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Step-by-step Solution
Step 1: Understand the given function f(x)
We have
$ f(x) = \ln\left(x + \sqrt{x^2 + 1}\right). $
One key property of this function is that
$ f(x) + f(-x) = 0. $
To see why, observe that
$ f(-x) = \ln\left(-x + \sqrt{x^2 + 1}\right). $
Hence,
$ f(x) + f(-x) = \ln\bigl(x + \sqrt{x^2 + 1}\bigr) + \ln\bigl(\sqrt{x^2 + 1} - x\bigr). $
Notice that
$ \bigl(x + \sqrt{x^2 + 1}\bigr)\,\bigl(\sqrt{x^2 + 1} - x\bigr) = (x)(\sqrt{x^2 + 1}) - (x)(x) + (\sqrt{x^2 + 1})^2 - x\sqrt{x^2 + 1} = x^2 + 1 - x^2 = 1. $
This means
$ \ln\bigl(x + \sqrt{x^2 + 1}\bigr) + \ln\bigl(\sqrt{x^2 + 1} - x\bigr) = \ln(1) = 0. $
Thus,
$ f(x) + f(-x) = 0. $
Step 2: Express g(t) in terms of a definite integral
We are given
$ g(t) = \int_{-\pi/2}^{\pi/2} \cos\!\Bigl(\frac{\pi}{4}t + f(x)\Bigr)\, dx. $
Step 3: Use the property f(x) + f(-x)=0 to simplify g(t)
Split the integral from $-\pi/2$ to $0$ and from $0$ to $\pi/2$, then combine terms using the symmetry:
$ g(t) = \int_{-\pi/2}^{\pi/2} \cos\!\Bigl(\frac{\pi}{4}t + f(x)\Bigr)\, dx
= \int_{0}^{\pi/2} \Bigl[\cos\!\bigl(\frac{\pi}{4}t + f(x)\bigr) + \cos\!\bigl(\frac{\pi}{4}t + f(-x)\bigr)\Bigr] dx. $
Since $f(-x) = -\,f(x)$, we have
$ \cos\!\Bigl(\frac{\pi}{4}t + f(-x)\Bigr) = \cos\!\Bigl(\frac{\pi}{4}t - f(x)\Bigr). $
Using the trigonometric identity
$ \cos(\alpha + \beta) + \cos(\alpha - \beta) = 2\,\cos(\alpha)\cos(\beta), $
we get
$ \cos\!\Bigl(\frac{\pi}{4}t + f(x)\Bigr) + \cos\!\Bigl(\frac{\pi}{4}t - f(x)\Bigr)
= 2 \cos\!\bigl(\tfrac{\pi}{4}t\bigr)\cos\!\bigl(f(x)\bigr). $
Hence,
$ g(t) = 2 \int_{0}^{\pi/2} \cos\!\Bigl(\frac{\pi}{4}t\Bigr)\,\cos\!\bigl(f(x)\bigr)\, dx
= 2\,\cos\!\Bigl(\frac{\pi}{4}t\Bigr) \int_{0}^{\pi/2} \cos\!\bigl(f(x)\bigr)\, dx. $
Step 4: Evaluate g(1) and g(0)
1) For $t=1$:
$ g(1) = 2\,\cos\!\Bigl(\frac{\pi}{4}\Bigr)\,\int_{0}^{\pi/2} \cos\!\bigl(f(x)\bigr)\, dx
= 2 \times \frac{\sqrt{2}}{2} \int_{0}^{\pi/2} \cos\!\bigl(f(x)\bigr)\, dx
= \sqrt{2} \int_{0}^{\pi/2} \cos\!\bigl(f(x)\bigr)\, dx. $
2) For $t=0$:
$ g(0) = 2\,\cos\!\Bigl(\frac{\pi}{4}\cdot 0\Bigr)\,\int_{0}^{\pi/2} \cos\!\bigl(f(x)\bigr)\, dx
= 2 \times 1 \times \int_{0}^{\pi/2} \cos\!\bigl(f(x)\bigr)\, dx
= 2 \int_{0}^{\pi/2} \cos\!\bigl(f(x)\bigr)\, dx. $
Step 5: Relate g(1) and g(0)
From the above, we see:
$ g(1) = \sqrt{2} \int_{0}^{\pi/2} \cos\!\bigl(f(x)\bigr)\, dx, $
$ g(0) = 2 \int_{0}^{\pi/2} \cos\!\bigl(f(x)\bigr)\, dx. $
Thus,
$ \sqrt{2}\,g(1) = \sqrt{2} \times \sqrt{2} \int_{0}^{\pi/2} \cos\!\bigl(f(x)\bigr)\, dx
= 2 \int_{0}^{\pi/2} \cos\!\bigl(f(x)\bigr)\, dx
= g(0). $
Therefore, the correct relation is
$ \sqrt{2}\,g(1) = g(0). $
