سوال ۲۸ کنکور ارشد علوم کامپیوتر ۱۴۰۵

Concavity of a parametric curve. For $x=x(t),\;y=y(t)$ the quantity that governs concavity is

$\frac{d^{2}y}{dx^{2}}=\frac{\dfrac{d}{dt}\!\left(\dfrac{dy}{dx}\right)}{\dfrac{dx}{dt}},$

not $\ddot{y}/\ddot{x}$.

First derivatives. From $x(t)=t^{2}+4$ and $y(t)=t^{3}-6t^{2}+5t$,

$\dot{x}=2t,\qquad \dot{y}=3t^{2}-12t+5,$

$\frac{dy}{dx}=\frac{\dot{y}}{\dot{x}}=\frac{3t^{2}-12t+5}{2t}=\frac{3t}{2}-6+\frac{5}{2t}.$

Second derivative. Differentiate with respect to $t$, then divide by $\dot{x}=2t$:

$\frac{d}{dt}\!\left(\frac{dy}{dx}\right)=\frac{3}{2}-\frac{5}{2t^{2}}=\frac{3t^{2}-5}{2t^{2}},\qquad \frac{d^{2}y}{dx^{2}}=\frac{3t^{2}-5}{4t^{3}}.$

Sign chart. Concavity is the sign of $\dfrac{3t^{2}-5}{4t^{3}}$, with breakpoints $t=0$ and $t=\pm\sqrt{5/3}$:

interval	$3t^{2}-5$	$4t^{3}$	$y''$	shape
$t<-\sqrt{5/3}$	$+$	$-$	$-$	concave down
$-\sqrt{5/3}<t<0$	$-$	$-$	$+$	concave up
$0<t<\sqrt{5/3}$	$-$	$+$	$-$	concave down
$t>\sqrt{5/3}$	$+$	$+$	$+$	concave up

The curve is concave down exactly where $y''<0$, i.e. on

$\left(-\infty,\,-\sqrt{\tfrac{5}{3}}\right)\cup\left(0,\,\sqrt{\tfrac{5}{3}}\right),$

which is option 4. (Restricting to the stated domain $t>0$, the concave-down part is $(0,\sqrt{5/3})$, which sits inside this set.)

Pitfall: Option 2 (concave down on the whole $(-\sqrt{5/3},\sqrt{5/3})$) is the trap. On $(-\sqrt{5/3},0)$ the sign flips to $+$ because $4t^{3}<0$ there, so the curve is concave up. Only option 4 is consistent with the full sign chart of $y''$.