1. Given the problem: If $x + \frac{1}{x} = 11$, find the value of $x^2 + \frac{1}{x^2}$. 2. Start by squaring both sides of the equation to involve $x^2$ and $\frac{1}{x^2}$: $$\left(x + \frac{1}{x}\right)^2 = 11^2$$ 3. Expand the left side using the identity $(a+b)^2 = a^2 + 2ab + b^2$: $$x^2 + 2 \cdot x \cdot \frac{1}{x} + \frac{1}{x^2} = 121$$ 4. Simplify the middle term, since $x \cdot \frac{1}{x} = 1$: $$x^2 + 2 + \frac{1}{x^2} = 121$$ 5. To isolate $x^2 + \frac{1}{x^2}$, subtract 2 from both sides: $$x^2 + \frac{1}{x^2} = 121 - 2 = 119$$ 6. Therefore, the value of $x^2 + \frac{1}{x^2}$ is $119$. Final answer: $119$ (option b).