سوال ۳۳ کنکور ارشد علوم کامپیوتر ۱۳۹۷

Split the $x$ out of the integrand so the parameter $x$ appears both in a limit and as a coefficient:

$g(x) = x\int_{1}^{e^{x}} f(t)\,dt \;-\; \int_{1}^{e^{x}} t\,f(t)\,dt = x\,A(x) - B(x).$

First derivative. With the Leibniz rule, $A'(x) = f(e^{x})\,e^{x}$ and $B'(x) = e^{x}f(e^{x})\,e^{x} = e^{2x}f(e^{x})$, so

$g'(x) = A(x) + x\,A'(x) - B'(x) = A(x) + \bigl(x e^{x} - e^{2x}\bigr)f(e^{x}) = A(x) + P(x)\,f(e^{x}),$

where $P(x) = xe^{x}-e^{2x}$.

Second derivative. Since $A'(x) = e^{x}f(e^{x})$,

$g''(x) = e^{x}f(e^{x}) + P'(x)\,f(e^{x}) + P(x)\,f'(e^{x})\,e^{x}.$

Evaluate at $x=0$. Here $e^{0}=1$, $P(0) = 0-1 = -1$, and $P'(x)=e^{x}+xe^{x}-2e^{2x}$ gives $P'(0) = 1+0-2 = -1$. Hence

$g''(0) = f(1) + (-1)f(1) + (-1)f'(1) = -f'(1).$

Compute $f'(1)$. By the Leibniz rule on $f(x)=\int_{1}^{\ln x} t\,e^{t^{4}}\,dt$,

$f'(x) = (\ln x)\,e^{(\ln x)^{4}}\cdot\frac{1}{x}.$

At $x=1$, $\ln 1 = 0$, so $f'(1) = 0\cdot e^{0}\cdot 1 = 0$.

Therefore $g''(0) = -f'(1) = 0$.

Pitfall: Don't drop the $\frac1x$ chain-rule factor when differentiating $\int_{1}^{\ln x}$, and note that the surviving term is $-f'(1)$, not $\pm f(1)$ — the $f(1)$ pieces cancel because $P(0)=P'(0)=-1$.