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英語 高校生

fについてです 解説が載っていなかったため質問しています、。 なぜ、③を選ぶことができるのでしょうか?

Long-s doctrin holds that we are protected from fungi not just by layered immune defenses but ( e ) we are mammals*, with core temperatures higher than fungi prefer. The cooler outer surfaces of our bodies are at risk of minor assaults-think of athlete's foot*, yeast infections, ringworm*-but in people with healthy immune systems, invasive* infections have been ( f ). That may have left us overconfident. "We have an enormous (g) spot," says Arturo Casadevall, a physician and molecular microbiologist at the Johns Hopkins Bloomberg School of Public Health. "Walk into the street and ask people what are they afraid of, and they'll tell you they're afraid of bacteria, they're afraid of viruses, but they don't fear dying of fungi." Ironically, it is our successes that made us vulnerable*. Fungi exploit damaged immune systems, but before the mid-20th century people with impaired immunity didn't live very long. Since then, medicine has gotten very good at keeping such people (h), even though their immune systems are compromised by illness or cancer treatment or age. It has also developed an array of therapies that deliberately suppress immunity, to keep transplant recipients healthy and treat autoimmune* disorders such as lupus* and rheumatoid arthritis*. ( i ) vast numbers of people are living now who are especially vulnerable to fungi. Not all of our vulnerability is the fault of medicine preserving life so successfully. Other ( j ) actions have opened more doors between the fungal world and our own. We clear land for crops and settlement and perturb* what were stable balances between fungi and their hosts. We carry goods and animals across the world, and fungi hitchhike on them. We drench crops in fungicides* and enhance the resistance of organisms residing nearby. (s) ELSE

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数学 高校生

xy平面上に2つの放物線 C1:y=8, C2:y=ーx^2ー4x+aがある. C1上の点P(t、t^2)(t>0)におけるC1の接線をlとする. (1)lの方程式をtを用いて表せ. (2)lがC2に接するとする。このとき、aをtを用いて表せ、また、lとC2の接点の... 続きを読む

■ 解答 とおく. f(x)=x2, g(x)=-x2-4x+α C:y=f(x) y P(t, t2) XxC (i) s=S__(h(x)-g(x)}dx =S__2(x+t+2)dx = -(x+t+: =1/(1+2). (Ⅱ) 直線 PQ の傾きは P²-a-1-4- t-0 -t-2 (ただし, a= -2t2-4t-4.) したがって, 直線 PQ の方程式は y=(1-1)x+a. t C2:y=g(x) (1) f'(x) = 2x より,P(t, f2) における Cの 接線の方程式は, よって, y-t=f(t)(x-t). y-t=2t(x-t). y=2tx-t². T = [ " [ f ( x ) = { (t − q ) x + a}]dx T= - {x² - (-)x−a}ax -lt- = =1/3/3-2/31 a 2 t- -ax (2) ① の右辺をh(x) とおく. y=h(x) と y=g(x) を連立し,yを消去すると, h(x)=g(x). 2tx-f=-x2-4x+a. x2+2(t+2)x-t-a=0. l が C2 に接する条件は, ②が重解をもつこ とであるから,②の判別式をDとすると, 01=(z+2)-1 (−f-a)=0. これより, a=-2t-4t-4. また、このとき②は重解 x= -(t+2) =-t-2 をもつ. 2 ---at 2 =-11³ -- 1/1/1at 6 =-11³-(-212-41-4)t 6 =cof+2t+2t. したがって, S-T=1/2(t+2)-(qt+2t+2t) そこで, 8 =-1713³ + 2 + 31315 F(t)=1/213+2t+10/23 == (3) よって, l と C2 の接点のx座標は, -t-2. とおくと, C:y=f(x) y l:y=h(x) F'(t) = - 3³t² +2 2tx-t² 2 2 -t-2 P(t, t2) t+ t- 2 √3 x 0 T よって, t>0 における F(t) の増減は次 のようになる. C2:y=g(x) S Q(0, a) 2 t (0) ... 3 F'(t) + 0 F(t) 7 極大 -16- 無断転載複製禁止/著作権法が認める範囲で利用してください。

解決済み 回答数: 1