Question 6

(a) The asymptotes pass through the centre with gradient $\pm\dfrac32$: $y=\pm\tfrac32(x+2)$.

(b) $y=\tfrac32(x+2)+\dfrac{\lambda-12}{2(x+2)}$, so

$\begin{aligned}&\frac{dy}{dx}\\ & =\frac32-\frac{\lambda-12}{2(x+2)^2}.\end{aligned}$

For $\dfrac{dy}{dx}>0$ we need $3(x+2)^2-(\lambda-12)>0$ for every $x\ne-2$. Since $(x+2)^2$ can be arbitrarily close to 0, this requires $-(\lambda-12)\ge0$, and $\lambda\ne12$, so $\lambda<12$.

Method 2 (the source’s quotient-rule alternative): $\begin{aligned}&\frac{dy}{dx}\\ & =\frac{2(x+2)(6x+12)-2(3x^2+12x+\lambda)}{[2(x+2)]^2}\\ & =\frac{3x^2+12x+24-\lambda}{2(x+2)^2}.\end{aligned}$ The numerator is $3(x+2)^2+12-\lambda$. For $\lambda<12$ it is positive everywhere, equivalently its discriminant is $12^2-4(3)(24-\lambda)=12(\lambda-12)<0$. The source’s line “for all $x\ne2$” should read $x\ne-2$. The excluded case $\lambda=12$ would also have positive derivative on its domain. Here “strictly increasing” means separately on $(-\infty,-2)$ and $(-2,\infty)$, as in the printed derivative argument: across the gap, values on the left tend to $+\infty$ while values on the right tend to $-\infty$, so there is no global increasing order on the disconnected domain.