which one of the following statements is false for these vector fields?

Question

If f is a scalar field and F is a vector field, Which one of the following statements is false?

1. $$\nabla^2(fg) = f\nabla^2g + g\nabla^2f + 2\nabla f \cdot \nabla g$$ for any twice differentiable scalar fields f and g

2. There is a vector field $\mathbf{F}$ such that $\nabla \times \mathbf{F} = (x,y,z)$

3. Any vector field of the form $\mathbf{F}(x,y,z)= (f(y,z), g(x,z), h(x,y))$ is solenoidal

4. Any vector field of the form $\mathbf{F}(x,y,z)= (f(x), g(x), h(x))$ is irrotational

Answer 1

1 is true by just applying the product rules for divergence and gradient: $$\nabla^2(fg)=\nabla\cdot(\nabla(fg))$$ $$=\nabla\cdot[f\nabla g+g\nabla f]$$ $$=f\nabla^2g+g\nabla^2f+\nabla f\cdot \nabla g+\nabla g\cdot\nabla f $$ $$=f\nabla^2g+g\nabla^2f+2\nabla f\cdot \nabla g$$

2 is false. This is because we know that: $$\nabla\cdot(\nabla \times \mathbf{F})=0$$ And: $$\nabla\cdot\langle x,y,z\rangle=3$$ 3 is also true. A function is solenoidal if it has zero divergence: $$\nabla\cdot\langle f(y,z),g(x,z),h(x,y) \rangle=\frac{\partial}{\partial x}f(y,z)+\frac{\partial}{\partial y}g(x,z)+\frac{\partial}{\partial z}h(x,y)$$ $$=0$$ Lastly, 4 is true is as well. A vector field is irrotational if it has zero curl: $$\nabla\times \mathbf{F}=\langle \frac{\partial}{\partial y}h(z)-\frac{\partial}{\partial z}g(y), \frac{\partial}{\partial z}f(x)-\frac{\partial}{\partial x}h(z), \frac{\partial}{\partial x}g(y)-\frac{\partial}{\partial y}f(x) \rangle$$ Which is clearly equal to the zero vector, therefore there is no curl.

If you did not make a typo on the last question then this question is also false and we have: $$\nabla\times\mathbf{F}=\langle \frac{\partial}{\partial y}h(x)-\frac{\partial}{\partial z}g(x), \frac{\partial}{\partial z}f(x)-\frac{\partial}{\partial x}h(x), \frac{\partial}{\partial x}g(x)-\frac{\partial}{\partial y}f(x) \rangle$$ $$=\langle 0,-h'(x),g'(x) \rangle$$ Which is not zero.