Compute $A^TA = 3I_3$ by equating columns to be orthogonal with magnitude $\sqrt{3}$.

Column 1 norm$^2$: $0+(2x)^2+(2x)^2=8x^2=3 \Rightarrow x^2=\frac{3}{8}$: 2 values of $x$.

Column 2 norm$^2$: $4y^2+y^2+y^2=6y^2=3 \Rightarrow y^2=\frac{1}{2}$: 2 values of $y$.

With $x\neq y$: we need to discard cases where $x=y$. Since $x^2=3/8$ and $y^2=1/2$, $x\neq\pm y$ (different magnitudes), so all $2\times2=4$ combinations are valid.

Total $= \mathbf{4}$

Let $v = x - y$, so $\frac{dv}{dx} = 1 - \frac{dy}{dx} = 1 - v^2$.

Separate: $\frac{dv}{1-v^2} = dx \Rightarrow \frac{1}{2}\ln\left|\frac{1+v}{1-v}\right| = x + C$

i.e. $\ln\left|\frac{1+x-y}{1-x+y}\right| = 2x + C'$

At $(1,1)$: $v=0$, $\ln(1)=2+C' \Rightarrow C'=-2$

Final: $-\ln\left|\dfrac{1-x+y}{1+x-y}\right| = 2(x-1)$ which matches option B.

Since $\log 2 \approx 0.693$ and $\sin 0=0$, we have $f(0)=\log 2 > 0$.

Near $x=0$: $f(x)=\log 2-\sin x$ (since $\log 2>\sin x$ locally). So $f$ is smooth near $x=0$.

$g(x) = f(f(x)) = |\log 2 - \sin(f(x))|$. At $x=0$: $f(0)=\log 2$, so $\sin(f(0))=\sin(\log 2)$. Since $\log 2 < 1 < \pi$, $\sin(\log 2)>0$.

$g(0) = \log 2 - \sin(\log 2) > 0$, so near $x=0$, $g(x)=\log 2 - \sin(f(x))$.

$g'(0)=-\cos(f(0))\cdot f'(0)=-\cos(\log 2)\cdot(-\cos 0)=\cos(\log 2)$

Ketones contain a carbonyl group (C=O) bonded to two carbon atoms. Aldehydes have -CHO (formyl group). Carboxyl = -COOH, amino = -NH₂.