Suppose f and g are two functions such that f,g: R `->`R `f(x)=ln(1+sqrt(1+x^2))` and `g(x)=ln(x+sqrt(1+x^2))` then find the value of `xe^(g(x))f(1/x)+g'(x)` at `x=1`

Options

A. `1`
B. `0`
C. `2`
D. `-1`

Answer

To solve the problem, we need to find the value of \( xe^{g(x)}f\left(\frac{1}{x}\right) + g'(x) \) at \( x = 1 \). Let's break this down step by step.

### Step 1: Define the functions
We have:
\[ f(x) = \ln(1 + \sqrt{1 + x^2}) \]
\[ g(x) = \ln(x + \sqrt{1 + x^2}) \]

### Step 2: Find \( g'(x) \)
To find \( g'(x) \), we differentiate \( g(x) \):
\[ g(x) = \ln(x + \sqrt{1 + x^2}) \]
Using the chain rule:
\[ g'(x) = \frac{1}{x + \sqrt{1 + x^2}} \cdot \left(1 + \frac{x}{\sqrt{1 + x^2}}\right) \]
Simplifying:
\[ g'(x) = \frac{1 + \frac{x}{\sqrt{1 + x^2}}}{x + \sqrt{1 + x^2}} = \frac{\sqrt{1 + x^2} + x}{(x + \sqrt{1 + x^2})\sqrt{1 + x^2}} = \frac{1}{\sqrt{1 + x^2}} \]

### Step 3: Find \( g(1) \) and \( g'(1) \)
Now, we calculate \( g(1) \):
\[ g(1) = \ln(1 + \sqrt{1 + 1^2}) = \ln(1 + \sqrt{2}) = \ln(1 + \sqrt{2}) \]
And \( g'(1) \):
\[ g'(1) = \frac{1}{\sqrt{1 + 1^2}} = \frac{1}{\sqrt{2}} \]

### Step 4: Find \( f\left(\frac{1}{x}\right) \)
Next, we need to find \( f\left(\frac{1}{1}\right) = f(1) \):
\[ f(1) = \ln(1 + \sqrt{1 + 1^2}) = \ln(1 + \sqrt{2}) = \ln(1 + \sqrt{2}) \]

### Step 5: Calculate \( xe^{g(x)}f\left(\frac{1}{x}\right) + g'(x) \) at \( x = 1 \)
Now we can substitute \( x = 1 \):
\[ xe^{g(x)}f\left(\frac{1}{x}\right) + g'(x) = 1 \cdot e^{g(1)} f(1) + g'(1) \]
Substituting the values we found:
\[ = e^{\ln(1 + \sqrt{2})} \cdot \ln(1 + \sqrt{2}) + \frac{1}{\sqrt{2}} \]
Using the property \( e^{\ln(a)} = a \):
\[ = (1 + \sqrt{2}) \cdot \ln(1 + \sqrt{2}) + \frac{1}{\sqrt{2}} \]

### Final Answer
Thus, the final value is:
\[ (1 + \sqrt{2}) \ln(1 + \sqrt{2}) + \frac{1}{\sqrt{2}} \]

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