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英語 高校生

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LESSON 9 Quome: Bryor 1 Choose the best answer to fill in the blanks. (1) (1) When I was a would (2) You've got ( 1 a few eggs child, I ( 2 should ) on your tie. 2 an egg ) often play baseball with my friends. 4 might 3 must (3) He has such a soft voice that I can ( hardly ℗ hard (4) She cannot speak English, ( nor better 2 nor less (5) The crowd watched the firefighter ( climbing 2 climbed (7) His arguments forced them ( 1 admit to admit Did you have fried eggs for breakfast? dime 3some egg 4 some eggs (9) His English essay was ( ). 1 superior than Carl's 3 superior to Carl's (11) He told me that he ( 1 had never been was never (12) Willy was surprised ( hear (13) The foreigner was used ( 1 handle ) hear him. 3 already ) French. (6) Let's stay home and watch a movie (Y) it's sunny tomorrow. 1 although as soon as 3 even if 4 when 2 to be heard 3 much better 2 handling 1) the ladder. 3 to climb ) he was right. 3 admitted (10) We then moved to Paris, () we lived for six years. 3 where 1 that 2 which ) to America before. ) the news. 4 admitting (8) It is not that I dislike my new job (___) that the working hours are too long. 1 so 2 with 3 for but (神戸学院 4 yet superior for Carl's 4 superior as Carl's 4 to have climbed much less 2 never comes 4 will never come 3 by hearing ) a pair of chopsticks. 3 to handle FERONE 4 what (センター 4 to hear (黒 to handling 2 (1 (2 (創 (名塩 RETESAHONE ( (学) (北海道 GR

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数学 高校生

この解説の丸つけたとこなんですけど、 α‬で極小値βで極大値をとる場合は考えられないのですか?

11/22 10/23 実力アップ問題 72 3次関数f(x)=x+ax²+2bxが, 0<x<2の範囲で極大値と極小値をもつ CHECK 1 CHECK 2 CHECK 3 |ような実数a,b の条件を求め, それを ab 座標平面上に図示せよ。 ( 千葉大*) ヒント! 3次関数f(x)が0<x<2の範囲に極大値・極小値をもつための条 件は, 2次方程式f'(x)=0の解の範囲の問題に帰着するんだよ。 軸x= 難易度 y=f(x)=x+ax²+2bx...…① ①をxで微分して, f'(x) = 3x2 +2ax+2b 3次関数y=f(x) 図1 が0<x<2の 範囲に極大値と 極小値をもつた めの条件は,図1 に示すように,2 次方程式f'(x)= 0が, 0<x<2の 範囲に, 相異なる 2 実数解をもつこ とである。 f'(x)=0 2次方程式 3x²+2ax+2b = 0 ….…② の判別式をDとおくと, この条件は, (i)=a²-3.2b>0 :. b</a² (ii)0<軸-1/3 <2 ∴-6<a<0 8----- 0a 9 極大 下に凸の 放物線 a 3 y=f'(x) B 2 y=f(x) 極小 x β 2 x (iii) ƒ´(0) = 2b>0 :. b>0 (iv) ƒ´(2) = 12 +4a+2b>0 ::b>-2a-6 以上 (i)~(iv)より,求める条件は b</a^² かつ -6<a<0 かつ 6 b > 0 かつb>-2a-6 ・・・ ( ) これらの条件をすべてみたす点(a,b) の i存在領域を 右図の網目 部で示す。 【境界はすべ て含まない。 ・ b=-2a-6 b=0 a²=-2a-6 b: -6 -3 a=-6 るので, b= 9², 60 10 参考 b= 1a²b=-2a-6から6を消 去して, a=0 a²+12a+36=0 (a+6)2=0 ∴a=-6 (重解) とな -=-a² ≥ b = -2a-6 l£ 6 上図のようにa=-6で接する。 109

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