2020-M2-09

Let $y=mx+c$ be an oblique asymptote.

$\begin{array}{rcl}
m & = & \dlim_{x\to\infty} \dfrac{f(x)}{x} \\
m & = & \dlim_{x\to\infty} \dfrac{(x+4)^3}{x(x-4)^2} \\
m & = & \dlim_{x\to\infty} \dfrac{(x+4)^3/x^3}{x(x-4)^2/x^3} \\
m & = & \dlim_{x\to\infty} \dfrac{(1+\frac{4}{x})^3}{(1-\frac{4}{x})^2} \\
m & = & \dfrac{(1+0)^3}{(1-0)^2} \\
m & = & 1
\end{array}$

Hence, we have

$\begin{array}{rcl}
c & = & \dlim_{x\to\infty} (f(x)-mx) \\
c & = & \dlim_{x\to\infty} \left(\dfrac{(x+4)^3}{(x-4)^2}-x\right) \\
c & = & \dlim_{x\to\infty} \dfrac{(x+4)^3-x(x-4)^2}{(x-4)^2} \\
c & = & \dlim_{x\to\infty} \dfrac{x^3+3x^2(4)+3x(4)^2+(4)^3-x(x^2-8x+16)}{(x-4)^2} \\
c & = & \dlim_{x\to\infty} \dfrac{x^3+12x^2+48x+64-x^3+8x^2-16x}{(x-4)^2} \\
c & = & \dlim_{x\to\infty} \dfrac{20x^2-32x+64}{(x-4)^2} \\
c & = & \dlim_{x\to\infty} \dfrac{(20x^2-32x+64)/x^2}{(x-4)^2/x^2} \\
c & = & \dlim_{x\to\infty} \dfrac{20-\frac{32}{x}+\frac{64}{x^2}}{(1-\frac{4}{x})^2} \\
c & = & \dfrac{20-0+0}{(x-0)^2} \\
c & = & 20
\end{array}$

Therefore, the equation of the oblique asymptote is $y=x+20$.

$\begin{array}{rcl}
f'(x) & = & 0 \\
\dfrac{(x+4)^2(x-20)}{(x-4)^3} & = & 0 \\
(x+4)^2(x-20) & = & 0
\end{array}$

$\therefore x=-4$ (repeated) or $x=20$.

$\begin{array}{|l|c|c|c|c|c|c|c|} \hline
x & x < -4 & x=-4 & -4 < x < 4 & 4 < x < 20 & x = 20 & x > 20 \\ \hline
f'(x) & +ve & 0 & +ve & -ve & 0 & +ve \\ \hline
f(x) & \text{increasing} & — & \text{increasing} & \text{decreasing} & \text{min. point} & \text{increasing} \\ \hline
\end{array}$

Therefore, there is only one turning point.

Hence, the claim is disagreed.

$\begin{array}{rcl}
f”(x) & = & 0 \\
\dfrac{384(x+4)}{(x-4)^4} & = & 0 \\
x+4 & = & 0
\end{array}$

$\therefore x=-4$.

$\begin{array}{|l|c|c|c|c|} \hline
x & x< -4 & x= -4 & -4 < x < 4 & x > 4 \\ \hline
f”(x) & -ve & 0 & +ve & +ve \\ \hline
f(x) & \text{concave downwards} & \text{pt of inflexion} & \text{concave upwards} & \text{concave upwards} \\ \hline
\end{array}$

$\therefore$ the point of inflexion of $H$ is $(-4,0)$.

$\begin{array}{cl}
= & \dint_{-4}^0 \dfrac{(x+4)^3}{(x-4)^2} dx
\end{array}$

Let $u=x-4$, then $du=dx$.

When $x=-4$, $u=-8$.

When $x=0$, $u=-4$.

Hence, the required area

$\begin{array}{cl}
= & \dint_{-8}^{-4} \dfrac{(u+8)^3}{u^2} du \\
= & \dint_{-8}^{-4} \dfrac{u^3+3u^2(8)+3u(8)^2+(8)^3}{u^2} du \\
= & \dint_{-8}^{-4} \left(u+24+\dfrac{192}{u}+\dfrac{512}{u^2}\right) du \\
= & \left[ \dfrac{1}{2}u^2+24u+192\ln |u| -\dfrac{512}{u} \right]_{-8}^{-4} \\
= & \dfrac{1}{2}(-4)^2+24(-4)+192\ln|-4|-\dfrac{512}{-4} -\dfrac{1}{2}(-8)^2-24(-8)-192\ln |-8|+\dfrac{512}{-8} \\
= & 136+192\ln 4-192\ln 8 \\
= & 136+192\ln (2^2) -192 \ln (2^3) \\
= & 136+382\ln 2-576\ln 2 \\
= & 136-192\ln 2
\end{array}$