سوال ۳۱ کنکور ارشد علوم کامپیوتر ۱۴۰۰

For $g$ to have a continuous second derivative at $x=0$, the two pieces must agree in value, first derivative, and second derivative at $0$ (a continuous $g''$ forces $g, g'$ continuous too, and the one-sided values of $g''$ must match).

Right piece $h(x)=3x^{2}-4x+7$:

$h(0)=7,\qquad h'(0)=-4,\qquad h''(0)=6.$

Left piece $\ell(x)=f(x)e^{x}$. Using the product rule:

$\ell'(x)=e^{x}\bigl(f(x)+f'(x)\bigr),$ $\ell''(x)=e^{x}\bigl(f(x)+2f'(x)+f''(x)\bigr).$

Evaluating at $0$ (where $e^{0}=1$):

$\ell(0)=f(0),\quad \ell'(0)=f(0)+f'(0),\quad \ell''(0)=f(0)+2f'(0)+f''(0).$

Match the three conditions:

Therefore

$\frac{f''(0)}{f(0)}=\frac{21}{7}=3.$

The answer is option 2.

Pitfall: Don't just match the function values. "Continuous second derivative" demands matching $g$, $g'$, and $g''$ — three equations — and you must differentiate the whole product $f(x)e^{x}$, not $f(x)$ alone.