Differentiation: RI 2024 Promo Q1

Method 1

$$\frac{\,\mathrm{d} y}{\,\mathrm{d} x}=a+\frac bx-\frac c{x^2}.$$ At $x=\dfrac32$, $\dfrac{\,\mathrm{d} y}{\,\mathrm{d} x}=0\Rightarrow a+\dfrac23b-\dfrac49c=0\quad----(1)$

The gradient of $D$ at $x=1$ is equal to the gradient of the line $y=x-5$. $$\text{At }x=1,\quad \frac{\,\mathrm{d} y}{\,\mathrm{d} x}=1\Rightarrow a+b-c=1\quad----(2)$$ $y$-coordinate of $D$ at $x=1$ is $1-5=-4$.

Substituting $x=1$ and $y=-4$ into equation of $D$, we get $$a+c=-4\quad----(3)$$ From GC, $a=2$, $b=-7$, $c=-6$.

Method 2

$$\frac{\,\mathrm{d} y}{\,\mathrm{d} x}=a+\frac bx-\frac c{x^2}.$$ At $x=\dfrac32$, $\dfrac{\,\mathrm{d} y}{\,\mathrm{d} x}=0\Rightarrow a+\dfrac23b-\dfrac49c=0\quad----(1)$

Gradient of $D$ at $x=1$ is $\left.\dfrac{\,\mathrm{d} y}{\,\mathrm{d} x}\right|_{x=1}=a+b-c$.

$y$-coordinate of $D$ at $x=1$ is $a+c$.

So, the equation of tangent to $D$ at $x=1$ is $$y-(a+c)=(a+b-c)(x-1)\Rightarrow y=(a+b-c)x-b+2c.$$ Comparing this line with $y=x-5$, we get $$a+b-c=1\quad----(2),\qquad -b+2c=-5\quad----(3).$$ From GC, $a=2$, $b=-7$, $c=-6$.