سوال ۲۸ کنکور ارشد علوم کامپیوتر ۱۴۰۴

اهمیت

۰/۵

فرض کنید تابع $f:\mathbb{R}\to\mathbb{R}$ دو بار مشتق‌پذیر باشد. اگر به‌ازای هر $x>۰$ ، $f'(x) > g'(x)$ ، آنگاه برای هر $x>۰$ ، کدام مورد درست است؟

پاسخ صحیح: گزینهٔ C — $f(x) - f(\circ) > g(x) - g(\circ)$

پاسخ تشریحی

Define the auxiliary function

$h(x) = f(x) - g(x).$

The hypothesis says that for every $x>0$ ,

$h'(x) = f'(x) - g'(x) > 0,$

so $h$ is strictly increasing on $[0,\infty)$ . By the Mean Value Theorem (or the increasing-function test), for every $x>0$ we have $h(x) > h(0)$ , that is

$f(x) - g(x) > f(0) - g(0).$

Rearranging the constants to opposite sides gives

$f(x) - f(0) > g(x) - g(0),$

which is exactly option 3.

Why the others fail.

Option	Claim	Status
1	$f(x) > g(x)$	False. We only control the change in $h$ , not its level. If $f(0) \ll g(0)$ , then $h$ can stay negative even while increasing. E.g. $f(x)=x-10,\ g(x)=0$ : here $f'=1>0=g'$ but $f(1)=-9 < 0 = g(1)$ .
2	$f''(x) > g''(x)$	False. A sign condition on the first derivative says nothing about the second derivative. Take $f(x)=x,\ g(x)=0$ : then $f'=1>0=g'$ but $f''=g''=0$ .
3	$f(x)-f(0) > g(x)-g(0)$	True — this is $h(x) > h(0)$ .
4	$f'(x)-f'(0) > g'(x)-g'(0)$	False. This would require $f'-g'=h'$ to be increasing, i.e. $h''>0$ , which is not given. With $f(x)=x,\ g(x)=0$ : LHS $=1-1=0$ and RHS $=0-0=0$ , so the strict inequality fails.

Exam trick: A pointwise sign on a derivative only constrains the integral (the net change) of that quantity, never its absolute value or its own derivative. Whenever you see " $f' > g'$ ", form $h=f-g$ and remember $h'>0 \Rightarrow h$ is increasing, giving $h(x) > h(0)$ .

#monotonicity#derivative-sign#mean-value-theorem#auxiliary-function#increasing-function