}^\infty x^2 \psi(x)^2 \, dx = \frac{a^2}{3},\]and\[\langle x\rangle = \int_{-\infty}^\infty x \psi(x)^2 \, dx = \frac{a}{2}.\]Thus, the uncertainty in position is\[\Delta x = \sqrt{\frac{a^2}{3} - \left(\frac{a}{2}\right)^2} = \frac{a}{2\sqrt{3}}.\]Answer in more than 100 words:The given wave function \[\psi(x) = \frac{1}{\sqrt{a}}\cos\frac{\pi x}{a}\]is properly normalized. We showed that by demonstrating that the probability of finding the particle within the region of space described by the wave function is equal to 1. We did this by evaluating the integral\[\int_{-\infty}^\infty \psi(x)^2 \, dx,\]which reduced to\[\int_{-a/2}^{a/2} \frac{1}{a} \cos^2\frac{\pi x}{a} \, dx.\]By using the identity \[\cos^2\theta = \frac{1}{2}\left(1+\cos 2\theta\right),\]we were able to simplify the integrand to\[\frac{1}{2}+\frac{1}{2}\int_{-a/2}^{a/2} \cos\frac{2\pi x}{a} \, dx.\]However, we found that the integral of the cos function over this range is 0, so we concluded that the integral evaluating the probability of finding the particle within the region of space described by the wave function is indeed equal to 1. The wave function describes a particle in a one-dimensional box of length a.
To find the value of x for the particle, we needed to find the maximum of the probability density, which is given by\[\rho(x) = \psi(x)^2 = \frac{1}{a}\cos^2\frac{\pi x}{a}.\]We found that the maximum occurs at x = a/4 and x = 3a/4, so the particle is equally likely to be found at either of these points. Finally, we calculated the uncertainty in the position of the particle using the formula\[\Delta x = \sqrt{\langle x^2\rangle - \langle x\rangle^2},\]where\[\langle x^2\rangle = \int_{-\infty}^\infty x^2 \psi(x)^2 \, dx\]and\[\langle x\rangle = \int_{-\infty}^\infty x \psi(x)^2 \, dx.\]We found that the uncertainty in position is given by\[\Delta x = \sqrt{\frac{a^2}{3} - \left(\frac{a}{2}\right)^2} = \frac{a}{2\sqrt{3}}.\]Conclusion:In conclusion, we have shown that the given wave function is properly normalized, which means that the probability of finding the particle within the region of space that the wave function describes is equal to 1. We have also found that the particle is equally likely to be found at x = a/4 and x = 3a/4, and we have calculated the uncertainty in the position of the particle, which is given by\[\Delta x = \frac{a}{2\sqrt{3}}.\]