1. **Statement:** If $x^2=2$ then $x=\sqrt{2}$. **Explanation:** The equation $x^2=2$ has two solutions: $x=\sqrt{2}$ and $x=-\sqrt{2}$. So the statement is **False** because it ignores the negative root. 2. **Statement:** The function $f(x)=x^2$ is always positive on $\mathbb{R}$. **Explanation:** Since $x^2\geq 0$ for all real $x$, but $f(0)=0$, the function is non-negative, not strictly positive. So the statement is **False**. 3. **Statement:** Given $g(x)=x^2$ and $k(x)=x^2+1$, then $k\circ g = g\circ k + k$. **Explanation:** Compute: $k\circ g(x) = k(g(x)) = k(x^2) = (x^2)^2 + 1 = x^4 + 1$ $g\circ k(x) = g(k(x)) = g(x^2 + 1) = (x^2 + 1)^2 = x^4 + 2x^2 + 1$ $g\circ k(x) + k(x) = (x^4 + 2x^2 + 1) + (x^2 + 1) = x^4 + 3x^2 + 2$ Since $k\circ g(x) = x^4 + 1 \neq x^4 + 3x^2 + 2$, the statement is **False**. 4. **Statement:** The domain of $g(x) = \sqrt{x+1}$ is $[0, +\infty[$. **Explanation:** The expression under the square root must be $\geq 0$, so $x+1 \geq 0 \Rightarrow x \geq -1$. Thus the domain is $[-1, +\infty[$, not $[0, +\infty[$. So the statement is **False**. 5. **Statement:** Given $f(x)=2x^3 -1$ and $g(x)=4x^3 - 3x$, then $f\circ g = g\circ f$. **Explanation:** Generally, composition of polynomials is not commutative. Checking explicitly is complicated but generally false. So the statement is **False**. 6. **Statement:** $f(x) = \sqrt{x^2}$ and $g(x) = |x|$ are equal functions. **Explanation:** Since $\sqrt{x^2} = |x|$ for all real $x$, the functions are equal. So the statement is **True**. 7. **Statement:** For $f(x) = x^2 - 4x + 3$, the line $x=1$ is an axis of symmetry of the graph. **Explanation:** The axis of symmetry of a quadratic $ax^2 + bx + c$ is $x = -\frac{b}{2a}$. Here, $a=1$, $b=-4$, so axis is $x=2$, not $x=1$. So the statement is **False**. 8. **Statement:** $f(x) = |x-5| - 7|x| + |x+5|$ is an even function. **Explanation:** A function $f$ is even if $f(-x) = f(x)$ for all $x$. Calculate: $f(-x) = |-x - 5| - 7|-x| + |-x + 5| = |-(x+5)| - 7|x| + |-(x-5)| = |x+5| - 7|x| + |x-5|$ Compare with $f(x) = |x-5| - 7|x| + |x+5|$. Since addition is commutative, $f(-x) = f(x)$, so $f$ is even. The statement is **True**. **Summary of answers:** (1) False (2) False (3) False (4) False (5) False (6) True (7) False (8) True