問題文全文(内容文)：
2⃣ $0 \leqq x \leqq \frac{1}{\sqrt 3}$
$f(x)=\int_x^{\sqrt 3 x} \sqrt{1-t^2} dt$
(1)f(x)の最大値
(2)$\displaystyle \lim_{ x \to \infty } \frac{f(x)}{x}$
2⃣ $0 \leqq x \leqq \frac{1}{\sqrt 3}$
$f(x)=\int_x^{\sqrt 3 x} \sqrt{1-t^2} dt$
(1)f(x)の最大値
(2)$\displaystyle \lim_{ x \to \infty } \frac{f(x)}{x}$
単元：
Warning: usort() expects parameter 1 to be array, bool given in /home/kaiketsudb/kaiketsu-db.net/public_html/wp-content/themes/lightning-child-sample/single.php on line 102
Warning: Invalid argument supplied for foreach() in /home/kaiketsudb/kaiketsu-db.net/public_html/wp-content/themes/lightning-child-sample/single.php on line 103
Warning: usort() expects parameter 1 to be array, bool given in /home/kaiketsudb/kaiketsu-db.net/public_html/wp-content/themes/lightning-child-sample/single.php on line 102
Warning: Invalid argument supplied for foreach() in /home/kaiketsudb/kaiketsu-db.net/public_html/wp-content/themes/lightning-child-sample/single.php on line 103
指導講師：
ますただ
問題文全文(内容文)：
2⃣ $0 \leqq x \leqq \frac{1}{\sqrt 3}$
$f(x)=\int_x^{\sqrt 3 x} \sqrt{1-t^2} dt$
(1)f(x)の最大値
(2)$\displaystyle \lim_{ x \to \infty } \frac{f(x)}{x}$
2⃣ $0 \leqq x \leqq \frac{1}{\sqrt 3}$
$f(x)=\int_x^{\sqrt 3 x} \sqrt{1-t^2} dt$
(1)f(x)の最大値
(2)$\displaystyle \lim_{ x \to \infty } \frac{f(x)}{x}$
投稿日：2020.12.08